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Topics In Analytic Number Theory

Thèse

Patrick Letendre

Doctorat en mathématiques

Philosophiæ doctor (Ph. D.)

Québec, Canada

Topics in Analytic Number Theory

Thèse

Patrick Letendre

Sous la direction de:

Jean-Marie De Koninck, directeur de recherche

Résumé

Le présent document est un compte-rendu de quatre présentations que j'ai faites au

congrès de Théorie des Nombres Québec-Maine entre 2013 et 2016. Au l des ans, j'ai

eectué quelques améliorations et corrections aux documents originaux. Le contenu,

l'esprit et l'organisation sont restés essentiellement inchangés. Les quatre sujets sont

fondamentalement distincts tout en étant dans un même cercle d'idées.

Le premier chapitre traite d'un certain nombre de sujets en relation avec le comporte-

ment moyen de certaines fonctions multiplicatives, dans un ensemble bien précis, qui

partagent plusieurs propriétés avec la fonction indicatrice des nombres libres de puis-

sance k-ième. En particulier, on y établit plusieurs estimations de la variance dans des

intervalles courts et dans des progressions arithmétiques.

Le deuxième chapitre étudie un problème du crible combinatoire. Il y est question

d'établir une majoration analogue à la célèbre inégalité de Brun-Titchmarsh, mais pour

les nombres libres de puissance k-ième. Après quelques remarques élémentaires, on

établit une nouvelle inégalité en supposant une conjecture forte en lien avec la densité

maximale d'une suite de nombres ayant un diviseur de la forme pk1pk2 où p1 et p2 sont

des nombres premiers qui satisfont certaines conditions. La méthode fournit aussi une

majoration eective pour le nombre de nombres libres de puissance k-ième dans un

intervalle [x+ 1, x+ h] lorsque h est petit par rapport à x.

Le troisième chapitre, écrit en collaboration avec Jean-Marie De Koninck, établit des

inégalités particulières pour la fonction τ(n) qui compte le nombre de diviseurs de n.

L'objectif est d'obtenir une majoration de τ(n) qui ne dépend pas des facteurs premiers

de n, mais seulement du nombre de facteurs premiers distincts de n et de son ordre de

grandeur, i.e. de log n. L'inégalité principale (Théorèmes 3.4 et 3.5) a nécessité un bon

volume de calcul sur ordinateur, et donc beaucoup de programmation avec Maple.

Finalement, le Chapitre 4 est le début d'une étude du nombre de points entiers près

iii

d'une courbe dans l'espace R3. Le problème peut aussi être vu comme celui du nombre

de points entiers près de deux courbes dans le plan Euclidien simultanément. L'objectif

principal est d'utiliser l'information des deux courbes de façon nontriviale, soit de faire

mieux que les meilleurs résultats connus pour une seule courbe. Étant donné la com-

plexité du problème déjà en deux dimensions et du nombre de méthodes disponibles, il

nous a semblé impossible de faire un traitement complet de la question. On s'est donc

concentré sur une méthode qui utilise des approximations linéaires. Cette dernière peut

sans doute être substantiellement améliorée.

iv

Contents

Résumé iii

Contents v

Remerciements viii

Avant-propos ix

Introduction 1

1 On a class of multiplicative functions 4

Résumé 4

Abstract 41.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Statement of the theorems . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Some notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 Preparatory Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.5 Proof of Theorem 1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.6 Proof of Theorem 1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 341.7 Proof of Theorem 1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 401.8 Some estimates in small intervals . . . . . . . . . . . . . . . . . . . . 48

Bibliography 58

2 A Brun-Titchmarsh inequality for k-free numbers 60

Résumé 60

Abstract 602.1 Introduction and notation . . . . . . . . . . . . . . . . . . . . . . . . 612.2 Preliminary results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632.3 Proof of Theorem 2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 642.4 Proof of Theorem 2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 672.5 Relation among the gk(h; q, a) . . . . . . . . . . . . . . . . . . . . . . 712.6 A sieve identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

v

2.7 Counting special pairs . . . . . . . . . . . . . . . . . . . . . . . . . . 752.8 The main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 842.9 Final remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

Bibliography 90

3 New upper bounds for the number of divisors function 91

Résumé 91

Abstract 913.1 Introduction and notation . . . . . . . . . . . . . . . . . . . . . . . . 923.2 Background results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 923.3 Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 953.4 Preliminary lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 963.5 Proof of Theorem 3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1063.6 Proof of Theorem 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1083.7 Proof of Theorem 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1103.8 Proof of Theorem 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1113.9 Proof of Theorem 3.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

3.9.1 Preliminary steps . . . . . . . . . . . . . . . . . . . . . . . . . 1163.9.2 A rst argumentation . . . . . . . . . . . . . . . . . . . . . . . 1173.9.3 A rst verication . . . . . . . . . . . . . . . . . . . . . . . . . 1193.9.4 Reducing the upper bound for δ′ . . . . . . . . . . . . . . . . 1203.9.5 The last verication . . . . . . . . . . . . . . . . . . . . . . . 123

3.10 Final remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

Bibliography 126

4 Lattice points close to a three-dimensional smooth curve 128

Résumé 128

Abstract 1284.1 Introduction and notation . . . . . . . . . . . . . . . . . . . . . . . . 1294.2 Statement of the Theorems . . . . . . . . . . . . . . . . . . . . . . . . 1314.3 Preparatory lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

4.3.1 Minor arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1354.3.2 Relationship between f and a general plane . . . . . . . . . . 136

4.4 Linear major arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1394.5 Major arcs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1524.6 Proof of Theorem 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

4.6.1 Setting a structure on S . . . . . . . . . . . . . . . . . . . . . 1594.6.2 The S2 term . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

4.6.2.1 The case s = 1 . . . . . . . . . . . . . . . . . . . . . 1634.6.2.2 The case s ≥ 2 . . . . . . . . . . . . . . . . . . . . . 165

vi

4.6.2.3 Another method . . . . . . . . . . . . . . . . . . . . 1684.6.3 The S1 term . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1724.6.4 The exceptional cases . . . . . . . . . . . . . . . . . . . . . . . 179

4.7 Proof of Theorem 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1804.8 Proof of Theorem 4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 1814.9 Examples and concluding remarks . . . . . . . . . . . . . . . . . . . . 182

Bibliography 184

Conclusion 186

vii

Remerciements

Je tiens d'abord à remercier l'université Laval pour m'avoir fourni un excellent en-

vironnement d'étude pendant plusieurs années. Ensuite, je remercie l'ensemble des

professeurs du département de mathématiques et de statistique qui ont participé à

mon développement en tant que mathématiciens et en tant qu'humains. Des remer-

ciements bien particuliers pour monsieur Claude Levesque qui a été mon directeur

pendant plusieurs années, notamment à la maîtrise. Je vais me rapeller de plusieurs

discussions sur diérents sujets de théorie des nombres. Et enn, je remercie mon

directeur monsieur Jean-Marie De Koninck, le célèbre mathématicien aux 1001 occu-

pations, pour sa façon naturelle de communiquer sa passion pour les mathématiques et

pour d'innombrables conseils de toutes sortes.

Je remercie mes véricateurs, i.e. Jean-Marie De Koninck, Nicolas Doyon, Sary Drap-

peau et Antonio Lei. Chacun d'eux m'a fourni une liste de suggestions qui m'ont aidé

à améliorer fortement la qualité du document.

Je remercie tous ceux qui ont contribué de près ou de loin à une idée ou chose qui

m'alimente encore aujourd'hui d'une quelconque façon.

Je remercie ma famille extraordinaire qui m'encourage depuis toujours, avec une pensée

très spéciale qui va à mes amis, ma famille rapprochée et la famille de mon frère. Pour

nir, je remercie ma mère pour tout.

viii

Avant-propos

Les 4 chapitres de cette thèse contiennent des résultats à publier sous une forme ou une

autre. Les Chapitres 1, 2 et 4 ont été entièrement conçus et écrits par l'auteur de cette

thèse. Le Chapitre 3 est un article écrit en collaboration avec Jean-Marie De Koninck.

La contribution de ce dernier est une preuve simpliée du Lemme 3.5 et une grande

quantité de commentaires et conseils judicieux. Aucun des chapitres n'a été soumis en

date du dépôt nal de cette thèse.

ix

Introduction

In Chapter 1, we consider the class of multiplicative functions f that can be written as

f(n) =∑dk|n

g(d)

where k ≥ 2 is xed and g is a multiplicative function for which we require that

|g(p)| = 1 and |g(pj)| ≤ 1 for j ≥ 2.

Such a function f is said to be in the class Ck.

One can show that for each such function f we have∑n≤x

f(n) = cfx+ Ef (x) with |Ef (x)| x1/k (x→∞)

where

cf :=∑n≥1

g(n)

nk.

We want to estimate how the average of a function f ∈ Ck behave in most intervals.

For this reason, we evaluate

M(h,N) :=N−1∑n=0

∣∣∣∣ h∑a=1

f(n+ a)− cfh∣∣∣∣2 (h,N ∈ N)

as well as

T (h,M ; β) :=M∑n=0

∣∣∣∣ h∑a=1

f(nh+ a+ β)− cfh∣∣∣∣2 (β = 0, . . . , h− 1).

Here, h, M and N are integers that satisfy h(M + 1) = N .

Another similar topic is the study of how the average of such a xed function f behave

in arithmetic progressions. Again, given an integer q ≥ 2, one can prove the estimate

Q(x; q, a, f) :=∑n≤x

n≡a mod q

f(n) := g(q, a, f)x+ E(x; q, a, f)

1

where

g(q, a, f) :=1

q

∏p-q

(∑i≥0

g(pi)

pik

)∏p|q

( ∑i≥0

(pik,q)|a

g(pi)(pik, q)

pik

).

We are thus led to consider the quantity

W(x; q, f) :=

q∑a=1

|E(x; q, a, f)|2.

In Chapter 2, we denote by µk the indicator function of the set of k-free numbers

(k ≥ 2). For any positive integers x and h, let

Qk(x) :=x∑

n=1

µk(n),

Gk(x, h) := Qk(x+ h)−Qk(x)

and

gk(h) := maxx≥0

Gk(x, h).

We are interested by gk(h) as h→∞.

In Chapter 3, we let τ(n) stand for the number of divisors of the positive integer n and

ω(n) stand for the number of prime factors of the positive integer n. We shall also be

using the functions

γ(n) :=∏p|n

p and b(n) :=∏p|n

1

log p.

In 1915, Ramanujan obtained the inequality

τ(n) ≤(

log(nγ(n))

ω(n)

)ω(n)

b(n) (n ≥ 2).

We want to nd an upper bound for the function τ(n) that does just depend on ω(n)

and the size of n, that is log n.

In Chapter 4, we let X := [X, 2X] be a xed interval, with X ≥ 2. We write f(x) :=

(x, f1(x), f2(x)), where f1 and f2 are xed functions in Cn(X ) for some n ≥ 1. Much

2

work has been done to count the number of integer points close or on a given curve

in R2. In this paper, we want to estimate the number of integer solutions close to the

curve f(x) when x ∈ X . That is, if we write

Λ := (v1, v2, v3) ∈ R3| ∃x ∈ X s.t. v1 = x, |v2 − f1(x)| < δ1 and |v3 − f2(x)| < δ2,

then we want to give good upper bounds for S := #(Λ ∩ Z3). It is assumed that

0 ≤ δ1, δ2 ≤ 1/4. We are ready to assume various conditions on derivatives of the

functions f1(x) and f2(x).

3

Chapter 1

On a class of multiplicative functions

Patrick Letendre

Résumé

Pour chaque fonction f dans un ensemble de fonctions multiplicatives précisé-

ment dénies, nous estimons la variance des sommes prises dans les intervalles de

longueur h et dans les progressions arithmétiques de raison q.

Abstract

For each function f in a set of precisely dened multiplicative functions, we es-

timate the variance of the sums taken in the intervals of length h and in the

arithmetic progressions of common dierence q.

4

1.1 Introduction

For each k ≥ 2, we note by Ck the class of multiplicative functions f that are supported

on k-th powers, that is

(1.1.1) f(n) =∑dk|n

g(d)

for some multiplicative function g for which we require that

|g(p)| = 1 and |g(pj)| ≤ 1 for j ≥ 2.

This class contains the indicator function of the k-free numbers, noted µk, that satises

(1.1.2) µk(n) :=∑dk|n

µ(d) (k ≥ 2)

and also the functions τ1,k(n) dened more generally by

(1.1.3) τj,k(n) :=∑

djek=n

1 (j, k ∈ N).

A large source of examples is provided by the choice of any additive function t : N→ R.Indeed, in this case we can consider the function f that is dened by g(d) = exp(2πit(d))

in (1.1.1).

The asymptotic behavior of functions in this class is well known. In fact, one can easily

show that for each such function f we have

(1.1.4)∑n≤x

f(n) = cfx+ Ef (x) with |Ef (x)| x1/k (x→∞)

where

(1.1.5) cf :=∑n≥1

g(n)

nk.

It turns out that the class Ck has received a great deal of attention in the literature,

often through one of the functions dened in (1.1.2) and (1.1.3). Here, we focus on

some estimates that we believe to be of some interest for the distribution of the values

of a xed function f ∈ Ck in small intervals and in arithmetic progressions. All the

techniques are elementary, while for certain particular functions, analytic methods will

provide better results.

5

One of the rst thing that comes to mind, for a given f ∈ Ck, is the variance

(1.1.6) M(h,N) :=N−1∑n=0

∣∣∣∣ h∑a=1

f(n+ a)− cfh∣∣∣∣2 (h,N ∈ N)

and the same with a higher power. Hall in [9] has evaluated (1.1.6) in the case where

the function f is the indicator function of the squarefree integers. As we will see, the

quantity

(1.1.7) T (h,M ; β) :=M∑n=0

∣∣∣∣ h∑a=1

f(nh+ a+ β)− cfh∣∣∣∣2 (β = 0, . . . , h− 1)

is related to the varianceM(h,N). Here, h, M and N are integers that satisfy h(M +

1) = N , in which case we clearly have the identity

(1.1.8)h−1∑β=0

T (h,M ; β) =M(h,N)

that will be used in the process of the evaluation of (1.1.7), namely in Theorem 1.2. We

also provide an estimate for a more general quantity with higher powers. The result is

not of the same quality but leads to better results to the question of how many intervals

have a large error term.

Another similar situation is when the partitioning of the interval [1, N ] is done with

arithmetic progressions. Again, given an integer q ≥ 2, one can easily prove the estimate

(1.1.9) Q(x; q, a, f) :=∑n≤x

n≡a mod q

f(n) := g(q, a, f)x+ E(x; q, a, f)

where

(1.1.10) g(q, a, f) :=1

q

∏p-q

(∑i≥0

g(pi)

pik

)∏p|q

( ∑i≥0

(pik,q)|a

g(pi)(pik, q)

pik

).

We are thus led to consider the quantity

(1.1.11) W(x; q, f) :=

q∑a=1

|E(x; q, a, f)|2.

An asymptotic result for a similar quantity is the main theme in a paper of Nunes [15].

There are many papers that study the distribution in arithmetic progressions with an

extra average over the values of q ≤ Q. Some examples of this are in Warlimont [20],

6

Vaughan [19] and in the paper of Brüdern, Granville, Perelli, Vaughan, Wooley [1].

The strongest general result that we are aware of is

(1.1.12) W(x; q, f) (2k)ω(q) log4 x

(x1/kq1−1/k +

x2/k

q1/k

)where ω(q) stands for the number of distinct prime factors of q. One can easily obtain

(1.1.12) with the methods used in the paper of Le Boudec [14]. In Theorem 1.3 we

obtain an asymptotic expression for W(x; q, f) by a combination of methods from the

papers [13], [2], [3] and the method that leads to Theorem 1.2.

We also present a series of estimates inspired by the famous Bombieri-Vinogradov

theorem for the distribution of primes in arithmetic progressions. We believe that this

idea rst originated from Orr in his thesis [16]. Here we present a version valid in a

small interval [x+ 1, x+ y] and obtain results in terms of the quantity

(1.1.13) Sk(x, y) := #d > y1/k| χ[x+1,x+y](d) = 1

where χ[x+1,x+y] is the indicator function of the integers d for which there exists an

integer m such that mdk ∈ [x + 1, x + y]. We present in Lemma 1.2 the simplest

nontrivial estimate of Sk(x, y).

1.2 Statement of the theorems

Fix an integer k ≥ 2. Given h and M in N, write

(1.2.1) γf :=∏p

(1− 1

p

)(∑a≥0

1

pa

(|g(pa)|2 + 2<

∑i≥1

g(pa)g(pa+i)

pik

)),

(1.2.2) `(ξ) :=

log 2h if ξ = 1/2,

1 otherwise

and

(1.2.3) `1(ξ) :=

log 2M if ξ = 1/2,

1 otherwise.

Theorem 1.1. Let f be a function in Ck. Then there exists a positive constant θk ≤ 1k+2

such that

M(h,N) = 2ζ(1/k − 1)

1/k − 1γfh

1/kN +O(Nh1/2k +Nhθk`(kθk)

)+O(N2/(k+1)h2−2/(k+1) logN + h2N1/k logN

).

7

In particular we have

M(h,N) = (1 + o(1))2ζ(1/k − 1)

1/k − 1γfh

1/kN

in the region 1 ≤ h ≤ H where H = o

(N (k−1)/(2k−1)

logk/(2k−1)N

)as H,N →∞.

Theorem 1.2. Let f be a function in Ck. Then for each β ∈ 0, . . . , h−1 there existsa positive constant θk ≤ 1

k+2such that

T (h,M ; β) = 2ζ(1/k − 1)

1/k − 1γfh

1/k(M + 1) +O

(Mhθk`(kθk)

)+O

(Mh1/2k

∏p|h

(1 +

1

p1/2+

1

p

) ∏pj‖h

(r−1)k<j≤rk

(2r − 1 + 2

pj

prk

))

+O

(h2M1/k logN + hM2/(k+1) logN

∏p|h

(1 +

2

pk/(k+1)

)).

In particular we have

T (h,M ; β) = (1 + o(1))2ζ(1/k − 1)

1/k − 1γfh

1/k(M + 1)

in the region 1 ≤ h ≤ H where H = o

(N (k−1)/(3k−2)

logk/(3k−2)N

)as H,N →∞.

Theorem 1.3. Let f be a function in Ck. Assume that x = qM where M is a real

number with M ≥ 1. Then there exists a positive constant θk ≤ 1k+2

such that

W(x; q, f) = 2ζ(1/k − 1)

1/k − 1γf (q)qM

1/k +R(x; q, f),

where

γf (q) :=∏p-q

((1− 1

p

)∑j≥0

1

pj

(|g(pj)|2 + 2<g(pj)

∑i≥1

g(pj+i)

pik

))

×∏pα‖q

(r−1)k<α≤rk

(1− 1

p

)( r−1∑j=0

1

pjk

(|g(pj)|2 + 2<g(pj)

∑i≥1

g(pj+i)

pik

)

+1

pα−α/k

∑i≥r

1

pi

(|g(pi)|2 + 2<g(pi)

∑l≥1

g(pi+l)

plk

))

8

and

R(x; q, f) 2ω(q)q∏p|q

(1 +

1

p

)+M

12k `1(kθk)q

∏p|q

(1 +

1

p1/2+

1

p

)

+M θkq∏p|q

(1 +

1

pkθk

)+

2ω(q)x1+1/k log x

q.

1.3 Some notation

The subscript (k) to a sum∑

(k)

t|q

means that it runs over the divisors t of q = pα11 · · · p

αjj

which are of the form t = pβ1

1 · · · pβjj where each βi ∈ 0, k, 2k, . . . , bαi/kck, αi. The

superscript ′ to a sum∑′

t|q

means that the sum runs over the squarefree divisors t of

q. A double sum like∑h

a=1

∑hb=1 is simply written as

∑ha,b=1.

For a xed value of k, we consider the function

(1.3.1) η(n) :=∏pα‖n

pdαk e.

It has the nice property that if (dk, q) = r then η(r) | d.

By m ∼M we mean that m ∈ [M, 2M).

1.4 Preparatory Lemmas

Lemma 1.1. Let (an) be a sequence of complex numbers. Then the identity

(1.4.1)N∑n=1

|an − T |2 =N∑n=1

|an|2 −1

N

∣∣∣∣ N∑n=1

an

∣∣∣∣2 +N |ε|2,

holds with

T :=1

N

N∑n=1

an + ε

with ε ∈ C.

Proof. By expanding the LHS of (1.4.1), we nd

N∑n=1

|an|2 − 2<

(T

N∑n=1

an

)+N |T |2.

The result then follows by using the denition of T .

9

Lemma 1.2. For each k ≥ 2, we have

Sk(x, h) :=∑d>h1/k

∑x<n≤x+h

dk|n

1 h1/k + x1k+2 (x, h ≥ 1).

Proof. We use the rst and second derivative estimate for the number of integers close

to a smooth curve as explained in the paper of Huxley and Sargos [11]. We consider

the function r(d) := xdk, so that we have

λ1 :=x

Mk+1, λ2 :=

x

Mk+2and δ :=

h

Mk

for d ∈ (M, 2M ]. Now, the rst test is

λ1M + δM + δ/λ1 + 1 = (λ1M + 1)(δ/λ1 + 1).

Such an inequality is easily obtained but not explicitly written in the cited paper. We

have that ∑M<d≤2M

∑x<n≤x+h

dk|n

1

is bounded both byx

Mk+

h

Mk−1+hM

x+ 1

and byx1/3

Mk−1

3

+h

Mk−1+h1/2M

x1/2+ 1.

Using these estimates, we nd that

Sk(x, h) blog((x+h)/h)/ log(2)c∑

j=0

h

(2jh1/k)k−1+ min

(2jh1/k,

x

(2jh1/k)k,

x1/3

(2jh1/k)k−1

3

) h1/k + x

1k+2 ,

while the other contributions are negligible or included in this estimate.

Lemma 1.3. Let a1, . . . , am and n1, . . . , nm be respectively a sequence of integers and

of positive integers. The number of solutions (mod [n1, . . . , nm]) to the system of

congruences

n ≡ ai (mod ni) i = 1, . . . ,m

is

(1.4.2)

1 if (ni, nj)|(ai − aj) for i 6= j,

0 otherwise.

10

Corollary 1.1. Let h ≥ 1 be an integer and let a1, . . . , am and n1, . . . , nm be re-

spectively a sequence of integers and of positive integers. The number of solutions

(mod [h,n1,...,nm]h

) to the system of congruences

hn ≡ ai (mod ni) (i = 1, . . . ,m)

is 1 if (ni, nj)|(ai − aj) for 1 ≤ i < j ≤ m and (h, ni)|ai for 1 ≤ i ≤ m,

0 otherwise.

Proof. We apply Lemma 1.3 to the system of congruences

t ≡ ai (mod ni) for 1 ≤ i ≤ m

t ≡ 0 (mod h).

Lemma 1.4. Given a set n1, . . . , nm, consider the functions

Sj := Sj(n1, . . . , nm) =∏

1≤i1<···<ij≤m

(ni1 , . . . , nij) (j = 1, . . . ,m).

Then, we have the identity

[n1, . . . , nm] =m∏j=1

S(−1)j+1

j .

Lemma 1.5. Let h ≥ 1 be an integer. Then the identity

h∑a,b=1t|a−b

1 =h2

t+ tP

(h

t

)

holds with

P (x) := x − x2

(here x stands for the fractional part of x).

11

Lemma 1.6. For k ≥ 2, we have

(1.4.3)∞∑d=1

P

(M

dk

)= 2

ζ( 1k− 1)

1k− 1

M1/k +O(M θk),

where

θk ≤

1

k+2for k ∈ 2, 3, 4, 5, 6,

67k+6

for k ∈ 7, 8, 9, 10, 11,

1k+3

for k ≥ 12.

Proof. Firstly, we have∑d>M1/k

P

(M

dk

)=

∑d>M1/k

M

dk− M2

d2k= M1/k

(1

k − 1− 1

2k − 1

)+O(1),

where we used the Euler-Maclaurin formula to show that

N∑n=1

1

nr= ζ(r) +

N1−r

1− r+

1

2N r+O

( r

N r+1

)and also that

ζ(s) =1

2+

1

s− 1+s(s+ 1)

2

∫ ∞1

P (x)

xs+2dx

holds in the region < s > −1. For the second part, we write

bM1/kc∑d=1

P

(M

dk

)=

∫ M1/k

1

P

(M

tk

)dt−

[M1/k]∑n=1

(ψ2

(M

nk

)−∫ n

n−1

ψ2

(M

tk

)dt

)+O(1),

where ψ2(x) := B2(x) is the second Bernoulli function

ψ2(x) = x2 − x+ 1/6 =1

2π2

∑n∈Z\0

e(nx)

n2

(here e(y) = e2πiy) that possesses an expansion as an absolutely convergent Fourier

series. We evaluate the rst integral by doing the change of variable x = Mtk, and obtain∫ M1/k

1

P

(M

tk

)dt =

M1/k

k

∫ M

1

P (x)

x1+1/kdx =

M1/k

k

∫ ∞1

P (x)

x1+1/kdx+O(1).

To evaluate the sum, we begin by splitting it into parts with a parameter N to be chosen

later. For the large values of n, we use the fact that ψ2 is continuous and satises

maxx1,x2∈[n,n+1]

∣∣∣∣ψ2

(M

xk1

)− ψ2

(M

xk2

)∣∣∣∣ kM

nk+1

12

and nd that ∑N<n≤M1/k

(ψ2

(M

nk

)−∫ n

n−1

ψ2

(M

tk

)dt

) M

Nk.

Then, we estimate the integral by using the rst derivative test (see [12] Lemma 8.10)

on each term of the Fourier series∫ N

1

ψ2

(M

tk

)dt =

1

2π2

∑n6=0

1

n2

∫ N

1

e

(nM

tk

)dt

∑n6=0

1

n2

(1

knM+Nk+1

knM

) Nk+1

kM.

Finally, we use Theorem 2.2 of [6] to get

2T∑n=T

ψ2

(M

nk

)=

1

2π2

∑n 6=0

1

n2

2T∑m=T

e

(nM

mk

)

∑n6=0

1

n2

(T

(k2nM

T k+2

) 12

+

(T k+2

k2nM

) 12

)

kM1/2

T k/2+T k/2+1

kM1/2

and nd the inequality

N∑n=1

ψ2

(M

nk

) W +

[log(N/W )/ log(2)]+1∑j=0

kM1/2

(2jW )k/2+

(2jW )k/2+1

kM1/2

W +kM1/2

W k/2+Nk/2+1

kM1/2M

1k+2 +

Nk/2+1

kM1/2

holds with W := (kM1/2)2/(k+2). So far we have showed that

∞∑d=1

P

(M

dk

)= 2

ζ( 1k− 1)

1k− 1

M1/k +O

(M

1k+2 +

Nk/2+1

kM1/2+Nk+1

kM+M

Nk

)after some simplications. The result follows with the choice N = (kM2)

12k+1 for k ≥ 4

and N = M3

3k+2 for k = 2, 3. The other results are obtained in the same manner this

time using Theorem 6.9 of [18]. Observe that better results can be obtained with the

theory of exponent pairs.

We now dene the constants

(1.4.4) γf (a, b) :=∞∑

d,e=1(dk,ek)|a−b

g(d)g(e)

[dk, ek](a, b ∈ 1, . . . , h).

13

Lemma 1.7. Let f be a function in Ck. Then, with cf , γf and θk respectively dened

in (1.1.5), (1.2.1) and Lemma 1.6, we have

F (h) :=h∑

a,b=1

γf (a, b) = |cf |2h2 + 2ζ(1/k − 1)

1/k − 1γfh

1/k +O(h1/2k + hθk) (h ∈ N).

On the other hand, if θk = 12k

then the error term becomes hθk log h (except if f satises

the additional condition |g(p2)| = 1, in which case it is h1/2k).

Proof. We have

F (h) =h∑

a,b=1

∞∑d,e=1

(dk,ek)|a−b

g(d)g(e)(dk, ek)

dkek

=∞∑

d,e=1

g(d)g(e)(dk, ek)

dkek

h∑a,b=1

(dk,ek)|a−b

1

=∞∑

d,e=1

g(d)g(e)(dk, ek)

dkek

(h2

(dk, ek)+ (dk, ek)P

(h

(dk, ek)

))

= h2|cf |2 +∞∑

d,e=1

g(d)g(e)(dk, ek)2

dkekP

(h

(dk, ek)

)= h2|cf |2 + F1(h),(1.4.5)

say, where we used Lemma 1.5.

The strategy is to split the arithmetic factor in F1(h) to obtain a linear combination of

the expression evaluated in Lemma 1.6. First observe that since f is in Ck,

1 + 2∑i≥1

<(g(pi))

pik6= 0

and

1 +∑i≥1

g(pi)

pik6= 0

for each prime number p.

Now, observe that

F1(h) =∞∑j=1

∞∑d,e=1

(d,e)=j

g(d)g(e)

dkekj2kP

(h

jk

)=∞∑j=1

∞∑d,e=1

(d,e)=1

g(dj)g(ej)

dkekP

(h

jk

)

14

and consider the function

ξ(j) :=∞∑

d,e=1(d,e)=1

g(dj)g(ej)

dkek(j ∈ N).

We easily show that

ξ(j) = βfθ(j) (j ∈ N),

where

βf :=∏p

(1 + 2

∑i≥1

<(g(pi))

pik

)and

θ(j) :=∏pa‖j

(1 + 2

∑i≥1

<(g(pi))

pik

)−1(|g(pa)|2 + 2

∑i≥1

<(g(pa)g(pa+i))

pik

)(j ∈ N).

Our task will be to evaluate

F1(h) =∑j≥1

ξ(j)P

(h

jk

)= βf

∑j≥1

θ(j)P

(h

jk

)(h ∈ R≥0).

By setting

ν(pi) := θ(pi)− θ(pi−1) (i ∈ N),

we obtain

F1(h) = βf∑j≥1

∑d|j

ν(d)P

(h

jk

)= βf

∑d≥1

ν(d)∑j≥1

P

(h

dkjk

)and from there, in light of Lemma 1.6,

∑n≥1

P( xnk

)=

2ζ(1/k − 1)

1/k − 1x1/k +O(xθk) if x ≥ 1,

O(x) otherwise.

Therefore, using the inequality z ≤ z1/k for 0 ≤ z ≤ 1, we nd that

F1(h) = 2βfζ(1/k − 1)

1/k − 1h1/k

∑d≥1

ν(d)

d+O

h1/k∑d≥h1/k

|ν(d)|d

+ hθk∑d≤h1/k

|ν(d)|dkθk

= 2βf

ζ(1/k − 1)

1/k − 1h1/k

∑d≥1

ν(d)

d+O

(h1/kE1(h) + hθkE2(h)

),(1.4.6)

say. We then use the estimates

(1.4.7) ν(p) ≤ 4

pk − 3and ν(pj) ≤ pk + 3

pk − 3for j ≥ 2

15

to show that ∑d≥1

ν(d)

d

converges absolutely and that the sum E1(h) in the error term is h−1/2k and that

hθkE2(h) is hθk + h1/2k if θk 6= 12k

and hθk log(h) otherwise (except if the function

satises the stronger condition |g(p2)| = 1, in which case it is h1/2k). Indeed, we use

the same method for each case: we write d = rs where r is squarefree and s powerful.

Clearly,

E1(h) ∑

1≤≤h1/k

|ν(r)|r

∑s≥h1/k/r

|ν(s)|s

+∑r>h1/k

|ν(r)|r

1

h1/2k

h1/k∑r=1

|ν(r)|r1/2

+1

h1/k

∏p

(1 +

4

pk − 3

) 1

h1/2k(1.4.8)

and similarly for E2(h). Using this in (1.4.6), the result follows from (1.4.5).

For a xed β, we dene the constants

(1.4.9) γf (h, a, b, β) :=∑d,e≥1

(h,dk)|a+β

(h,ek)|b+β(dk,ek)|a−b

g(d)g(e)h

[h, dk, ek].

Lemma 1.8. Let f be a function in Ck and consider

(1.4.10)

G(h) := sup(β1,β2)∈0,...,h−12

∣∣∣∣∣h∑

a,b=1

γf (h, a, b, β1)−h∑

a,b=1

γf (h, a, b, β2)

∣∣∣∣∣ (h ∈ N).

Then,

G(h) h1/2k∏p|h

(1 +

1

p1/2+

1

p

) ∏pj‖h

(r−1)k<j≤rk

(2r − 1 + 2

pj

prk

).

Proof. We rst write

h∑a,b=1

γf (h, a, b, β) =∑d,e≥1

hg(d)g(e)

[h, dk, ek]

h∑a,b=1

(dk,ek)|a−b(h,dk)|a+β

(h,ek)|b+β

1.

16

Using Lemma 1.4 to evaluate [h, dk, ek], we nd that

G(h) ≤∑

(k)

w|h

∑(k)

r,s|h(r,s)=w

∑t≥1

(tk,h)=w

∑d,e≥1

(d,e)=t

(h,dk)=r

(h,ek)=s

rstk

dkekw

∣∣∣∣ h∑a,b=1tk|a−br|a+β1

s|b+β1

1−h∑

a,b=1tk|a−br|a+β2

s|b+β2

1

∣∣∣∣.

We easily show that

(1.4.11)∑d,e≥1

(d,e)=t

(h,dk)=r

(h,ek)=s

1

dkek η2k(w)

t2kηk(r)ηk(s),

which yields

(1.4.12) G(h)∑

(k)

w|h

∑(k)

r,s|h(r,s)=w

∑t≥1

(tk,h)=w

rsη2k(w)

tkwηk(r)ηk(s)


s|b+β1

1−h∑


s|b+β2

1

∣∣∣∣.

To estimate the absolute value appearing in (1.4.12), we will be using three distinct

arguments. Firstly, we observe that (tk, r) = (tk, s) = w and thus, after a change of

variable, that (tkηk(w)/w, r) = (tkηk(w)/w, s) = 1. So it is enough to estimate the

error term of the double sum

Σ :=A+N∑a=A+1

B+M∑b=B+1l|ua+vb

1

with integers A,B,M,N, l, u, v ≥ 1 such that (u, l) = (v, l) = 1. For this, we write

Σ =1

l

l∑j=1

A+N∑a=A+1

B+M∑b=B+1

el(j(ua+ vb))

=MN

l+ E

and then

E :=1

l

l−1∑j=1

A+N∑a=A+1

B+M∑b=B+1

el(j(ua+ vb)) l

l−1∑j=1

1

‖ju‖l1

‖jv‖l l,

where the last inequality follows from the Cauchy-Schwarz inequality.

Also, one can see that the solutions in our specic case, counted by one of the two

expressions in absolute value in (1.4.12), represent points in a square of h by h points in

17

Z2 where the coordinates have the form (ir, js) and the condition tk|a− b is interpretedas the fact that these points are on lines of equation y = x + mtk. In view of this

representation, we see that the main diagonal (m = 0) brings no contribution to G(h),

and we deduce that the range of the sum on t in (1.4.12) is 1 ≤ t < h1/k. Also, for each

t, the error is no larger than the number of non-main diagonals since the solutions are

in arithmetic progression on each diagonal line. So the estimate htk

is always true.

Using the above explanations, we see that the sum over t appearing in (1.4.12) is

=h1/k∑t≥1

(tk,h)=w

1

tk


s|b+β1

1−h∑


s|b+β2

1

∣∣∣∣

=1

ηk(w)

h1/k/η(w)∑t≥1

(tkηk(w)/w,h/w)=1

1

tk

∣∣∣∣ h∑a,b=1

tkηk(w)|a−br|a+β1

s|b+β1

1−h∑

a,b=1tkηk(w)|a−br|a+β2

s|b+β2

1

∣∣∣∣

1

w

(hw)1/2k/η(w)∑t=1

1 +1

η2k(w)

∑t≥(hw)1/2k/η(w)

h

t2k

h1/2k

η(w)w1−1/2k.

Then, we easily verify that∑(k)

w|h

w12k η2k−1(w)

w2

∑(k)

r,s|h(r,s)=w

rs

ηk(r)ηk(s)∏p|h

(1 +

1

p12

+1

p

) ∏pj‖h

(r−1)k<j≤rk

(2r − 1 + 2

pj

prk

),

which shows in particular that G(h) h1/2k+ε as requested, thus completing the proof.

Lemma 1.9. Let f be a function in Ck and consider the expression

S(q) :=∑e,d≥1

g(d)

dkg(e)

ek(dk, ek)2

(dk, ek, q)P

(M(dk, ek, q)

(dk, ek)

)(q ∈ N).

It satises

S(q) = 2ζ(1/k − 1)

1/k − 1γf (q)M

1/k +R(q),

where γf (q) has been dened in the statement of Theorem 1.3 and where

(1.4.13) R(q) 2ω(q)∏p|q

(1+

1

p

)+M

12k `1(kθk)

∏p|q

(1+

1

p1/2+

1

p

)+M θk

∏p|q

(1+

1

pkθk

).

18

Moreover, the last term in (1.4.13) can be omitted if kθk ≤ 1/2.

Proof. We write t = (dk, ek, q), d→ dη(t), e→ eη(t) and (d, e) = r

S(q) =∑

(k)t|q

1

t

∑r≥1

(r,q/t)=1

r2k∑e,d≥1

(d,e)=r

g(η(t)d)

dkg(η(t)e)

ekP

(Mt

ηk(t)rk

)

because if we write s = rη(t), then we have

(dkη(t)k, ekη(t)k, q) = t ⇔(sk

t,q

t

)= 1⇔

((s

η(t)

)k,q

t

)= 1⇔

(rk,

q

t

)= 1

⇔(r,q

t

)= 1.

We thus have

S(q) =∑

(k)

t|q

1

t

∑r≥1

(r,q/t)=1

P

(Mt

ηk(t)rk

) ∑e,d≥1

(d,e)=1

g(η(t)rd)

dkg(η(t)re)

ek

=∑

(k)

t|q

1

t

∑r≥1

(r,q/t)=1

ξ(η(t)r)P

(Mt

ηk(t)rk

)

= βf∑

(k)t|q

1

t

∑r≥1

(r,q/t)=1

θ(η(t)r)P

(Mt

ηk(t)rk

),

where we used the notation of Lemma 1.7. Pursuing along these lines, we obtain that

S(q) = βf∑

(k)t|q

1

t

∑s|q/t

µ(s)∑r≥1

θ(η(t)sr)P

(Mt

η(t)kskrk

)

= βf∑

(k)

t|q

1

t

∑s|q/t

µ(s)∑r≥1

∑u|η(t)sr

ν(u)P

(Mt

η(t)kskrk

)

= βf∑

(k)t|q

1

t

∑s|q/t

µ(s)∑u≥1

ν(u)∑r≥1

P

(Mt

[u, η(t)s]krk

).

From there, we use Lemma 1.6 as we did in Lemma 1.7. Firstly, regarding the main

term, we have to evaluate

(1.4.14)∑

(k)t|q

t1/k

tη(t)

∑s|q/t

µ(s)

s

∑u≥1

ν(u)

u(u, η(t)s)

19

in the range Mt ≥ [u, η(t)s]k. To do so, we complete the sum and consider as part of

the error term the sum on Mt < [u, η(t)s]k. We easily obtain that

∑u≥1

ν(u)

u(u, η(t)s) =

∏p

(∑i≥0

ν(pi)

pi

) ∏pα‖η(t)s

(∑i≥0

θ(pα+i)

pi

)(∑i≥0

θ(pi)

pi

)−1

,

where we used the fact that∑i≥0

ν(pi)

pi=

(1− 1

p

)∑i≥0

θ(pi)

pi6= 0

for each prime p ≥ 2 and f ∈ Ck. Using this in (1.4.14), we obtain

∏p-q

(∑i≥0

ν(pi)

pi

) ∏pα‖q

(r−1)k<α≤rk

(1− 1

p

)( r−1∑i=0

θ(pi)

pik+

1

pα−α/k+r

(∑i≥0

θ(pr+i)

pi

))

which yields for the main term of S(q) the expression

2ζ(1/k − 1)

1/k − 1M1/k

∏p-q

((1− 1

p

)∑j≥0

1

pj

(|g(pj)|2 + 2<g(pj)

∑i≥1

g(pj+i)

pik

))

×∏pα‖q

(r−1)k<α≤rk

(1− 1

p

)( r−1∑j=0

1

pjk

(|g(pj)|2 + 2<g(pj)

∑i≥1

g(pj+i)

pik

)

+1

pα−α/k

∑i≥r

1

pi

(|g(pi)|2 + 2<g(pi)

∑l≥1

g(pi+l)

plk

)).

It remains to show that the error term R(q) is (1.4.13). Firstly, for n = pα11 · · · p

αii , we

compute ∑u>z

|ν(nu)|u

=∑

a1,...,ai≥0

|ν(npa11 · · · p

aii )|

pa11 · · · p

aii

∑upa11 ···p

aii >z

(u,n)=1

|ν(u)|u

.

Using the upper bounds (1.4.7) we evaluate the above inner sum using the unique

decomposition u = ab where µ(a)2 = 1 and b powerful, thereby allowing us to write

∑u≥w

|ν(u)|u

w∑a=1

|ν(a)|a

∑b>w/a

1

b+∑a>w

|ν(a)|a 1

w1/2

w∑a=1

|ν(a)|a1/2

+∑a>w

4ω(a)

ak+1

1

w1/2+

log3w

wk 1

w1/2.

20

It follows that

∑u>z

|ν(nu)|u

1

z1/2

∑a1,...,ai≥0

|ν(npa11 · · · p

aii )|

pa1/21 · · · pai/2i

1

z1/2

∏p‖n

(|ν(p)|+ |ν(p2)|

p1/2+|ν(p3)|p

+|ν(p4)|p3/2

)

×∏pα‖nα≥2

(|ν(pα)|+ |ν(pα+1)|

p1/2+|ν(pα+2)|

p

)

1

z1/2

∏p‖n

(1

p1/2+

1

p+

1

p3/2

) ∏pα‖nα≥2

(1 +

1

p1/2+

1

p

)

=

(1

zγ1(n)

)1/2∏p|n

(1 +

1

p1/2+

1

p

),

where γ1(n) :=∏p‖n

p. In the same way, we have

∑u≤z

|ν(nu)|uξ

1

γξ1(n)

∏p|n

(1 +

1

pξ

)if 1/2 < ξ < 1

l(ξ)z1/2−ξ

γ1(n)1/2

∏p|n

(1 +

1

p1/2+

1

p

)if 0 < ξ ≤ 1/2

and ∑u≥1

|ν(nu)|u

1

γ1(n)

∏p‖n

(1 +

5

p

)∏p2|n

(1 +

1

p

).

Gathering the estimates, for Mt < [u, η(t)s]k we have

∑(k)

t|q

t1/k

tη(t)

∑′

s|q/t

1

s

∑u≥1

Mt<[u,η(t)s]k

|ν(u)|u

(u, η(t)s)

∑

(k)t|q

t1/k

tη(t)

∑′

s|q/t

1

s

∑j|η(t)s

j∑u≥1

u>(Mt)1/kj/η(t)s

|ν(u)|u

=∑

(k)

t|q

t1/k

tη(t)

∑′

s|q/t

1

s

∑j|η(t)s

∑u≥1

u>(Mt)1/k/η(t)s≥1

|ν(ju)|u

21

Further let

W :=∑q∈I

v(q) =∑u∈S

r(u).

Finally, let τI(n) :=∑q|nq∈I

1 for each n ∈ N,

z := maxu1 6=u2ui∈S

τI(u1 − u2)

and

Xn :=∑

(q1,...,qn)∈Inqi 6=qj for i6=j

1

[q1, . . . , qn].

Then, by noting S := #S and I := #I, we have the estimates

(1.4.15) W ≤

2I + (2z)1/2I1/2S,

n!S + n!1/nS1−1/n(hXn + In)1/n (n ≥ 2).

Proof. For the rst inequality we write

W =∑q∈I

v(q) ≤ I1/2

(∑q∈I

v(q) +∑q∈I

v2(q)− v(q)

)1/2

.

Then we notice that if W ≥ 2I,

2IW ≤ W 2 ≤ I

(∑q∈I

v2(q)

),

that is ∑q∈I

v2(q) ≥ 2W = 2∑q∈I

v(q),

which is ∑q∈I

v2(q)− v(q) ≥∑q∈I

v(q).

In this case we nd

W =∑q∈I

v(q) ≤ I1/2

(2∑q∈I

v2(q)− v(q)

)1/2

.

We see that v2(q)−v(q) is the number of pairs (u1, u2) with u1 6= u2 for which q|u1−u2.

Then we have ∑q∈I

v2(q)− v(q) =∑

u1,u2∈Su1 6=u2

τI(u1 − u2) ≤ zS2,

23

implying that

W ≤ 2I + I1/2(2zS2)1/2.

For the second inequality we write∑u∈S

r(u) ≤(∑u∈S

1

)1−1/n(∑u∈S

r(u)n)1/n

= S1−1/n

(∑u∈S

gn(r(u)) +∑u∈S

r(u)n − gn(r(u))

)1/n

where gn(w) :=∏n−1

j=0 (w − j) is a polynomial of degree n, which we write as

gn(w) :=n∑k=0

s(n, k)wk.

Clearly, s(n, n) = 1 and |gn(−1)| =∑n

k=0 |s(n, k)| = n!. The s(n, k) are called Stirling

numbers of the rst kind. We use these for a combinatorial purpose: we want our

n-tuples (q1, . . . , qn) to contain pairwise distinct numbers. We distinguish two cases:

(i)∑u∈S

gn(r(u)) ≥ K∑u∈S

r(u)n − gn(r(u))

(ii)∑u∈S

gn(r(u)) < K∑u∈S

r(u)n − gn(r(u))

for a positive constant K to be chosen later. In the rst case we deduce that∑u∈S

r(u)n =∑u∈S

gn(r(u)) +∑u∈S

r(u)n − gn(r(u)) ≤(

1 +1

K

)∑u∈S

gn(r(u))

and we estimate the last sum by noticing that∑u∈I

gn(r(u)) =∑

(q1,...,qn)∈Inqi 6=qj for i 6=j

|(u ∈ S| u ≡ ai (mod qi) for i = 1, . . . , n|

≤∑

(q1,...,qn)∈Inqi 6=qj for i 6=j

h

[qi, . . . , qn]+ 1 = hXn + In,

where we have used Lemma 1.3. In the second case we nd∑u∈S

r(u)n < (K + 1)∑u∈S

r(u)n − gn(r(u)) < (K + 1)(n!− 1)∑u∈S

r(u)n−1.

Now, we show by induction using the Cauchy-Schwarz inequality that for any j ∈0, . . . , n− 1,∑

u∈S

r(u)n < C∑u∈S

r(u)n−1 =⇒∑u∈S

r(u)j+1 < C∑u∈S

r(u)j,

24

where C > 0. We deduce that in this case we have

W < (K + 1)(n!− 1)S.

We have thus shown that

W ≤ (K + 1)(n!− 1)S +

(1 +

1

K

)1/n

S1−1/n(hXn + In)1/n

and the result follows with the choice K := 1n!−1

.

We say that an integer vector u = (u1, . . . , un) is primitive if (u1, . . . , un) = 1. The

following lemma can be seen as a generalization of Lemma 5 of [10].

Lemma 1.11. Let a1, . . . , am be a sequence of real numbers with each ai ∈ [A, 2A].

Assume that A,B,∆ > 0 are real numbers satisfying

(1.4.16)∆

A B.

Then the number of non zero integer solutions to the system of inequalities

(1.4.17) |a1b1 − aibi| ∆ (i = 2, . . . ,m)

in primitive vectors b = (b1, . . . , bm) of length ≤ B is

(1.4.18) 1 +m−1∑j=1

B

(∆

A

)j.

Proof. We write

vi := (a1, 0, . . . , 0,−ai, 0, . . . , 0) for i = 2, . . . ,m.

The hypotheses imply that the vectors vi's span a vector space V of dimension m− 1.

Let v be a unitary vector that spans V ⊥. The set v, v2, . . . , vm is a base of Rm. Letb be any vector that satises (1.4.17). Setting

b := αv +m∑i=2

αivi,

we deduce from (1.4.17) that, for each i ∈ 2, . . . ,m,

(1.4.19) 〈b, vi〉 =m∑j=2

αj(a21 + aiajδi,j) = θi∆,

25

where θi is a real number no larger that 1 in absolute value and δi,j stands for the

Kronecker symbol. We deduce a linear system Mx = c where

M :=

a2

1 + a22 a2

1 · · · a21

a21 a2

1 + a23 · · · a2

1...

......

...

a21 a2

1 · · · a21 + a2

m

, x :=

α2

α3

...

αm

and c :=

θ2∆

θ3∆...

θm∆

.

We show that the family of matrices dened by

An(x, z1, . . . , zn) :=

x+ z1 x · · · x

x x+ z2 · · · x...

......

...

x x · · · x+ zn

satises the recurring relation

(1.4.20)

det An(x, z1, . . . , zn) = z1 det An−1(x, z2, . . . , zn) + z2 det An−1(x, 0, z3, . . . , zn).

For this, we expand det An(x, z1, . . . , zn) with the rst row after having subtracted the

second row. By using the identity M−1 det M = adj M , (1.4.20) and a trivial estimate

for the entries in adj M , we deduce that M−1 := (ci,j) satisfy

|ci,j| 1

A2for all 1 ≤ i, j ≤ m− 1.

We deduce from x = M−1c that |αi| ∆

A2for 2 ≤ i ≤ m. Also,

‖b− αv‖ =

(〈m∑i=2

αivi,m∑j=2

αjvj〉

)1/2

=

(m∑i,=2

αiαj(a21 + aiajδi,j)

)1/2

∆

A.

This proves that the solutions b to the system (1.4.17) are close to the line λv with

λ ∈ R. More precisely, we write

K := z ∈ Rn| ‖z − λv‖ ≤ C1∆

Aand |λ| ≤ B

for a large enough constant C1. Note that K is then a convex symmetric body. Let

1 ≤ j ≤ m be the exact dimension of the linear space L spanned by all the vector

26

solutions of the system (1.4.17) in K. If there is no solution then (1.4.18) holds. If

j = 1 then there are at most two primitive solutions in K and the result follows.

If 2 ≤ j ≤ m, then we write K := K ∩ L. The integer points in L form a lattice

Γ, in which case the number of solutions in K is at most j·j!Vol(K)det Γ

by a theorem of

Blichfeldt (see [8] p. 62 (12)). Now, the result follows from the fact that det Γ ≥ 1

and from Vol(K) B(

∆A

)j−1, which in turn follows from inequality (1.4.16) and the

fact that for any two orthogonal vectors u1, u2 in K, with ‖u1‖ ≥ ‖u2‖, we must have

‖u2‖ ∆A.

Lemma 1.12. The estimate

Υ(h, r, s,m) :=2h∑

a,b=1r|a,s|bm|a−b

1 h2(r, s)

rsm+h(r, s)

rs

holds when (r, s)|m and [r, s] h.

Proof. It is enough to consider, for a xed k, the equation

ar1 − bs1 = km1 (r1 = r/(r, s), s1 = s/(r, s), m1 = m/(r, s)).

It is easy to show that two consecutive solutions a1 and a2 of this equation are distant

by exactly s1 and that |k| h/m. We deduce that

Υ(h, r, s,m)(h

rs1

+ 1

)(h

m+ 1

)and the result follows the hypothesis.

Lemma 1.13. Let f be a function in Ck. Then,∣∣∣∣∣q∑

a=1

g(q, a, f)E(x; q, a, f)

∣∣∣∣∣ 2ω(q)x1/k

q(q ≥ 1).

Proof. We have

Σ :=

q∑a=1

g(q, a, f)E(x; q, a, f) =∑s|q

g(q, s, f)

q∑a=1

(a,q)=s

E(x; q, a, f),

so that we are led to consider the sum∑n≤x

(n,q)=s

f(n) =∑n≤x/s

(n,q/s)=1

f(ns)

27

=∑

(k)r|s

∑d≤zr,s

(dk,s)=r

g(d)∑n≤x/s

(n,q/s)=1

dk|ns

1 +∑

(k)r|s

∑d>zr,s

(dk,s)=r

g(d)∑n≤x/s

(n,q/s)=1

dk|ns

1

= Σ1 + Σ2,

say. First of all,

Σ1 =∑

(k)r|s

∑d≤zr,s

(dk,s)=r

(dk/r,q/s)=1

g(d)∑

m≤xr/sdk(m,q/s)=1

1

=∑

(k)

r|s

∑d≤zr,s

(dk,s)=r

(dk/r,q/s)=1

g(d)

(φ(q/s)xr

qdk+O(2ω(q/s))

).

From there, we write

φ(q/s)x

q

∑(k)

r|s

r∑d

(dk,s)=r

(dk/r,q/s)=1

g(d)

dk+O

(x

s

∑(k)

r|s

r∑d>zr,s

(dk,s)=r

(dk/r,q/s)=1

1

dk+ 2ω(q/s)

∑(k)

r|s

zr,sη(r)

)

= φ(q/s)g(q, s, f)x+O

(x

s

∑(k)

r|s

r

η(r)zk−1r,s

+ 2ω(q/s)∑

(k)

r|s

zr,sη(r)

).(1.4.21)

On the other hand,

Σ2 =∑

(k)r|s

∑d>zr,s

(dk,s)=r

(dk/r,q/s)=1

∑m≤xr/sdk(m,q/s)=1

1

x

s

∑(k)

r|s

r

η(r)k

∑d>zr,s/η(r)

1

dk

x

s

∑(k)

r|s

r

η(r)zk−1r,s

.(1.4.22)

Notice that

(1.4.23) |g(q, a, f)| τ(η((a, q)))

q

holds for all f ∈ Ck, as can be seen directly from (1.1.10). Now, by using (1.4.21),

(1.4.22) and (1.4.23), we nd that

|Σ| 1

q

∑s|q

2ω(q/s)τ(η(s))∑

(k)r|s

zr,sη(r)

28

+x

q

∑s|q

τ(η(s))

s

∑(k)

r|s

r

η(r)zk−1r,s

so that with the choice zr,s =

(xr

s2ω(q/s)

)1/k

we obtain that

|Σ| x1/k

q

∑s|q

2ω(q/s)(1−1/k) τ(η(s))

s1/k

∑(k)

r|s

r1/k

η(r)

x1/k

q

∑s|q

2ω(q/s)(1−1/k) τ2(η(s))

s1/k

x1/k

q

∏pj‖q

(21−1/k

(1 +

∑i≥1

(i+ 1)2

pi/k

))

x1/k

q

∏pj‖q

(21−1/k

(1 +

4p2/k

(p1/k − 1)3

)) 2ω(q)x1/k

q.

Lemma 1.14. Let q ≥ 2 be an integer and χ1 a multiplicative character modulo q. The

number of solutions χ to the equation χk = χ1, if any such solution exists, that is, if

χ1 can be written as a k-th power of a character, is given by the multiplicative function

g(k, q) :=∏pj‖q

g(k, pj)

dened by

g(k, pj) :=

(k, φ(pj)) if p ≥ 3,

1 if pj = 2,

(k, 2) if pj = 4,

(k, 2)(k, φ(2j−1)) if p = 2 and j ≥ 3.

In particular, g(k, q) ≤ 2kω(q).

Lemma 1.15. Let q ≥ 2 be an integer. Then∑χ 6=χ0

∣∣∣∣ a+B∑n=a+1

χ(n)

∣∣∣∣4 φ(q)B2 log2 q2ω(q).

Proof. This is the main result of the amazing paper of T. Cochrane and S. Shi [3]. The

upper bound is obtained by a further smoothing in the case when B is small enough

while careful computations are carried throughout.

29

1.5 Proof of Theorem 1.1

We refer to Lemma 1.1 with an :=∑h

a=1 f(n+ a) and T := cfh. We then have

(1.5.1)1

N

N−1∑n=0

h∑a=1

f(n+a) =1

N

(h−1∑n=1

nf(n) + h

N∑n=h

f(n) +N+h−1∑n=N+1

(N + h− n)f(n)

).

Using the Abel summation, we nd that

h−1∑n=1

nf(n) = cfh2 +O(hEf (h))−

∫ h

1

cf t+ Ef (t)dt

=1

2cfh

2 +O

(hEf (h) +

∫ h

1

|Ef (t)|dt)

=1

2cfh

2 +O(h1+1/k).

Similarly,h−1∑n=1

(h− n)f(N + n) =1

2cfh

2 +O(hN1/k).

Using these last two estimates along with (1.1.4) the sum in (1.5.1) can be written as

1

N

N−1∑n=0

h∑a=1

f(n+ a) =1

N

(cfhN +O(hN1/k)

)= cfh+O(hN1/k−1),

so that ε hN1/k−1 (by again referring to Lemma 1.1). We deduce from (1.4.1) that

(1.5.2) M(h,N) =N−1∑n=0

h∑a,b=1

f(n+ a)f(n+ b)− |cf |2h2N +O(h2N1/k).

We then have

N−1∑n=0

h∑a,b=1

f(n+ a)f(n+ b) =h∑

a,b=1

N−1∑n=0

f(n+ a)f(n+ b)(1.5.3)

=h∑

a,b=1

N−1∑n=0

∑dk|n+a

g(d)∑ek|n+b

g(e)

=h∑

a,b=1

∑1≤d,e≤(N+h)1/k

g(d)g(e)N−1∑n=0

dk|n+a

ek|n+b

1.

30

We focus our attention on the sum over d, e and we introduce a parameter z to split

the sum in two parts. The rst is

Σ1(a, b) :=∑

[d,e]≤z

g(d)g(e)N−1∑n=0

dk|n+a

ek|n+b

1

and the second is

Σ2(a, b) :=∑

[d,e]>z

g(d)g(e)N−1∑n=0

dk|n+a

ek|n+b

1.

By using Lemma 1.3, we get that

Σ1(a, b) = N∑

dk,ek≥1(dk,ek)|a−b

g(d)g(e)

[dk, ek]+O

( ∑[d,e]≤z

(dk,ek)|a−b

1 +N∑

[d,e]>z

(dk,ek)|a−b

1

[dk, ek]

)

where the rst sum is what we have dened as γf (a, b) in (1.4.4). Now, by summing

over a and b, we obtain

(1.5.4)h∑

a,b=1

Σ1(a, b) = Nh∑

a,b=1

γf (a, b) +O

(h2z logN + hz log2N +

h2N logN

zk−1+hN log2N

zk−1

).

Recall that the evaluation of the above sum was done in Lemma 1.7.

On the other hand,

(1.5.5)h∑

a,b=1

Σ2(a, b) =h∑

a,b=1

∑z<[d,e]<(N+h)1/k

N−1∑n=0

dk|n+a

ek|n+b

1 ≤ h#S,

where

S := (m,n, d, e) : |mdk − nek| < h, mdk, nek < N + h, [d, e] > z.

We shall now describe a way to estimate S := #S. At rst, we count the solutions

(m,n, d, e) of

mdk = nek, mdk, nek < N + h, [d, e] > z.

We see that it is

(1.5.6) ∑

[d,e]>z

N

[d, e]k N log2N

zk−1.

31

Then we count the remaining solutions that satisfy

(m,n) = r and (d, e) = s

for some xed r and s. After a change of variables, we see that it corresponds to the

set of solutions (m,n, d, e) of

0 < |mdk − nek| < h

rsk, (m,n) = 1, (d, e) = 1, de >

z

s, 1 ≤ mdk, nek N

rsk.

In particular, rsk < h. So we x an X satisfying h X N and count the solutions

with

0 < |mdk − nek| < h

rsk, (m,n) = 1, (d, e) = 1, de >

z

s,X

rsk mdk, nek X

rsk.

We do a dyadic cut for each ot the variables d and e, so that we assume that d ∈ [L, 2L)

and e ∈ [M, 2M) for some xed L and M that satisfy LM zs. We apply the same

argument in two ways. First of all, we have

0 <

∣∣∣∣mn − ek

dk

∣∣∣∣ < h

rskndk hMk

XLk.

For (d, e) xed, we have mnin an interval of length hMk

XLk, and since (m,n) = 1, we

have a distance of at least M2kr2s2k

X2 between two solutions. We deduce that the

number of solutions is at most

(1.5.7)h

LkX

Mkr2s2k+ 1 =

hX

LkMkr2s2k+ 1.

By summing (1.5.7) on (d, e) we obtain

(1.5.8)hX

Lk−1Mk−1r2s2k+ LM.

Also, there is no lost of generality in assuming that L ≥M and then by using the mean

value theorem to the function f1(w) = w1/k, we have

0 <

∣∣∣∣m1/k

n1/k− e

d

∣∣∣∣ hM

XL.

On the other hand, for xed (m,n), we nd that two distinct fractions edare separated

by at least 1L2 , so that the number of solutions is at most

(1.5.9)hM

XLL2 + 1 =

hLM

X+ 1.

32

By summing (1.5.9) on (m,n) we obtain

(1.5.10)hX

Lk−1Mk−1r2s2k+

X2

LkMkr2s2k.

We sum over the domain LM zsthe common term in the two estimates (1.5.8) and

(1.5.10). This leads tohX logN

zk−1r2sk+1.

Summing over X and then over rsk < h for a total of

hN logN

zk−1.

For the two other terms in (1.5.8) and (1.5.10), we consider two cases depending whether

LM is larger or smaller than X2/(k+1)

r2/(k+1)s2k/(k+1) . We have

∑LM X2/(k+1)

r2/(k+1)s2k/(k+1)

X2

LkMkr2s2k X2/(k+1) logN

r2/(k+1)s2k/(k+1)

and ∑zsLM X2/(k+1)

r2/(k+1)s2k/(k+1)

LM X2/(k+1) logN

r2/(k+1)s2k/(k+1).

We then sum over X and then over rsk < h for a total contribution of

h1−2/(k+1)N2/(k+1) logN

elements in the set S. It remains to consider the case where 1 ≤ mdk, nek hrsk

. We

simply neglect the inequality and estimate with the number of elements (m,n, d, e) to

nd ∑de>z/s

h2

r2s2kdkek h2 logN

r2sk+1zk−1

and, by summing over rsk < h, we get to a total contribution of

h2 logN

zk−1

which less than hN logNzk−1 . So far, we showed that

(1.5.11) hS h2N logN

zk−1+hN log2N

zk−1+ h2−2/(k+1)N2/(k+1) logN.

It remains to put together (1.5.2), (1.5.4), Lemma 1.7 and (1.5.11) with the choice

z = N1/k to complete the proof of Theorem 1.1.

33


Proceeding as in the proof of Theorem 1.1, we consider Lemma 1.1 with an :=∑h

a=1 f(nh+

a+ β) and T := cfh. We then have

1

M + 1

M∑n=0

h∑a=1

f(nh+ a+ β) =1

M + 1

N+β∑m=β+1

f(m)

=1

M + 1

(cfN +O(N1/k)

)= cfh+O(hN1/k−1),

from which it follows that

(1.6.1) T (h,M ; β) =h∑

a,b=1

M∑n=0

f(nh+ a+ β)f(nh+ b+ β)− |cf |2hN +O(hN1/k).

We now proceed to the evaluation of the sum in the RHS of (1.6.1). We x a, b and

write

M∑n=0

f(nh+a+β)f(nh+b+β) =M∑n=0

∑dk|nh+a+β

ek|nh+b+β

g(d)g(e)

=∑

(k)

w|h

∑(k)

r,s|h

∑t≥1

(h,tk)=w

∑[d,e]≤zr,s,w

(dk,h)=r

(ek,h)=s(d,e)=t

g(d)g(e)M∑n=0

dk|nh+a+β

ek|nh+b+β

1

+∑

(k)w|h

∑(k)

r,s|h

∑t≥1

(h,tk)=w

∑[d,e]>zr,s,w

(dk,h)=r

(ek,h)=s(d,e)=t

g(d)g(e)M∑n=0

dk|nh+a+β

ek|nh+b+β

1

= Σ1(a, b) + Σ2(a, b),

say. We split Σ1 to get

Σ1(a, b) = (M + 1)γf (h, a, b, β) + E1(a, b) + (M + 1)E2(a, b),

34

where γf (h, a, b, β) is the constant dened in (1.4.9) and

E1(a, b)∑

(k)

w|h

∑(k)

r,s|hr|a+βs|b+β

∑t≥1

(h,tk)=w

tk|a−b

∑[d,e]≤zr,s,w

(dk,h)=r

(ek,h)=s(d,e)=t

1

while

E2(a, b)∑

(k)

w|h

∑(k)

r,s|hr|a+βs|b+β

∑t≥1

(h,tk)=w

tk|a−b

∑[d,e]>zr,s,w

(dk,h)=r

(ek,h)=s(d,e)=t

rstk

dkekw.

We will also write

Ei :=h∑

a,b=1

Ei(a, b) (i = 1, 2).

We verify that (t, η(r)) = (t, η(s)) = η(w), and also that [t, η(r)]|d and [t, η(s)]|e. Then,we use the upper bound in Lemma 1.12 and obtain

E1 ∑

(k)

w|h

∑(k)

r,s|h

∑1≤t≤zr,s,w(h,tk)=w

∑[d,e]≤zr,s,w

(dk,h)=r

(ek,h)=s(d,e)=t

h2

[r, s]tk+

h

[r, s]

logN∑

(k)

w|h

∑(k)

r,s|h(r,s)=w

∑1≤t≤zr,s,w(h,tk)=w

h2wη2(w)zr,s,wrη(r)sη(s)tk+1

+hwη2(w)zr,s,wrη(r)sη(s)t

∑

(k)w|h

∑(k)

r,s|h(r,s)=w

h2wzr,s,wrη(r)sη(s)η(w)k−1

logN +hwη(w)zr,s,wrη(r)sη(s)

log2N.(1.6.2)

For E2 we distinguish the diagonal case a = b from the non-diagonal one. Hence we get

h∑a,b=1a6=b

E2(a, b) ∑

(k)w|h

∑(k)

r,s|h

∑1≤t≤h1/k

(h,tk)=w

∑[d,e]>zr,s,w

(dk,h)=r

(ek,h)=s(d,e)=t

h2

dkek

logN∑

(k)w|h

∑(k)

r,s|h(r,s)=w

∑1≤t≤h1/k

(h,tk)=w

η2(w)h2

η(r)η(s)tk+1zk−1r,s,w

35

logN∑

(k)

w|h

∑(k)

r,s|h(r,s)=w

h2

η(r)η(s)ηk−1(w)zk−1r,s,w

and

h∑a=1

E2(a, a) ∑

(k)

w|h

∑(k)

r,s|h

∑t≥1

(h,tk)=w

∑[d,e]>zr,s,w

(dk,h)=r

(ek,h)=s(d,e)=t

tkh

dkek

∑

(k)w|h

∑(k)

r,s|h(r,s)=w

∑t≥1

(h,tk)=w

η2k(w)h

ηk(r)ηk(s)tkmin

(1,ηk−1(r)ηk−1(s)tk−1

η2k−2(w)zk−1r,s,w

logN

)

∑

(k)w|h

∑(k)

r,s|h(r,s)=w

∑1≤t≤ zr,s,wη

2(w)

η(r)η(s)

(h,tk)=w

η2(w)h

η(r)η(s)tzk−1r,s,w

logN

+∑

(k)

w|h

∑(k)

r,s|h(r,s)=w

∑t>

zr,s,wη2(w)

η(r)η(s)

(h,tk)=w

η2k(w)h

ηk(r)ηk(s)tk

∑

(k)

w|h

∑(k)

r,s|h(r,s)=w

η(w)h

η(r)η(s)zk−1r,s,w

log2N.

We now turn to the contribution of all the sums Σ2(a, b). We write

h∑a,b=1

Σ2(a, b) h∑

a,b=1

∑(k)

w|h

∑(k)

r,s|h

∑t≥1

(h,tk)=w

∑[d,e]>zr,s,w

(dk,h)=r

(ek,h)=s(d,e)=t

M∑n=0

dk|nh+a+β

ek|nh+b+β

1

∑

(k)w|h

∑(k)

r,s|h(r,s)=w

∑1≤t≤2N1/k

(h,tk)=w

S(r, s, t, w),

where S(r, s, t, w) is the number of 4-tuples (m,n, d, e) that satisfy

|mdk − nek| < h, [d, e] > zr,s,w,

(d, e) = t, (h, dk) = r, (h, ek) = s, 1 ≤ mdk, nek N.

36

The evaluation of this quantity is done by essentially the same method that we have

used for the evaluation of the quantity that appears in (1.5.5). First we need an upper

bound for the number S0 of solutions to

mdk = nek, [d, e] > zr,s,w, (d, e) = t, (h, dk) = r, (h, ek) = s, 1 ≤ mdk, nek N.

We nd that

S0 ∑

[d,e]>zr,s,w(d,e)=t

(h,dk)=r

(h,ek)=s

N

[d, e]k η2(w)N logN


.

Then, we count the number of solutions with (m,n) = j, noticing that jtk < h. We

denote this quantity by S(r, s, t, w, j). We thus x a parameter X which satises

h X N . After a change of variables we have

0 <

∣∣∣∣m ηk(r)

ηk(w)dk − n η

k(s)

ηk(w)ek∣∣∣∣ < h

jtk, (d, e) = (m,n) = 1,

X

jtk m

ηk(r)

ηk(w)dk, n

ηk(s)

ηk(w)ek X

jtk

in the region de > η2(w)zr,s,wη(r)η(s)t

. Now, we make a dyadic cut in each of the variables d and

e assuming that d ∈ [L, 2L) and e ∈ [M, 2M) in the region LM η2(w)zr,s,wη(r)η(s)t

. Again,

we use twice the same argument in two dierent ways. On the one hand, we have

0 <

∣∣∣∣mn − ηk(s)ek

ηk(r)dk

∣∣∣∣ hηk(s)Mk

Xηk(r)Lk

and then, for (d, e) xed we have that two fractions mnare distant by j2t2kη2k(s)M2k

X2η2k(w).

The total number of solutions is therefore at most

(1.6.3)hηk(s)Mk

Xηk(r)LkX2η2k(w)

j2t2kη2k(s)M2k+ 1 =

hXη2k(w)

j2t2kηk(r)ηk(s)LkMk+ 1.

Summing (1.6.3) on (d, e), we get to

(1.6.4)hXη2k(w)

j2t2kηk(r)ηk(s)Lk−1Mk−1+ LM

which is the rst wanted inequality. On the other hand, there is no lost in generality

in assuming that Lη(r) ≥ Mη(s). Using the mean value theorem to the function

g(w) = w1/k, we obtain

0 <

∣∣∣∣m1/k

n1/k− η(s)e

η(r)d

∣∣∣∣ hMη(s)

XLη(r)⇒ 0 <

∣∣∣∣m1/kη(r)

n1/kη(s)− e

d

∣∣∣∣ hM

XL.

37

For (m,n) xed we have that two fractions edare distant by 1

L2 and thus the total

number of solutions is at most

(1.6.5)hM

XLL2 + 1 =

hLM

X+ 1.

Summing (1.6.5) on (m,n) yields

(1.6.6)hXη2k(w)

j2t2kηk(r)ηk(s)Lk−1Mk−1+

X2η2k(w)

j2t2kηk(r)ηk(s)LkMk

which is the second inequality.

We now consider the term that both estimates (1.6.4) and (1.6.6) have in common.

Summing on LM η2(w)zr,s,wη(r)η(s)t

we nd

hXη2(w) logN

j2tk+1η(r)η(s)zk−1r,s,w

.

We then sum over X and then over j < htk

to get

hNη2(w) logN

tk+1η(r)η(s)zk−1r,s,w

.

For the two remaining terms in (1.6.4) and (1.6.6), we split according whether if LM

is smaller or larger than

z :=X2/(k+1)η2k/(k+1)

j2/(k+1)t2k/(k+1)ηk/(k+1)(r)ηk/(k+1)(s).

Proceeding in this way, we nd that∑LMz

X2η2k(w)

j2t2kηk(r)ηk(s)LkMk X2/(k+1)η2k/(k+1)(w) logN

j2/(k+1)t2k/(k+1)ηk/(k+1)(r)ηk/(k+1)(s)

and ∑η2(w)zr,s,wη(r)η(s)t

LMz

LM X2/(k+1)η2k/(k+1)(w) logN

j2/(k+1)t2k/(k+1)ηk/(k+1)(r)ηk/(k+1)(s).

Summing over X and then over j ≤ htk

we arrive at

hM2/(k+1)η2k/(k+1)(w) logN

tkηk/(k+1)(r)ηk/(k+1)(s).

It remains to treat the case where

0 <

∣∣∣∣m ηk(r)

ηk(w)dk − n η

k(s)

ηk(w)ek∣∣∣∣ < h

jtk, (d, e) = (m,n) = 1, m

ηk(r)

ηk(w)dk, n

ηk(s)

ηk(w)ek h

jtk

38

in the region de > η2(w)zr,s,wη(r)η(s)t

. For that, we just ignore the inequality and we estimate

by the number of such elements (m,n, d, e). We nd∑de>

η2(w)zr,s,wη(r)η(s)t

h2η2k(w)

j2t2kηk(r)ηk(s)dkek h2η2(w) logN

j2tk+1η(r)η(s)zk−1r,s,w

and by summing over j < htk

we get

h2η2(w) logN


which is less than hNη2(w) logN


. So far we have showed that

Σ2 ∑

(k)

w|h

∑(k)

r,s|h(r,s)=w

∑1≤t≤2N1/k

(h,tk)=w

(η2(w)N logN


+hNη2(w) logN


+hM2/(k+1)η2k/(k+1)(w) logN

tkηk/(k+1)(r)ηk/(k+1)(s)

)= S1 + S2 + S3,

say. First of all,

S1 + S2 ∑

(k)

w|h

∑(k)

r,s|h(r,s)=w

η(w)N log2N

η(r)η(s)zk−1r,s,w

+hN logN

η(r)η(s)ηk−1(w)zk−1r,s,w

which amounts to the same contribution that we had obtained for the term with E2.

Then,

S3 ∑

(k)w|h

∑(k)

r,s|h(r,s)=w

hM2/(k+1) logN

η(k2−k)/(k+1)(w)ηk/(k+1)(r)ηk/(k+1)(s)(1.6.7)

hM2/(k+1) logN∏p|h

(1 +

2

pk/(k+1)

).

We now choose

zr,s,w =

(Mrs

w

)1/k

.

On the one hand we have

(1.6.8) S1 hM1/k log2N∑

(k)w|h

∑(k)

r,s|h(r,s)=w

w1−1/kη(w)

r1−1/kη(r)s1−1/kη(s) hM1/k log2N

39

and on the other hand

(1.6.9)

S2 h2M1/k logN∑

(k)

w|h

∑(k)

r,s|h(r,s)=w

w1−1/k

r1−1/kη(r)s1−1/kη(s)ηk−1(w) h2M1/k logN.

We obtain the same estimate for (1.6.2) as for S1 + S2.

Gathering (1.6.7), (1.6.8) and (1.6.9) in (1.6.1), we obtain the estimate

T (h,M ; β) = (M + 1)h∑

a,b=1

γf (h, a, b, β)− |cf |2hN

+O

(h2M1/k logN + hM2/(k+1) logN

∏p|h

(1 +

2

pk/(k+1)

)).(1.6.10)

Recalling (1.1.8) and Theorem 1.1, we obtain that

(M + 1)

(h−1∑β′=0

h∑a,b=1

γf (h, a, b, β′)− |cf |2h3 − 2

ζ(1/k − 1)

1/k − 1γfh

1+1/k

)

= O(Nh1/2k +Nhθk`(kθk) +N2/(k+1)h2−2/(k+1) logN + h2N1/k logN

)+O

(h3M1/k logN + h2M2/(k+1) logN

∏p|h

(1 +

2

pk/(k+1)

))Letting M →∞, we nd the estimate∣∣∣∣∣

h−1∑β′=0

h∑a,b=1

γf (h, a, b, β′)− |cf |2h3 − 2

ζ(1/k − 1)

1/k − 1γfh

1+1/k

∣∣∣∣∣ h1+1/2k + h1+θk`(kθk).

Finally, we use Lemma 1.8 at each term in the sum over β′ to get only terms with β

and some error bounded by G(h). Then, we divide by h to get∣∣∣∣∣h∑

a,b=1

γf (h, a, b, β)− |cf |2h2 − 2ζ(1/k − 1)

1/k − 1γfh

1/k

∣∣∣∣∣ h1/2k + hθk`(kθk) +G(h).

Theorem 1.2 follows by inserting this last inequality into (1.6.10).


First observe that

W(x; q, f) =

q∑a=1

|Q(x; q, a, f)− g(q, a, f)x|2

40

x2 log x

q

∑(k)

w|q

∑(k)

r,s|q(r,s)=w

∑t≥1

(tk,q)=w

t≤x1/k

wη2(w)


x2 log x

q

∑(k)

w|q

∑(k)

r,s|q(r,s)=w

w

η(w)k−1η(r)η(s)zk−1r,s,w

.(1.7.7)

Now, we turn to the evaluation of the total contribution of the terms Σ2(j). We write

∑0<|j|≤x/q

Σ2(j) ≤∑

0<|j|≤x/q

∑(k)

w|q

∑(k)

r,s|q(r,s)=w

∑t≥1

(tk,q)=w

∑[d,e]>zr,s,w

(dk,q)=r

(ek,q)=s(d,e)=t

x∑n=1

1≤n+jq≤xdk|n,ek|n+jq

1(1.7.8)

=∑

(k)w|q

∑(k)

r,s|q(r,s)=w

∑t≥1

(tk,q)=w

S(r, s, t, w)

where

S(r, s, t, w) := #(m,n, d, e)| 1 ≤ mdk, nek ≤ x,mdk ≡ nek mod q,

[d, e] > zr,s,w, (d, e) = t, (tk, q) = w, (dk, q) = r, (ek, q) = s.

For each 4-tuple (r, s, t, w) we may write

(1.7.9) S(r, s, t, w) =∑u|q

T (r, s, t, w, u)

where T (r, s, t, w, u) counts the elements counted by S(r, s, t, w) restricted to the case

where (mdk, q) = u. After a change of variable, we notice that T (r, s, t, w, u) counts

the points (m,n, d, e) that satisfy

1 ≤ mdk ≤ xrη(w)k

uηk(r)tk, 1 ≤ nek ≤ xsη(w)k

uηk(s)tk, de >

zr,s,wη2(w)

tη(r)η(s)

ηk(r)tk

rη(w)kmdk ≡ ηk(s)tk

sη(w)knek mod

q

u,

(ηk(r)tk

rη(w)kmdk,

q

u

)= 1.

Also, we notice that [r, s] | u. A simple computation, not taking into account the

congruence, shows that the estimate

T (r, s, t, w, u) x2rsη(w)2 log x

tk+1u2η(r)η(s)zk−1r,s,w

43

is valid in each case, in particular when u = q. For the general case, we use Lemma

1.15 and obtain the estimate

(1.7.10)

T (r, s, t, w, u) g(k, q/u)x2rsη(w)2 log x

φ(q/u)tk+1u2η(r)η(s)zk−1r,s,w

+g1/2(k, q/u)2ω(q/u)x(rs)1/2η(w) log3 x

t(k+1)/2uη1/2(r)η1/2(s)z(k−1)/2r,s,w

.

To prove it, we perform a dyadic cut for each variable d and e, assuming that d ∈ [L, 2L)

and e ∈ [M, 2M) for some xed L and M such that LM ≥ zr,s,wη(w)2

tη(r)η(s). In each such

region, we count the number of solutions with the characters modulo j = q/u, obtaining

that∣∣∣∣ 1

φ(j)

∑χ

∑d∼L,e∼Mmx1/Lk

nx2/Mk

χ(αβ−1mdkn−1e−k)

∣∣∣∣ x1x2g(k, j)

φ(j)Lk−1Mk−1

+1

φ(j)

∑χk 6=χ0

∣∣∣∣ ∑d∼L,e∼Mmx1/Lk

nx2/Mk

χ(mdkn−1e−k)

∣∣∣∣where we have set

x1 :=xrη(w)k

uηk(r)tk, x2 :=

xsη(w)k

uηk(s)tk, α :=

ηk(r)tk

rη(w)k, β :=

ηk(s)tk

sη(w)k

and used Lemma 1.14. Now, by using the Cauchy-Schwarz's inequality three times (or

the generalized Hölder's inequality) and Lemma 1.15, we estimate the second term with

x1/21 x

1/22 g1/2(k, j)

L(k−1)/2M (k−1)/2log2 q2ω(j).

It remains to sum this last inequality in the domain LM zr,s,wη(w)2

tη(r)η(s)to obtain the

expected estimate (1.7.10). Now, observe that∑u|q

[r,s]|u

g(k, q/u)

φ(q/u)u2 g(k, v)

[r, s]2φ(v)

∏p|v

(1 +

1

p

)

and ∑u|q

[r,s]|u

2ω(q/u)g(k, q/u)12

u 2ω(v)g(k, v)

12

[r, s]

∏p|v

(1 +

1

p

),

where we have set v := q/[r, s]. We deduce from (1.7.9) that

S(r, s, t, w) g(k, v)x2wη(w)2 log x

qtk+1η(r)η(s)zk−1r,s,w

∏p|v

(1 +

2

p

)(1.7.11)

44

+g1/2(k, v)2ω(v)xwη(w) log3 x

t(k+1)/2(rs)1/2η1/2(r)η1/2(s)z(k−1)/2r,s,w

∏p|v

(1 +

1

p

).

By using the estimate (1.7.11) in (1.7.8), we nally obtain∑0<|j|≤x/q

Σ2(j) ∑

(k)w|q

∑(k)

r,s|q(r,s)=w

g(k, v)x2w log x

qη(w)k−1η(r)η(s)zk−1r,s,w

∏p|v

(1 +

2

p

)(1.7.12)

+g1/2(k, v)2ω(v)xw log3 x

η(w)(k−1)/2(rs)1/2η1/2(r)η1/2(s)z(k−1)/2r,s,w

∏p|v

(1 +

1

p

).

We now want to optimize our choice of zr,s,w in (1.7.6), (1.7.7) and (1.7.12). A good

one is given by

zr,s,w =

x1/k2ω(qw/rs)

∏p|q

(1 +

3

p

)−1

if k = 2,

x1/k1.9ω(qw/rs) if k ≥ 3.

Thus, using this in (1.7.12), we nd that

(1.7.13)

∑0<|j|≤x/q

Σ2(j)

2ω(q)x1+1/k log x

q+ 2ω(q)x(k+1)/2k log3 x

∏p|q

(1 +

5

2p

)if k = 2,

2ω(q)x1+1/k log x

q+ 2ω(q)x(k+1)/2k log3 x if k ≥ 3.

We have seen from (1.7.5) that the main term of W1 is

(1.7.14) Σ =∑d,e≥1

g(d)g(e)

[dk, ek]

∑|j|≤bx−1

qc

(dk,ek)

(dk,ek,q)|j

(x− |j|q).

We write x = qM + a with 1 ≤ a < q + 1 and verify that∑|j|≤bx−1

qc

m|j

(x− |j|q) = qmP

(M

m

)+Mx

m+aM

m+ a− 2a

M

m

.

Using it in (1.7.14), we get to

Σ = q∑d,e≥1

g(d)g(e)(dk, ek)2

dkek(dk, ek, q)P

(M(dk, ek, q)

(dk, ek)

)+Mx

∑d,e≥1

g(d)g(e)(dk, ek, q)

dkek+O(aM).

45

Now, since

W2 =x2

q2

q∑a=1

∑d,e≥1

[(dk,q),(ek,q)]|a

g(d)g(e)(dk, q)(ek, q)

dkek

=x2

q

∑d,e≥1

g(d)g(e)(dk, ek, q)

dkek,

we obtain that

(1.7.15) Σ−W2 = q∑d,e≥1

g(d)g(e)(dk, ek)2

dkek(dk, ek, q)P

(M(dk, ek, q)

(dk, ek)

)+O

(ax

q+ aM

).

In Lemma 1.9 we evaluated

S(q) :=∑e,d≥1

g(d)

dkg(e)

ek(dk, ek)2

(dk, ek, q)P

(M(dk, ek, q)

(dk, ek)

).

We put all together to get (1.7.1), that is: W3, the error term of (1.7.3), (1.7.13) and

Lemma 1.9 in (1.7.15), we can state the following result:

For x = Mq + a with 1 ≤ a ≤ q we have

(1.7.16) W(x; q, f) = γf (q)qM1/k +O(R(x; q, a, f)),

where

R(x; q, a, f) := 2ω(q)q∏p|q

(1+

1

p

)+`1(kθk)M

12k q∏p|q

(1+

1

p12

+1

p

)+M θkq

∏p|q

(1+

1

pkθk

)

+aM +2ω(q)x1+1/k log x

q+

2ω(q)x(k+1)/2k log3 x

∏p|q

(1 +

5

2p

)if k = 2,

2ω(q)x(k+1)/2k log3 x k ≥ 3.

Firstly, we verify that the last term is always dominated by the second or by the fth.

Secondly, we notice that for x = qM + ξ with 1 ≤ ξ ≤ q and z = qM + 1 we have

Q(x; q, a, f)−Q(z; q, a, f) =

f(qM + a) if 1 ≤ a ≤ ξ,

0 otherwise

= g(q, a, f)(ξ − 1) + E(x; q, a, f)− E(z; q, a, f)

46

from which we deduce the inequality

(1.7.17) |E(x; q, a, f)− E(z; q, a, f)| f(qM + a) + τ(η((a, q)))

(see (1.4.23)).

Setting

δ(x; q, a, f) := E(x; q, a, f)− E(z; q, a, f),

we may then write

W(x; q, f) =

q∑a=1

|E(z; q, a, f) + δ(x; q, a, f)|2

= W(z; q, f) +O(E1/2W(z; q, f)1/2 + E

),(1.7.18)

where

E :=

qM+q∑n=qM+1

|f(n)|2 +

q∑a=1

τ(η((a, q)))2

after a use of the Cauchy-Schwarz inequality and of (1.7.17).

We verify thatq∑

a=1

τ(η((a, q)))2 q∏p|q

(1 +

3

p

)and we have shown that∑

n≤N

|f(n)|2 = N∑d,e≥1

g(d)g(e)

[dk, ek]+O(N1/k log2N).

Thus we have

E

q∏p|q

(1 +

3

p

)si q x1/2 log2 x

qxε si q x1/2 log2 x

for each ε > 0 since |f(n)| xε. We see that E is negligible compared to R(x; q, 1, f).

Also, in the region q > x1/2 log2 x we verify that E1/2W(z; q, f)1/2 also is negligible.

Finally, in the region q < x1/2 log2 x, since the fth term dominate we see that it is

larger than E1/2W(z; q, f)1/2 x(k+1)/2k+ε. All of these comments in (1.7.18) shows

that

W(x; q, f) =W(z; q, f) +O(R(z; q, 1, f))

which, along with (1.7.16), completes the proof of the theorem.

47

1.8 Some estimates in small intervals

Let x, h ∈ R with x ≥ 1 and h ≥ 1. We consider a family of intervals I := (Ij) that

satisfy |Ij| ∈ [h, 2h], Ij ∈ [x+ 1, 2x] and Ij1 ∩ Ij2 = ∅ if j1 6= j2. For 2h ≤ x, we let

(1.8.1) R(m)(x; f) := maxI

∑Ij∈I

∣∣∣∣∑n∈Ij

f(n)− cf |Ij|∣∣∣∣m,

where the maximum is taken over all the possible families I satisfying the above con-

ditions.

Proposition 1.1. Let f be a function in Ck. Assume that I is a family as previously

dened. Then we have

(1.8.2) R(m)(x; f) xhm−kk logc1(m)+m−1 x+ x

mk+1 logc2(m)+m−1 x

for k + 1 ≤ m ≤ 2k, where

c1(m) :=

0 if k + 1 ≤ m ≤ 2k − 1,

1 if m = 2k

and

c2(m) :=

1 if m = k + 1,

0 if k + 2 ≤ m ≤ 2k.

Moreover, for any xed ε > 0, we have

(1.8.3) R(m)(x; f)ε xhm−kk + x

mk+1

+ε

for k + 1 ≤ m ≤ 2k − 1.

Proof. We begin with a small reduction. A direct computation and use of Lemma 1.2

imply that (1.8.2) is trivial for any integer m ≥ 1 in the region h ≥ xk/(k+2). So we will

assume throughout the proof that h < xk/(k+2). Observe that

(1.8.4) R(m)(x; f)∑Ij∈I

∣∣∣∣∣h1/k +∑

2h1/k<d<2x1/k

dk|n,n∈Ij

1

∣∣∣∣∣m

xhm−kk +

∑Ij∈I

∣∣∣∣∣ ∑2h1/k<d<2x1/k

rj(d)

∣∣∣∣∣m

,

where we used the fact that |I| x/h and the notation rj(d) = 1 only if dk divides

an integer in Ij and rj(d) = 0 otherwise. We split the sum over d into intervals Li :=

48

[2ih1/k, 2i+1h1/k) with i = 1, . . . , dlog((x/h)1/k)/ log 2e =: u and we write Li = |Li|. We

further expand the last term in (1.8.4) to get

∑Ij∈I

∣∣∣∣∣u∑i=1

∑d∈Li

rj(d)

∣∣∣∣∣m

=∑Ij∈I

∣∣∣∣∣u∑i=1

1

ρi

∑d∈Li

ρirj(d)

∣∣∣∣∣m

≤∑Ij∈I

(u∑i=1

1

ρm/(m−1)i

)m−1 u∑i=1

(∑d∈Li

ρirj(d)

)m

=

(u∑i=1

1

ρm/(m−1)i

)m−1 u∑i=1

ρmi∑Ij∈I

∑d1,...,dm∈Li

rj(d1) · · · rj(dm)(1.8.5)

where ρi = (1 + ε1)i(1−1/m), for some ε1 ≥ 0, so that the rst sum in (1.8.5) is min(logm−1 x, 1

εm−11

). The two outer sums are bounded by

#(d1, . . . , dm) ∈ Li| |b1dk1 − bjdkj | ≤ 2h such that x ≤ bjd

kj ≤ 2x for j = 2, . . . ,m,

a quantity that we denote Jm(Li, h, x). We evaluate it with the help of Lemma 1.11

in two ways. First, this lemma counts only primitive points. So we assume that

(b1, . . . , bm) = r and that (d1, . . . , dm) = s and the system becomes, after a change in

notation, the new system

(1.8.6) |b1dk1 − bjdkj | ≤

2h

rsk(j = 2, . . . ,m, (b1, . . . , bm) = 1, (d1, . . . , dm) = 1).

We deduce that we either have rsk ≤ 2h or equality everywhere. In the later case, the

number of solution is bounded by

∑d1,...,dm∈Li

x

[d1, . . . , dm]k x log2m−2 Li

Lk−1i

by using the theory of Theorem 3.5 and Corollary 3.6 of [18], since the relation

gm(n) :=∑

[d1,...,dm]=n

1 =∏pα‖n

((α + 1)m − αm)

holds. We turn to the case where rsk ≤ 2h. For each such xed pair (r, s), we denote

by Jm(Li, h, x, r, s) the number of solution to the system (1.8.6). We will establish

two distinct upper bounds for this quantity. Following Lemma 1.3, we x the vector

(d1, . . . , dm), we have

∆1 :=h

rsk, A1 :=

L

s, and B1 :=

x

Lkr

49

and nd the estimate 1 + xhL2ki r2 since ∆1/A1 1. Thus we get that

(1.8.7) Jm(Li, h, x, r, s)Lmism

+xh

L2k−mi r2sm

.

Also, by using the mean theorem with the function h(w) := w1/k on each term in system

(1.8.6), we get the equivalent system

(1.8.8)

|b1k1 d1 − b

1kj dj|

h

r1k sx1− 1

k

(j = 2, . . . ,m, (b1, . . . , bm) = 1, (d1, . . . , dm) = 1).

So that, by xing the vector (b1, . . . , bm) with

∆2 :=h

r1/ksx1−1/k, A2 :=

x1/k

Lr1/k, and B2 :=

L

s,

we nd the estimate 1 + hL2

xs2since ∆2/A2 1 holds in the region h < xk/(k+2).

Summing up, we nd

(1.8.9) Jm(Li, h, x, r, s)xm

Lkmi rm+

xm−1h

Lkm−2i s2rm

so that we have shown that

Jm(Li, h, x) x log2m−2 Li

Lk−1i

+∑rsk≤2h

Jm(Li, h, x, r, s)

x log2m−2 Li

Lk−1i

+∑rsk≤2h

min

(Lmism

+ xh

L2k−mi

r2sm, xm

Lkmi

rm+ xm−1h

Lkm−2i

s2rm

).(1.8.10)

We permute the sum on i with the sum over r, s in (1.8.5). We take the rst term of

the minimum in the region Li ≤(xsr

)1/(k+1)in two ways. First, with ε1 = 0, to get∑

rsk≤2h

∑Li≤(xsr )

1k+1

Lmism

+xh

L2k−mi r2sm

∑rsk≤2h

( x

rsk

) mk+1

+xh logc1(m) x

h(2k−m)/kr2sm

xmk+1 logc2(m) x+ xh

m−kk logc1(m) x

and∑rsk≤2h

∑Li>(xsr )

1k+1

xm

Lkmi rm+

xm−1h

Lkm−2i s2rm

∑rsk≤2h

( x

rsk

) mk+1

+

(xm−k+1

rm+2sk(m+2)

) 1k+1

h

xmk+1 logc2(m) x+ x

m−k+1k+1 h.

This is the rst result. The second result is obtained by the choice of a small enough

value of ε1 to get a xed ε.

50

For an integer q ≥ 2, we dene

(1.8.11) R2(x, y; q, f) :=

q∑a=1

|E(x, y; q, a, f)|,

where

(1.8.12) Q(x, y; q, a, f) :=

x+y∑n=x+1

n≡a mod q

f(n) := g(q, a, f)y + E(x, y; q, a, f).

Proposition 1.2. Let f be a function in Ck. Then,

R2(x, y; q, f) y1/kq1−1/k + Sk(x, y).

Proof. For a xed a, we write

Q(x, y; q, a, f) =∑

(k)

t|qt|a

∑d≤zt

(dk,q)=t

g(d)

x+y∑n=x+1

n≡a mod qdk|n

1 +∑

(k)

t|qt|a

∑d>zt

(dk,q)=t

g(d)

x+y∑n=x+1

n≡a mod qdk|n

1

= Σ1(a) + Σ2(a),

say. Clearly,

Σ1(a) =∑

(k)t|qt|a

∑d≤zt

(dk,q)=t

g(d)

(y

[dk, q]+O(1)

)

=y

q

∑(k)

t|qt|a

t∑d≤zt

(dk,q)=t

g(d)

dk+O

(∑(k)

t|qt|a

∑d≤zt

(dk,q)=t

1

).

Notice that the condition (dk, q) = t signies that η(t)|d, it follows that

Σ1(a) = g(q, a, f)y +O

(y

q

∑(k)

t|qt|a

t

ηk(t)

∑d>zt/η(t)

1

dk+∑

(k)

t|qt|a

ztη(t)

)

= g(q, a, f)y +O

(y

q

∑(k)

t|qt|a

t

η(t)zk−1t

+∑

(k)t|qt|a

ztη(t)

)(1.8.13)

On the other hand,

q∑a=1

Σ2(a) =∑

(k)

t|q

∑d>zt

(dk,q)=t

x+y∑n=x+1dk|n

1

51

∑

(k)t|q

∑d>zt

(dk,q)=t

y

dk+ Sk(x, y)

y∑

(k)

t|q

1

η(t)zk−1t

+ Sk(x, y).(1.8.14)

By using (1.8.13) and (1.8.14) in (1.8.11), we obtain that

R2(x, y; q, f) y∑

(k)t|q

1

η(t)zk−1t

+ q∑

(k)t|q

zttη(t)

+ Sk(x, y)

y1/kq1−1/k∑

(k)

t|q

t1/k

tη(t)+ Sk(x, y)

y1/kq1−1/k + Sk(x, y),

where we have chosen

zt =

(yt

q

)1/k

.

Given a xed a, further set

(1.8.15) R3(x, y; q, f) :=∑q≤Q

|E(x, y; q, a, f)|.

Consider also the two functions

τq(n) :=∑d|n

(d,q)=1

1

and

τ1,k(n, z) :=∑dk|nd>z

1.

Proposition 1.3. Let f be a function in Ck and assume that δ > 0 xed. Then, for

xδ y ≤ x, Q ≤ y and 0 6= |a| x,

R3(x, y; q, f) y1/kQ1−1/k log(y) + (Sk(x, y) + y1−θ(1−1/k) +Q)xε,

for any xed θ < 1 and ε > 0. The implicit constant depends on ε, δ, θ and k.

52

= S1 + S2 + S3 + S4,(1.8.17)

say. We then have

S3 =∑

(k)

t|a

∑y1/k<dt|dk

χ[x+1,x+y](d)∑q≤Q

(dk,q)=t


q|mdk−amdk−a6=0

1

Sk(x, y)xε,

also

S2 ∑

(k)t|a

∑yθ/k<d≤y1/k

t|dk


mdk−a6=0

τ(mdk − a)

yxε∑

(k)

t|a

∑yθ/k<d≤y1/k

t|dk

1

dk

y1−θ(1−1/k)xε

and nally

S1 ∑L

∑(k)

t|a

∑zt,L/η(t)<d≤yθ/k/η(t)

∑q∼L/t

(η(t)kdk/t,q)=1

(x+y)/η(t)kdk∑m=(x+1)/η(t)kdk

q|mη(t)kdk/t−a/tmη(t)kdk/t−a/t6=0

1

∑L

∑(k)

t|a



mη(t)kdk/t−a/t6=0

τη(t)kdk(mη(t)kdk/t− a/t)

∑L

∑(k)

t|a



mη(t)kdk−a6=0

τη(t)kdk

(mη(t)kdk − a

)

∑L

∑(k)

t|a


y log(y)

φ(η(t)kdk)

y log(y)∑L

∑(k)

t|a

1

φ(η(t)k)


1

φ(dk)

y log(y)∑L

∑(k)

t|a

η(t)k−1

φ(η(t)k)zk−1t,L

54

y1/k log(y)∑L

L1−1/k∑

(k)t|a

η(t)k−1

t1−1/kφ(η(t)k)

Q1−1/ky1/k log(y),

where we used a result of Shiu [17] and the fact that the result is trivial if Q < y1−θ+ε.

We gather the above estimates for S1, S2 and S3 into (1.8.17). The proof follows by

summing the estimates (1.8.16) and (1.8.17) by the denition (1.8.15).

Let J be a set of non zero integers. We write J for #J and set

(1.8.18) R4(x, y; q, f) :=∑q≤Q

maxa∈J|E(x, y; q, a, f)|.

Proposition 1.4. Let f be a function in Ck and assume that Q ≤ y ≤ x. Then,

R4(x, y; q, f) = (y1/kQ1−1/kJ1/k +Q+ JSk(x, y))xε

for each ε > 0.

Proof. We have

Q(x, y; q, a, f) =∑d≤z

g(d)

x+y∑n=x+1

n≡a mod qdk|n

1 +∑d>z

g(d)

x+y∑n=x+1

n≡a mod qdk|n

1 = Σ1(a, q) + Σ2(a, q),

say. It is clear that, for each a < q,

Σ1(a, q) = g(q, a, f)y +O

(z +

y

q

∑d>z

(dk, q)

dk

).

Thus, the contribution arising from the error term to (1.8.18) is∑q≤Q

z +y

q

∑d>z

(dk, q)

dk zQ+ y

∑q≤Q

1

q

∑(k)

e|q

e

ηk(e)min

(1,ηk−1(e)

zk−1

) zQ+

y

zk−1

∑q≤Q

∑(k)

e|q

e

η(e)

zQ+y logQ

zk−1

∑e≤Q

1

η(e) zQ+

yQε

zk−1.(1.8.19)

On the other hand,∑q≤Q

maxa∈J

Σ2(a, q) ∑q≤Q

maxa∈J

∑z<d≤y1/k

x+y∑n=x+1

n≡a mod qdk|nn6=a

1 +∑q≤Q

maxa∈J

∑d>y1/k

x+y∑n=x+1

n≡a mod qdk|nn6=a

1

55

+Qxε

Jyxε

zk−1+ JSk(x, y)xε +Qxε.(1.8.20)

We get the result by adding (1.8.19) and (1.8.20) with the choice z :=

(Jy

Q

)1/k

.

Consider the function

ψ(x; q, a) :=∑n≤x

n≡a (mod q)

Λ(n)

where

Λ(n) :=

log p if n = pa (a ≥ 1),

0 otherwise

is the classical von Mangoldt function.

Theorem (Bombieri-Vinogradov). Let A > 0 be xed. Then∑q≤Q

maxy≤x

maxa

(a,q)=1

∣∣∣∣ψ(y; q, a)− y

φ(q)

∣∣∣∣ x1/2Q log5 x

provided that x1/2 log−A x ≤ Q ≤ x1/2.

Proposition 1.4 is a kind of interpolation between Proposition 1.3 and a kind of theorem

similar to the Bombieri-Vinogradov's theorem, which is the subject of Proposition 1.5.

We let

R5(x, y; q, f) :=∑q≤Q

maxa∈[1,q]

|E(x, y; q, a, f)|.

Proposition 1.5. Let f be a function in Ck and assume that Q ≤ y ≤ x. Then,

R5(x, y; q, f)

y1/3Q+ X if y1/2 log3/2 y Q y2/3

y2/3Q1/3 log y + X if Q y1/2 log3/2 yif k = 2,

y1/k+εQ1−1/2k + X if k ≥ 3,

where

X := min

(yεQ1/2Sk(x, y), Sk(x, y) +Q(Sk(x, y))1/2 + y1/2(Sk(x, y))1/2 log3/2Q

).

56

Proof. By examining the proof of Proposition 1.4, we nd that

R5(x, y; q, f) zQ+yQε

zk−1+∑q≤Q

maxa∈[1,q]

∑d>z

x+y∑n=x+1

n≡a mod qdk|n

1.

Observe that

∑q≤Q

maxa∈[1,q]

∑d>z

x+y∑n=x+1

n≡a mod qdk|n

1 =∑q≤Q

maxa∈[1,q]

∑z<d≤y1/k

x+y∑n=x+1

n≡a mod q

dk|n

1 +∑q≤Q

maxa∈[1,q]

∑d>y1/k

x+y∑n=x+1

n≡a mod q

dk|n

1

= Σ1 + Σ2,

say. In Σ1, there are Q congruences and about yzk−1 points while in Σ2 there are

Sk(x, y) points. Using Lemma 1.10, we nd that (1.4.15) leads to the estimates

Σ1 min

(Q+ yεQ1/2 y

zk−1,y

zk−1+

Qy1/2

z(k−1)/2+

y

z(k−1)/2log3/2Q

)and

Σ2 X .

The result follows from the choice z =y1/k

Q1/2kand the rst possibility in (1.4.15) for

k ≥ 3 and respectively the choices z = y1/(k+1) and z =y2/(k+1)

Q2/(k+1)log3/(k+1) Q with the

second possibility for k = 2.

57

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153 (2015), 1− 36.

[16] R. C. Orr, REMAINDER ESTIMATES FOR SQUAREFREE INTEGERS IN

ARITHMETIC PROGRESSIONS, Thesis (Ph.D.)-Syracuse University. 1969, 62

pp.

[17] P. Shiu, A Brun-Titchmarsh theorem for multiplicative functions, J. Reine Angew.

Math. 313 (1980), 161− 170.

[18] G. Tenenbaum, Introduction à la théorie analytique et probabiliste des nombres,

3e édition, Belin, 2008.

[19] R. C. Vaughan, A variance for k-free numbers in arithmetic progressions. Proc.

London Math. Soc. (3) 91 (2005), no. 3, 573− 597.

[20] R. Warlimont, On squarefree numbers in arithmetic progressions, Monatsh.

Math. 73 (1969), 433− 448.

59

Chapter 2

A Brun-Titchmarsh inequality for

k-free numbers

Patrick Letendre

Résumé

Nous obtenons une inégalité de Brun-Titchmarsh pour les nombres libres de puis-

sances k-ièmes qui améliore sur ce qui peut être obtenu avec le crible combinatoire

standard.

Abstract

We obtain a Brun-Titchmarsh inequality for k-free numbers that improves upon

results that can be done using standard combinatorial sieve.

60

2.1 Introduction and notation

Let φ stand for the Euler function and π(x; q, a) stand for the number of primes p ≤ x

such that p ≡ a (mod q). The Brun-Titchmarsh inequality, as proven by Montgomery

and Vaughan in [7], says that

π(x+ y; q, a)− π(x; q, a) <2y

φ(q) log(y/q)(a, q) = 1, x > 0, y > q.

Here we prove an analogue of the Brun-Titchmarsh inequality for k-free numbers.

Given an integer k ≥ 2, let µk stand for the indicator function of the set of k-free

numbers. Let also

Qk(x, z) := #n ≤ x : pk|n⇒ p > z,

so that Qk(x) := Qk(x, x1k ) is the usual counting function for the set of k-free numbers

not exceeding x. One can easily show that

Qk(x) =x

ζ(k)+O(x

1k ).

Some numerical computations lead us to suggest that there exists a small constant

c > 0 for which ∣∣∣∣∣Qk(x, z)− x∏p≤z

(1− 1

pk

)∣∣∣∣∣ π(z)

holds uniformly in the region z x1−ck . However this is still an open problem.

This paper deals with simpler related questions. Let a and q be positive integers

satisfying

(2.1.1) a ∈ [1, q] and gcd(a, q) = 1.

For any positive integers x and h, let

Qk(x; q, a) :=x−1∑n=0

µk(qn+ a),

Gk(x, h; q, a) := Qk(x+ h; q, a)−Qk(x; q, a)

and

gk(h; q, a) := maxx≥0

Gk(x, h; q, a).

61

We also write Gk(x, h) := Gk(x, h; 1, 1) and gk(h) := gk(h; 1, 1) and dene the function

ck(q) :=∏p|q

(1− 1

pk

)−1

.

In [2], Filaseta & Trifonov proved that

Gk(x, h) =h

ζ(k)(1 + o(1))

given thath

x1

2k+1 log x→∞ when h, x→∞.

They also showed that the largest gap between any two k-free integers is x1

2k+1 log x.

These results have not been improved since.

In the opposite direction, a result of Erd®s (see [1]) says that there exists gaps between

k-free numbers of length

ζ(k)

k

log x

log log x(1 + o(1)) (x→∞).

In [9], Tsang showed how to obtain estimates for certain tuples of squarefree numbers.

Here we use the same argument to give an upper bound for gk(h; q, a). Also, we build a

strategy that exploits the well known fact that the Buchstab-Rosser sieve gives better

results for the lower bound than for the upper bound when it comes to counting numbers

not divisible by a small k-th prime power. This leads to Corollary 2.1 where the formula

reveals that obtaining any improvement on traditional results for Gk(x, h; q, a), with h

small enough when compared to x, is equivalent to a non trivial upper bound for the

cardinality of the sets

S(M1,M2) := n ∈ [x+ 1, x+ h] : ∃ (p1, p2) ∈M1 ×M2 with p1 6= p2 such that pk1pk2 | n

whereMi (i = 1, 2) are sets of primes that we will dene later. Bounding |S(M1,M2)|can be done by using ideas from analytic number theory such as exponential sums

or more eciently by using various estimates for the number of integer points close

to a smooth curve (see [2], [4], [5] and [6]). The main novelty of this paper is the

determinants that are used that give limitations on some structured patterns that can

occur within S(M1,M2), which suggest that they can in fact be avoided. Then, with

a simple combinatorial argument, we show that under certain conditions a good upper

bound can be obtained. This result is given in Theorem 2.3. We believe that the

method we use to obtain this upper bound is of independent interest.

62

2.2 Preliminary results

We rst observe that, for any given pair of integers (a, q) satisfying (2.1.1),

(2.2.1) gk(m+ n; q, a) ≤ gk(m; q, a) + gk(n; q, a) (m,n ∈ Z≥0).

On the other hand, using the sieve of Eratosthenes, one can construct the following

table.

h g2(h)⌈

hζ(2)

⌉g3(h)

⌈hζ(3)

⌉5 4 4 5 5

6 5 4 6 5

7 6 5 7 6

8 6 5 7 7

9 7 6 8 8

10 8 7 9 9

15 12 10 14 13

16 12 10 14 14

17 12 11 15 15

20 14 13 18 17

25 18 16 22 21

30 21 19 27 25

40 27 25 35 34

50 34 31 44 42

h g2(h)⌈

hζ(2)

⌉g3(h)

⌈hζ(3)

⌉60 40 37 52 50

70 47 43 61 59

80 53 49 69 67

90 60 55 77 75

100 65 61 86 84

150 98 92 128 125

200 128 122 169 167

214 138 131 182 179

215 138 131 182 179

216 138 132 182 180

250 161 152 212 208

300 192 183 254 250

400 253 244 337 333

500 316 304 421 416

Indeed, to do so, we consider a xed set P of prime numbers and dene

P (z) :=∏p<zp∈P

p.

By taking P to be the set of all the prime numbers, we easily get

gk(h) = maxx∈1,... ,Pk(z)

∑d|P (z)

µ(d)

([x+ h

dk

]−[x

dk

])(2.2.2)

= h∏p<z

(1− 1

pk

)+ Ek(h)

where

Ek(h) := maxx∈1,... ,Pk(z)

∑d|P (z)

µ(d)

(x

dk

−x+ h

dk

).

Here y stands for the fractional part of y. Then choosing z large enough, one can see

that (2.2.2) holds. Details are given in the proof of Theorem 2.2. For large values of h,

63

one can speed up the process by nding the list of best congurations with only small

primes explicitly. Then we place the larger primes (less than z) directly in an optimal

way. This last idea can be generalized, but quickly becomes very complicated.

We see that the function gk satises ∆k(n) := gk(n + 1) − gk(n) ∈ 0, 1 for each n.The inequality (2.2.1) tells us that ∆k(n) = 1 can hold for at most 2k − 1 consecutive

integers. We do not know if ∆k(n) = 0 is possible for an arbitrary long sequence of

consecutive numbers.

We are interested by gk(h; q, a) as h→∞. A basic answer to this question is given by

the next two theorems.

Theorem 2.1. For every integer k ≥ 2 and a pair of integers (a, q) satisfying (2.1.1),

there exists a positive constant c1,k such that

gk(h; q, a)− ck(q)h

ζ(k)< c1,k

h2/(k+1)

(log h)2k/(k+1).

Theorem 2.2. Fix C1, C2 > 0. For every k ≥ 2 and a pair of integers (a, q) satisfying

(2.1.1) and q ≤ C1hC2,


ζ(k)> c2,k

h1/k

(log h)1/k(log log h)1−1/k

for some positive constant c2,k which depends on C1 and C2 only.

Our main motivation during the construction of this paper has been to improve Theorem

2.1, because we think that the truth should be close to the following conjecture.

Conjecture 2.1. For every integer k ≥ 2 and a pair of integers (a, q) satisfying (2.1.1),

we have

(2.2.3) gk(h; q, a) ≤ ck(q)h

ζ(k)+ c(ε, k)h1/k+ε

for every ε > 0.


We will construct explicitly an arithmetic progression for which the number of k-free

numbers is large. To do so, rst observe that setting Q = pk1 · · · pkr with (Q, q) = 1,

Q∑m=1

h∑j=1

µk((Q, q(m+ j) + a)) =

Q∑m=1

h∑j=1

∑dk|(Q,q(m+j)+a)

µ(d)

64

= hQ∑dk|Q

µ(d)

dk= hQ

∏p|Q

(1− 1

pk

).

From this, it follows that there exists at least one interval of length h with at least

h∏p|Q

(1− 1

pk

)

integers x with µk((Q, qx + a)) = 1. We will use this result with Q =∏p<zp-q

pk for some

z to be chosen later. Also, for any given set of distinct integers a1, . . . , al such that

ai ≡ a (mod q) for i = 1, . . . , l, one can show using the sieve of Eratosthenes that

(2.3.1)x∑

n=1

µk(qn+ a1) · · ·µk(qn+ al) = (1 + o(1))∏p-q

(1− η(p)

pk

)x (x→∞),

where

(2.3.2) η(p) := #ai (mod pk), i = 1, . . . , l.

This last statement says that any given type of l-tuple of k-free numbers of the form

(qn + a1, . . . , qn + al) exists if and only if it satises η(p) < pk for each prime p not

dividing q.

For any integers a and q satisfying (2.1.1), if p - q, the congruence

qn+ a ≡ 0 (mod pk)

has only one solution n in [0, pk − 1]. This implies that the set of integer solutions of

this congruence is an arithmetic progression of common dierence pk. It follows from

this observation that each prime p with pk > h can be positioned to divide none of

the integers in an arithmetic progression of length h. Therefore, from now on we can

assume that each prime p is less than or equal to h1/k. We write q1 < q2 < . . . for the

sequence of prime numbers that do not divide q.

From the above, we deduce that if qk1 +h

2≤ pk ≤ h, then pk divides at most one number,

say n1, in an arithmetic progression of length h and we can manage to have qk1 | n1.

Assume now that we have qk1 ·qk2 +h

3≤ pk ≤ h. In this case, pk divides at most two

numbers, say n1 and n2, and since we can solve for n the system of congruencesqn+ a ≡ 0 (mod qk1)

q(n+ pk) + a ≡ 0 (mod qk2),

65

it follows that we can manage to have qk1 | n1 and qk2 | n2. In general, when we have∏rj=1 q

kj+h

r+1≤ pk ≤ h (see Remark 2.1), we can have at most r multiples of pk, say

n1, . . . , nr. We solve for n the system of r congruencesqn+ a ≡ 0 (mod qk1)

q(n+ pk) + a ≡ 0 (mod qk2)...

q(n+ (r − 1)pk) + a ≡ 0 (mod qkr )

to ensure that qkj | nj (j = 1, . . . , r). We can choose the largest r satisfying

r < cklog h

log log h

with ck small enough to have ∏rj=1 q

kj + h

r + 1<

2h

r + 1.

This is possible, since by assumption we have q hO(1) and therefore ω(q) log hlog log h

.

With this choice of r, we can take z = βk(h log log h

log h

)1/kwith a suitable constant βk that

depends only on C1 and C2. We have thus found an interval of length h with more than

h∏

p<βk(h log log hlog h )

1/k

p-q

(1− 1

pk

)≥ ck(q)

h

ζ(k)+ c1,k

h1/k


k-free numbers, thereby completing the proof of Theorem 2.2.

Remark 2.1. An easy way to prove this inequality, along with the announced property,

is to consider an increasing sequence of∏r

j=1 qkj + h consecutive numbers of the shape

qm+ a. Then, we set the number corresponding to q(n− pk) + a (mod∏r

j=1 qkj ) in the

rst∏r

j=1 qkj elements of the sequence to be a multiple of pk. It only remains to verify

that with all these choices we cannot have r + 2 multiples of pk in the whole sequence.

If the condition q ≤ C1 exp(C2h1k ) holds, then it is easy to deduce from the proof of

Theorem 2.2 that

gk(h; q, a) ≥ h∏p<h

1k

p-q

(1− 1

pk

)= ck(q)

h

ζ(k)+c3,kh

1k

log h

where c3,k is positive and depends on C1 and C2.

66

Moreover, there exists absolute constants ck (k ≥ 2) for which∑0≤i≤ckz

∆k(h+ i) ≥ 1

where z := (h log log h)1k

logk+1k h

. Indeed, one can simply choose an optimal conguration with h

consecutive numbers, in which case, the worst possible setup is when each number near

this interval is sieved by the primes p ≤ 2βk(h log log h

log h

)1/k=: v. We cannot sieve more

than π(v) integers with these primes. To prove this, we proceed by contradiction

assuming that there are cπ(v) consecutive integers each of which is a multiple of a k-th

power of a prime p ≤ v. For each such prime p, there are at most cπ(v)pk

+ 1 multiples

of pk implying

cπ(v)−∑p≤v

(cπ(v)

pk+ 1

)≤ 0,

which is clearly false for c suciently large. The result then follows.


We provide a proof for the sake of completeness. It is based on standard sieve tech-

niques. First, for a xed sequence A of integers and a xed sequence of prime numbers

P , we letAd := n ∈ A : d|n

and

(2.4.1) S(A,P , z) :=∑n∈A

(n,P (z))=1

1.

When P is clear from the context, we simply write S(A, z). Usually, we write |Ad|for the cardinality of the set Ad, also we write |A| for |A1|. For d = p1p2 · · · pr withp1 > p2 > · · · > pr, we dene the arithmetic function

Pj(d) := pj

for each j ∈ 1, . . . , r. Finally, by Pj we mean a set of j-tuples of prime numbers

from P that verify some properties to be dened soon.

To obtain a sieve with the desired property, we make repeated use of the Buchstab

formula

(2.4.2) S(A,P , z) = |A| −∑p<zp∈P

S(Ap,P , p)

67

(see [3] p. 59 for further details). Then, we see that, if we set λ+(d) := µ(d)θ+(d) (where

θ+ : N → 0, 1 is an indicator function) with d = p1p2 · · · pj and p1 > p2 > · · · > pj,

we can choose a function θ+ supported according to the list of inequalities

p1 p1 ∈ P1

p2 < p1 p1 ∈ P1, p2 ∈ P

p3 < p2 < p1 (p3, p2, p1) ∈ P3(2.4.3)

p4 < p3 < p2 < p1 (p3, p2, p1) ∈ P3, p4 ∈ P

p5 < p4 < p3 < p2 < p1 (p5, p4, p3, p2, p1) ∈ P5

...

with the additional condition that

(2.4.4) (p2j+1, . . . , p1) ∈ P2j+1 =⇒ (p2j−1, . . . , p1) ∈ P2j−1

for each j ≥ 1. We then choose P2j+1 to be the set of (2j + 1)-tuples (p2j+1, . . . , p1)

with

p2j+1 · · · p1 < z2j+1 and p2j+1 < · · · < p1

for some sequence

z1, z3, . . . , z2j+1, . . .

to be chosen later, keeping in mind the additional condition (2.4.4). With the same

idea, we see that, if we set λ−(d) := µ(d)θ−(d) (again θ− : N → 0, 1 is an indicator

function) with d = p1p2 · · · pj and p1 > p2 > · · · > pj, then we can choose a function

θ− supported according to the list of inequalities

p1 p1 ∈ P0

p2 < p1 (p2, p1) ∈ P2

p3 < p2 < p1 (p2, p1) ∈ P2, p3 ∈ P

p4 < p3 < p2 < p1 (p4, p3, p2, p1) ∈ P4

p5 < p4 < p3 < p2 < p1 (p4, p3, p2, p1) ∈ P4, p5 ∈ P...

with the additional condition that

(2.4.5) (p2j+2, . . . , p1) ∈ P2j+2 =⇒ (p2j, . . . , p1) ∈ P2j

68

for each j ≥ 1 and p1 ∈ P0. We have to take P0 = p : p | P (z) and we choose P2j to

be the set of 2j-tuples (p2j, . . . , p1) with

p2j · · · p1 < z2j and p2j < · · · < p1

for some

z2, z4, . . . , z2j, . . .

to be chosen later, keeping in mind the additional condition (2.4.5).

Now we are ready to start with the argument with P = p : p - q. Let

I(dk) := #n ∈ [x+ 1, x+ h] : dk|qn+ a

so that we may write

I(dk) :=

h

dk+ r(dk) if (d, q) = 1,

0 otherwise,

where r(dk) is dened implicitly and |r(dk)| ≤ 1. We then have

x+h∑n=x+1

µk(qn+ a) ≤x+h∑

n=x+1

∑dk|qn+a

λ+(d)

=∑d≥1

µ(d)θ+(d)I(dk)

= h∑d≥1

(d,q)=1

µ(d)θ+(d)

dk+∑d≥1

(d,q)=1

µ(d)θ+(d)r(dk)

= ck(q)h

ζ(k)+ h

∑d≥1

(d,q)=1

µ(d)(θ+(d)− 1)

dk+∑d≥1

(d,q)=1

µ(d)θ+(d)r(dk)(2.4.6)

=: S1 + S2 + S3,

say. On the one hand,

S3 ≤∑j≥12-j

∑pj<···<p1pj ···p1<zj

1 +∑j≥12|j

∑pj<···<p1

pj−1···p1<zj−1

1

∑j≥12-j

zj +z2

1

log2 z1

+∑j≥32|j

∑pj<···<p1

pj ···p1<zjj−1j−1

1

69

z21

log2 z1

+∑j≥32-j

zj+1j

j .

The remaining terms in S2 are relevant only if θ+(d) = 0 and µ(d) = −1. We then

order the terms according to the rst unsatised line of (2.4.3). Assuming that it is

the rst one which is not satised, we then have

−h∑

P1(d)=p1

(d,q)=1p1≥z1

µ(d)

dk= h

∑p1≥z1

(p1,q)=1

1

pk1

∑d≥1

(d,q)=1P1(d)<p1

µ(d)

dk= h

∑p1≥z1

(p1,q)=1

1

pk1

∏p<p1

(p,q)=1

(1− 1

pk

) h

zk−11 log z1

.

Assuming that it is the j-th (j odd) line which is not satised, we have

−h∑

P1(d)=p1,... ,Pj(d)=pj(d,q)=1p1<z1

p3p2p1<z3...

pj ···p1≥zj

µ(d)

dk= h

∑P1(d)=p1,... ,Pj(d)=pj

(d,q)=1p1<z1

p3p2p1<z3...

pj ···p1≥zj

1

(p1 · · · pj)k∑d≥1

(d,q)=1P1(d)<pj

µ(d)

dk

= h∑

P1(d)=p1,... ,Pj(d)=pj(d,q)=1p1<z1

p3p2p1<z3...

pj ···p1≥zj

1

(p1 · · · pj)k∏p<pj

(p,q)=1

(1− 1

pk

)

h∑

pj<···<p1

pj ···p1≥zj

1

(p1 · · · pj)k h

zk−1j

.

Hence, by gathering these last estimates in (2.4.6), we obtain


ζ(k) z2

1

log2 z1

+h

zk−11 log z1

+∑j≥32-j

zj+1j

j +∑j≥32-j

h

zk−1j

,

where j log hlog log h

. Choosing z1 := (h log h)1k+1 and zj := h

jjk+1 otherwise, it follows

that


ζ(k) h

2k+1

(log h)2kk+1

+ h4

3k+1 log h,

thus completing the proof of Theorem 2.1.

70

2.5 Relation among the gk(h; q, a)

For each pair of integers (a, q) satisfying (2.1.1) we consider the sets H := h1, . . . , hland

Ξ(x,H; q, a) := q(x+ hi) + a : hi ∈ H for i = 1, . . . , l.

Recalling the denition of η(p) given in (2.3.2), we say that Ξ(x,H; q, a) is admissible

if for each prime p not dividing q we have η(p) < pk. We have seen in the proof of

Theorem 2.2 that there exists innitely many l-tuples of k-free numbers of the form

(q(x+ h1) + a, . . . , q(x+ hl) + a) if and only if Ξ(x,H; q, a) is admissible.

Proposition 2.1. For any two pairs of integers (a1, q) and (a2, q) satisfying (2.1.1),

we have

gk(h; q, a1) = gk(h; q, a2).

Proof. It suces to show that, for any xed H, if Ξ(x,H; q, a1) is admissible then so is

Ξ(x,H; q, a2). For this it suces to show that for each prime p that does not divide q

we have η1(p) = η2(p), where ηi refers to the system Ξ(x,H; q, ai) (i = 1, 2). This is an

easy consequence of the fact that

q(x+ hi) + a1 ≡ q(x+ q−1(a1 − a2) + hi) + a2 (mod pk).

Let

γ(n, z) :=∏p|np<z

p (n ∈ N, z ∈ R≥0).

Proposition 2.2. Fix any two pairs of integers (a1, q1) and (a2, q2) satisfying (2.1.1)

and q1, q2 ≤ C1hC2. Then for z = βk

(h log log h

log h

)1/k(see the proof of Theorem 2.2) we

have

γ(q1, z) = γ(q2, z) =⇒ gk(h; q1, a1) = gk(h; q2, a2).

Proof. It suces to show that, for any xed H, if Ξ(x,H; q1, a1) is admissible then so

is Ξ(x,H; q2, a2). We have seen in the proof of Theorem 2.2 how to show that large

primes do not aect gk(h; q, a). For each other xed prime p that does not divide q1 or

q2 we have an element r such that q1 ≡ rq2 (mod pk) and (r, pk) = 1. We deduce that

q2(x+ hi) + a2 ≡ 0 (mod pk) ⇔ rq2(x+ hi) + ra2 ≡ 0 (mod pk)

71

⇔ q1(x+ hi) + ra2 ≡ 0 (mod pk)

⇔ q1(x+ q−11 (ra2 − a1) + hi) + a1 ≡ 0 (mod pk).

We conclude that η1(p) = η2(p) where ηi refers to the system Ξ(x,H; qi, ai) (i = 1, 2)

and the result follows.

Remark 2.2. The last proof also tells us that if we set

µk(n; q) :=

1 pk | n⇒ p|q0 otherwise,

Qk(x; q) :=x∑

n=1

µk(n; q),

Gk(x, h; q) := Qk(x+ h; q)−Qk(x; q)

and

gk(h; q) := maxx≥0

Gk(x, h; q),

then we have the relation

gk(h; q) = gk(h; q, a)

for every integers a and q satisfying (2.1.1).

2.6 A sieve identity

As a rst step toward Conjecture 2.1, we prove the following lemma.

Lemma 2.1. Let A be a nite sequence of integers and P be a set of distinct prime

numbers. For any real z ≥ 1 and real numbers zp satisfying 1 ≤ zp ≤ z for each prime

p that divides P (z), there exists ϑ with |ϑ| ≤ 1 such that

S(A,P , z) = |A| −∑p|P (z)

∑n∈Ap

if p′|(n,P (z)/p)

then

p′<p or p′≥zp if p<zp

p′<zp if p≥zp

1 + ϑ∑n∈A

ξ(n),

where

ξ(n) := #p : ∃ p′ ∈ [zp, z) s.t. pp′ | (n, P (z)).

Proof. We write

S(A,P , z) = |A| −∑p|P (z)

∑n∈Ap

if p′|(n,P (z)/p)then p′<p

1(2.6.1)

72

= |A| −∑p|P (z)

∑n∈Ap

if p′|(n,P (z)/p)

then


p′<zp if p≥zp

1 + S(2.6.2)

where zp are real variables satisfying 1 ≤ zp ≤ z and where S is dened implicitly as

we move from (2.6.1) to (2.6.2). In fact,

S =∑p|P (z)

( ∑n∈Ap

if p′|(n,P (z)/p)

then


p′<zp if p≥zp

1−∑n∈Ap

if p′|(n,P (z)/p)then p′<p

1

)

=:∑p|P (z)

∑n∈Ap

χp(n),

say. One can verify that

χp(n) =

1 if ∃ p′ ∈ [zp, z) s.t. pp′ | (n, P (z)) when p < zp,

−1 if ∃ p′ ∈ [zp, p) s.t. pp′ | (n, P (z)) when p ≥ zp,

0 otherwise.

This implies that

|S| =∣∣∣∣ ∑p|P (z)

∑n∈Ap

χp(n)

∣∣∣∣ ≤ ∑p|P (z)

∑n∈Ap

|χp(n)| ≤∑n∈A

ξ(n),

thereby completing the proof of Lemma 2.1.

Corollary 2.1. Let a and q be integers satisfying (2.1.1) and P be the set of prime

numbers that are coprime to q. For any integer h ≥ 2 and z ≥ 32we have

Gk(x, h; q, a) ≤ ck(q)h

ζ(k)+c1(h log log h)

32k+1

(log h)6k−32k+1

+c2h

zk−1 log z

+c3z

log z+ c4

∑p|P (z)p<zp

zplog zp

+x+h∑

n=x+1

ν(n),(2.6.3)

where c1, c2, c3 and c4 are positive absolute constants, 32≤ zp ≤ z are variables to be

chosen for each prime that divides P (z) and

ν(n) := #p : ∃ p′ ∈ [max(zp, p), z) s.t. pkp′k | (qn+ a, P k(z)).

73

Proof. Let

γk(n) :=∏pk|n

p (n ∈ N)

and

A := γk(qn+ a) : n ∈ [x+ 1, x+ h].

Our starting point is the formula (2.6.2) written in the form

(2.6.4) S(A,P , z) = |A| −∑p|P (z)

∑n∈Ap

if p′|(n,P (z)/p)

then


p′<zp if p≥zp

1 +∑p|P (z)

∑n∈Ap

χp(n).

Consider the sets

P1 := p < z : p - q and p < zp and P2 := p < z : p - q and p ∈ [zp, z).

We have to evaluate the innermost sum in the second term of the RHS of (2.6.4). We

x a prime p. If p belongs to P1 then, by using an argument similar as in the proof of

Theorem 2.1, one can show that

(2.6.5)∑n∈Ap

if p′|(n,P (z)/p)then p′<p or p′≥zp

1 ≥ h

pk

∏p1>pp1-q

(1− 1

pk1

)− c4

zplog zp

− c5(h log log h)

32k+1

p3k

2k+1 (log h/pk)6k−32k+1

while if p belongs to P2, we have∑n∈Ap

if p′|(n,P (z)/p)then p′<zp

1 ≥ h

pk− 1−

∑n∈Ap

∃ p′ s.t. p′|(n,P (z)/p)and p′∈[zp,z)

1

≥ h

pk

∏p1>pp1-q

(1− 1

pk1

)− 1−

∑n∈Ap


1.(2.6.6)

Using inequalities (2.6.5) and (2.6.6), we can expand the RHS of (2.6.4). There is a

remarkable cancellation between the third term in the RHS of (2.6.4) and the sum over

p ∈ P2 of the third term in (2.6.6), that is∑p|P (z)

∑n∈Ap

χp(n) +∑p∈P2

∑n∈Ap


1 =∑n∈A

ν(n).

A routine computation concludes the proof.

74

Remark 2.3. One can see that (2.6.3) is also true when Gk(x, h; q, a) is replaced by

Gk(x, h; q) and ν(n) is replaced by

ν ′(n) := #p : ∃ p′ ∈ [max(zp, p), z) s.t. pkp′k | (n, P k(z)).

2.7 Counting special pairs

Let us consider the matrix F := ((t+ hi,j)q + a)J×J , that is

(2.7.1) F =

(t+ h1,1)q + a (t+ h1,2)q + a · · · (t+ h1,J)q + a

(t+ h2,1)q + a (t+ h2,2)q + a · · · (t+ h2,J)q + a...

......

(t+ hJ,1)q + a (t+ hJ,2)q + a · · · (t+ hJ,J)q + a

where we assume that 1 ≤ a ≤ q are positive integers, that h1,1 = 0, that hi,j are

pairwise distinct (i, j = 1, . . . , J) and that t ∈ x, . . . , x + h − 1 as well as each

t + hi,j. We also assume that there are 2J powers of distinct primes qi := pki which

satisfy

(2.7.2)

qi | (t+ hj,i)q + a when i ∈ [1, J ] and j ∈ [1, J ]

qi | (t+ hi−J,j)q + a when i ∈ [J + 1, 2J ] and j ∈ [1, J ].

We see that the determinant of F is equal to the determinant of the matrix

F ′ :=

tq + a h1,2q · · · h1,Jq

h2,1q (h2,2 − h1,2 − h2,1)q · · · (h2,J − h1,J − h2,1)q...

......

hJ,1q (hJ,2 − h1,2 − hJ,1)q · · · (hJ,J − h1,J − hJ,1)q

and, by expanding the determinant of F ′ along the rst row, we nd that

detF = detF ′ = A(tq + a) +B

where

A := det

(h2,2 − h1,2 − h2,1)q · · · (h2,J − h1,J − h2,1)q

......

(hJ,2 − h1,2 − hJ,1)q · · · (hJ,J − h1,J − hJ,1)q

and

B := det

0 h1,2q · · · h1,Jq

h2,1q h2,2q · · · h2,Jq...

......

hJ,1q hJ,2q · · · hJ,Jq

.

75

We see that q1 · · · q2J divides detF and we obtain an upper bound for the size of the

product provided that the determinant is not 0. This idea is made rigorous in the

following lemma.

Lemma 2.2. Let F be as in (2.7.1). Assume also that (q, a) = 1 and that J ≥ 2 is an

integer. If the condition

(2.7.3) h ≤ (xq)1/J

√J

is satised, then we have

(2.7.4) q1 · · · q2J ≤ 2JJJ/2xqhJ−1.

Proof. We observe that the rank of the matrix F is at least 2. Indeed, we have

Γ := det

(tq + a (t+ h1,2)q + a

(t+ h2,1)q + a (t+ h2,2)q + a

)= (tq + a)q(h2,2 − h1,2 − h2,1)− h1,2h2,1q

2.

If Γ = 0 then we deduce from (q, a) = 1 that q|h2,2−h1,2−h2,1. Since h1,1 = 0, we have

h1,2h2,1 6= 0 by hypothesis. It follows that Γ = 0 would imply that h2 > |h1,2h2,1| ≥tq ≥ xq, in contradiction with (2.7.3). Now, if the rank of F is J then the result follows

easily using Hadamard's inequality for determinants. We can assume that the rank

of F is an integer r ∈ [2, J − 1]. By permuting the rows and columns of the matrix

((t + hi,j)q + a)J×J we may assume that the lower right determinant is not 0. We x

s ∈ [1, J − r] and extract the only (r + 1) × (r + 1) submatrix which contains the

diagonal term (t + hs,s)q + a and the lower right r × r submatrix. By hypothesis, the

determinant of this matrix is 0. By doing the translation t 7→ t − hs,s and redening

the variables hi,j, we get the matrixtq + a (t+ h1,2)q + a · · · (t+ h1,r+1)q + a

(t+ h2,1)q + a (t+ h2,2)q + a · · · (t+ h2,r+1)q + a...

......

(t+ hr+1,1)q + a (t+ hr+1,2)q + a · · · (t+ hr+1,r+1)q + a

.

Again, the determinant of this matrix is of the form (detC) · (tq + a) +D with

C :=

(h2,2 − h1,2 − h2,1)q · · · (h2,r+1 − h1,r+1 − h2,1)q

......

(hr+1,2 − h1,2 − hr+1,1)q · · · (hr+1,r+1 − h1,r+1 − hr+1,1)q

76

and

D := det

0 h1,2q · · · h1,r+1q

h2,1q h2,2q · · · h2,r+1q...

......

hr+1,1q hr+1,2q · · · hr+1,r+1q

.

Using the fact that (detC) · (tq+ a) +D = 0 and (q, a) = 1 we deduce that qr+1| detC.

Then since h ≤ (xq)1/J√J

we must have detC = D = 0. We focus on the matrix C as

being a submatrix of

M :=

tq + a h1,2q · · · h1,r+1q

h2,1q (h2,2 − h1,2 − h2,1)q · · · (h2,r+1 − h1,r+1 − h2,1)q...

......

hr+1,1q (hr+1,2 − h1,2 − hr+1,1)q · · · (hr+1,r+1 − h1,r+1 − hr+1,1)q

,

a matrix of determinant 0. By permuting columns if necessary, using the notation

detC =: det(v2, . . . , vr+1), we may assume that we have a linear combination

(2.7.5) vr+1 = λ2v2 + · · ·+ λrvr.

We then consider the matrix

M ′ :=

tq + a h1,2q · · · (h1,r+1 − λ2h1,2 − · · · − λrh1,r)q

h2,1q (h2,2 − h1,2 − h2,1)q · · · 0...

......

hr+1,1q (hr+1,2 − h1,2 − hr+1,1)q · · · 0

which has the same determinant. We then have two possibilities: the lower left subma-

trix, say E, has its determinant equal to 0 or else

h1,r+1 = λ2h1,2 + · · ·+ λrh1,r.

In this latter case, we see that each r×r submatrix on the right inM has its determinant

equal to 0. Now assume that we are in the case when detE = 0 and consider any r× rsubmatrix ofM in the last r rows which is not C or E. This matrix contains the vector

vr+1. By inserting the linear combination (2.7.5) and simplifying the determinant for

each λi in the matrix we get that this determinant is ±λj detE for the missing j. This

is clearly 0. Overall we have shown that either each r× r submatrix of M , at the right

or at the bottom, has its determinant equal to 0.

We therefore have two possibilities and the proof of one is essentially the transpose of

the other. We will assume that each r× r submatrix at the right inM has determinant

77

0. With each such matrix, dierent from C, we add the rst row to all other rows. By

doing so, we maintain the fact that their determinant is equal to 0. Now, consider the

matrix

M ′′ :=

tq + a h1,2q · · · h1,r+1q

(t+ h2,1)q + a (h2,2 − h2,1)q · · · (h2,r+1 − h2,1)q...

......

(t+ hr+1,1)q + a (hr+1,2 − hr+1,1)q · · · (hr+1,r+1 − hr+1,1)q

which has its determinant equal to 0. By expanding with the rst column we nd that

c := det

(h2,2 − h2,1)q · · · (h2,r+1 − h2,1)q

......

(hr+1,2 − hr+1,1)q · · · (hr+1,r+1 − hr+1,1)q

= 0

since tq + a 6= 0. Using the fact that qsqs+J |tq + a, we nd qs|hj,1 for j = 2, . . . , r + 1.

Moreover,

0 = c ≡ det

h2,2q · · · h2,r+1q...

...

hr+1,2q · · · hr+1,r+1q

(mod qs).

This shows that, dening

F := det

(t+ h2,2)q + a · · · (t+ h2,r+1)q + a

......

(t+ hr+1,2)q + a · · · (t+ hr+1,r+1)q + a

and using qs|tq + a, we have qs|F . By hypothesis, we have F 6= 0 and

qr−1qJ−r+1 · · · qJq2J−r+1 · · · q2J |F.

Now we have learned that qs (or qs+J) can be added to this product. We repeat this

argument for s = 1, . . . , J − r and we get

qJ−1q1 · · · q2J |FqJ−rq1+α1J · · · qJ−r+αJ−rJ ≤ 2JJJ/2xq(hq)J−1

where again we used Hadamard's inequality, the fact that qj ≤ h and that αj ∈ 0, 1from j ∈ 1, . . . , J − r. This completes the proof.

Now we provide an example of how Lemma 2.2 can be used.

78

Lemma 2.3. Fix the positive integers x and h with h ≤ x, integers L1 and L2, integers

q and a that satisfy (2.1.1), an integer k ≥ 2 and an integer J ≥ 2. Let Mi (i = 1, 2)

be sets of prime numbers with smallest elements at least Li (i = 1, 2). Let Mi := #Mi

(i = 1, 2). Assume that (2.7.3) holds, and moreover that

(2.7.6) LJk1 LJk2 ≥ 2JJJ/2xqhJ−1,

(2.7.7) Lk1Lk2 > h,

that

(2.7.8) J ≤ M1

5M1/J2

are satised. Further set

S(M1,M2) := n ∈ [x+ 1, x+ h] : ∃ (p1, p2) ∈M1 ×M2 with p1 6= p2 such that pk1pk2 | qn+ a.

Then we have

(2.7.9) #S(M1,M2) ≤ 5M1M1−1/J2 .

Proof. For each n in S(M1,M2) we keep only one pair (p1, p2) with p1 6= p2 and

pk1pk2 | qn+ a and we introduce the set

S ′(M1,M2) := (p1, p2, n) : n ∈ S(M1,M2) and (p1, p2) is the chosen pair for n.

We have #S(M1,M2) = #S ′(M1,M2). A pair (p1, p2) denes the value of n in

S(M1,M2) from (2.7.7). We dene w(p) for each p ∈M2 by

w(p) := p1 ∈M1 : ∃ n such that (p1, p, n) is in S ′(M1,M2).

We have the identity

#S ′(M1,M2) =∑p∈M2

#w(p).

Now, all the conditions are met to apply Lemma 2.2. The hypothesis (2.7.6) shows

that (2.7.4) is not satised. This implies that we cannot have the structured pattern

that appears in the matrix F . We can interpret this fact by saying that any set of J

primes inM1 determines at most J − 1 primes inM2, that is any J-tuplet of distinct

primes inM1 is in at most J−1 sets w(p) with p ∈M2. Hypothesis (2.7.8) guarantees

that there is at least one such J-tuplet. Let z ≥ 1 be an integer to be chosen later

79

and cover the interval [z,M1] with dyadic segments [2jz, 2j+1z), j = 0, . . . , s with

s := blog(M1/z)/ log 2c. For each xed j, we consider the subset of primes p ∈M2 for

which #w(p) ∈ [2jz, 2j+1z). There are(M1

J

)subsets of J distinct primes fromM1 and

each such set w(p) has at least(

2jzJ

)of these subsets. It follows that

#S ′(M1,M2) ≤ (z − 1)M2 + (J − 1)s∑j=0

2j+1zM1(M1 − 1) · · · (M1 − J + 1)

2jz(2jz − 1) · · · (2jz − J + 1)

≤ (z − 1)M2 + 2(J − 1)MJ1

s∑j=0

1

(α2jz)J−1

≤ (z − 1)M2 + 2(J − 1)MJ

1

(αz)J−1(1− 21−J)

where we assumed that

(2.7.10) z − J ≥ αz for α := 5/6.

We choose

z := dZe, where Z :=

(2(J − 1)2

αJ−1(1− 21−J)

)1/JM1

M1/J2

.

Since J ≤ (1− α)Z ⇒ z − J ≥ αz, we may verify that for every J ≥ 2 we have

1

5≤ (1− α)

(2(J − 1)2

αJ−1(1− 21−J)

)1/J

and thus (2.7.10) is guaranteed by the hypothesis (2.7.8). We complete the proof by

writing

#S ′(M1,M2) ≤ ZM2 + 2(J − 1)MJ

1

(αZ)J−1(1− 21−J)

≤ M1M1− 1

J2

((2(J−1)

αJ−1(1−21−J )

) 1J

((J − 1)

1J + 1

(J−1)J−1J

))≤ 4.4M1M

1− 1J

2

for every J ≥ 2.

Remark 2.4. Lemma 2.3 is to be compared with the trivial upper bound M1M2. It is

possible to relax the conditions (2.7.6) and (2.7.3) at the price of a lost in the quality

of the result (2.7.9). One simply split the interval [x, x + h − 1] into shorter intervals

and apply Lemma 2.3 to each of them.

It is not clear what the optimal upper bound for #S(M1,M2) is. Assume thatMi is the

set of all the primes in [Li, 2Li] for i = 1, 2 which are coprime to q, with (L1L2)k > h.

80

It is natural to expect that (p1, p2) yields a value of n in S(M1,M2) with a probabilityh

(p1p2)kand this heuristic leads to

∑pi∈Mii=1,2p1 6=p2

h

(p1p2)k h

(L1L2)k−1 logL1 logL2

h

(L1L2)kM1M2.

The above cannot hold for L1L2 too large. Indeed, by writingM1 := p11 , p12 , . . . andM2 := p21 , p22 , . . . , it is possible to solve for n the system

q(n+ 1) + a ≡ 0 (mod (p11p21)k)

q(n+ 1 + pk11) + a ≡ 0 (mod (p11p22)k)

q(n+ 1 + 2pk11) + a ≡ 0 (mod (p11p23)k)

...

q(n+ 2) + a ≡ 0 (mod (p12p2l+1)k)

q(n+ 2 + pk12) + a ≡ 0 (mod (p12p2l+2

)k)

q(n+ 2 + 2pk12) + a ≡ 0 (mod (p12p2l+3

)k)

...

where l := min(M1,

⌊h−1pk11

⌋)and where the inequality j + upk1j ≤ h (j, u ≥ 1) holds.

We assumed that M1 ≤ M2 and it is understood that each line are considered in order

of appearance. That is, if a number n+ j + upk1j has been dened already, then we skip

this line and go to the next one and so on. This argument should leads to

#S(M1,M2) min

(M2,

hM1

Lk1, h

).

It seems hard to improve this last construction since in general the admissible systems

are extremely complex. Indeed, one easily proves the next lemma by induction.

Lemma 2.4. Let the integers h1, . . . , hs ∈ Z and r1, . . . , rs ∈ N be xed. Then, the

number of solutions mod [r1, . . . , rs] in x to the system of congruences

x ≡ hi (mod ri) (i = 1, . . . , s)

is 1 if (ri, rj)|(hi − hj) for i 6= j (i, j = 1, . . . , s),

0 otherwise.

81

Using this lemma with all the modulus pk1pk2, where (p1, p2) ∈M1×M2 with p1 6= p2 (if

(p1, p2) is chosen, we do not consider (p2, p1)), we can see how the relations among the

hi have to be structured to ensure the feasibility of an arrangement with #S(M1,M2)

large. We feel safe to formulate the following conjecture.

Conjecture 2.2. Let h be a positive integer and (q, a) be a pair that satises (2.1.1).

Assume thatMi is the set of all the primes in [Li, 2Li] for i = 1, 2 that are coprime to

q with (L1L2)k > h. Let S(M1,M2) be the set dened in Lemma 2.3. Then for each

xed ε > 0 we have

(2.7.11) #S(M1,M2)(M1 +M2 +

h

(L1L2)kM1M2

)hε.

Perhaps we overlooked a simple argument but so far we found no way to properly address

Conjecture 2.2. At least, we can show that the conjecture does not depend much on the

values taken by the integers q and a.

Lemma 2.5. Let a and q be integers that satisfy (2.1.1). Let H := h1, . . . , hs be a

set of integers and consider the set

R := r : r = pk1pk2 with (p1, p2) ∈M1 ×M2 and p1 6= p2

where eachMi is a set of prime numbers which are coprime to q. Assume that #R = s.

Then the system

q(n+ hj) + a ≡ 0 (mod rj) (j = 1, . . . , s),

is solvable in n if and only if the system

n+ hj ≡ 0 (mod rj) (j = 1, . . . , s)

is solvable in n.

Proof. Given an integer α, we denote any element β ∈ Z/αZ by (β)α. We consider

the criterion in Lemma 2.4. There are two cases to examine. The rst one is when

r1 := pk1pk2 and r2 := pk3p

k4, so that (r1, r2) = 1 and the conditions (r1, r2)|h1 − h2 and

(r1, r2)|h1 − h2 + a(q−1)r1 − a(q−1)r2 are therefore trivially veried. The second case is

when r1 := pk1pk2 and r2 := pk1p

k3, in which case (r1, r2) = pk1 and we therefore only need

to verify that

(2.7.12) pk1 | h1 − h2 ⇐⇒ pk1 | h1 − h2 + a(q−1)r1 − a(q−1)r2 .

82

From the identities

(q−1)r1 ≡ (q−1)pk1pk2(p−k2 )pk1 + (q−1)pk2p

k1(p−k1 )pk2 (mod r1)

and

(q−1)r2 ≡ (q−1)pk1pk3(p−k3 )pk1 + (q−1)pk3p

k1(p−k1 )pk3 (mod r2),

we deduce that (2.7.12) is true since pk1 | a(q−1)r1 − a(q−1)r2 .

Remark 2.5. Lemma 2.5 has an important consequence: it shows that Conjecture 2.2

does not depend on the value of q. To see this, we assume that we have two sets Hand R as dened in Lemma 2.5, relatively to the integer 1 and we assume that the sets

involved in R satisfy the hypothesis of Conjecture 2.2. If the system

n+ hj ≡ 0 (mod rj) (j = 1, . . . , s)

is solvable in n, then we consider the subset Rq of elements in R that are coprime to q

(we assume that #Rq = s′); it corresponds to a subset of H, say Hq. The system

q(n+ hj) + a ≡ 0 (mod rj) (j = 1, . . . , s′)

is solvable in n by Lemma 2.5. This shows that #Hq ≤ #H. In particular, we may

have R = Rq from the start and then Conjecture 2.2 is true for q.

In the opposite direction, assume that we have a set Rq of s′ elements, as dened in

Lemma 2.5, relatively to the integer q and assume that the sets involved in Rq satisfy

the hypothesis of Conjecture 2.2. Also, assume that Hq is a set such that 1 ≤ hj ≤ h

for each hj ∈ Hq. If the system

q(n+ hj) + a ≡ 0 (mod rj) (j = 1, . . . , s′)

is solvable in n, then so is the system

n+ hj ≡ 0 (mod rj) (j = 1, . . . , s′).

Now observe that any set H is obtained by adding some integers to a set Hq and it

corresponds to some new elements added to Rq to get R as well. The key point is that

the total number of solutions that we add by doing this is at most ω(q)(M1 +M2 +ω(q)).

This shows that #H ≤ #Hq + ω(q)(M1 + M2 + ω(q)). We have thus proved that

Conjecture 2.2 does not depend on q if we restrict our attention to the values less than

exp(chε) (for some xed c > 0), say.

83

2.8 The main results

Theorem 2.3. Assume that the pair (a, q) satises (2.1.1). For 2 ≤ J ≤[k2

]and

(qx)k+1

(J+1)k+1−J (log qx)Jk(k−1)

(J+1)k+1−J h (qx)1J , we have

(2.8.1) Gk(x, h; q, a) ≤ ck(q)h

ζ(k)+O

((qxhJ−1)1/Jk

log h

).

Proof. Fix a value of J and apply formula (2.6.3) which contains six terms on its RHS.

We begin with a simplication of the last term in (2.6.3), that is

x+h∑n=x+1

ν(n) Jx+h∑

n=x+1

ρ(n)

where ρ(n) ∈ 0, 1 and ρ(n) = 1⇔ ν(n) ≥ 1. Then set

W := J(qxhJ−1)1Jk , z := W

J2J−1 , w := W

J−12J−1 .

We can writex+h∑

n=x+1

ρ(n) ≤∑

Mi (i=1,2)

#S(M1,M2)

where we refer to the notation of Lemma 2.3 in the RHS and where the sets will soon

be dened. We estimate the fth term in (2.6.3) as follows

∑′

2≤p≤z

zplog zp

≤∑

2≤p≤w

zplog zp

+s∑j=0

∑2jw<p≤2j+1w

zplog zp

=:∑

2≤p≤w

z

log z+

s∑j=0

∑2jw<p≤2j+1w

z/2j

log z/2j

zw

logw W

log h

with s := blog(z/w)/ log 2c and where we have implicitly dened the variables zp.

The contribution of the primes p ≤ w to the last sum in (2.6.3) is reduced to 0.

Also, if we x a value of j, we observe that since for each prime p in [2jw, 2j+1w)

we took zp = z/2j we have that in this setting we can apply the result of Lemma

2.3 to all the remaining primes belonging to the interval (z/2j, z]. That is we let

M1,j := p ∈ [2jw, 2j+1w) : p - q andM2,j := p ∈ (z/2j, z] : p - q and verify that

each condition of Lemma 2.3 is satised. For J = 2, there is some concern with (2.7.8).

We may use the result #S(M1,j,M2,j) 2jwz1/2

(log h)3/2 only for j ≥ log log h2 log 2

+c1 with c1 large

84

enough. Otherwise, we use the trivial estimate #S(M1,j,M2,j) 2jwz(log h)2 until (2.7.8)

is fullled. We then get

∑Mi (i=1,2)M1≤M2

#S(M1,M2) =

log log h2 log 2

+c1∑j=0

#S(M1,j,M2,j) +s∑

j> log log h2 log 2

+c1

#S(M1,j,M2,j)

∑

0≤j< log log h2 log 2

+c1

2jwz

(log h)2+

∑log log h2 log 2

+c1≤j≤s

2jwz1/2

(log h)3/2

wz

(log h)3/2+

z3/2

(log h)3/2 W

(log h)3/2.

For J ≥ 3 the condition (2.7.8) is satised and we have

∑Mi (i=1,2)

#S(M1,M2) =s∑j=0

#S(M1,j,M2,j)

s∑j=0

2jwz1−1/J

(log h)2−1/J

z2−1/J

(log h)2−1/J=

W

(log h)2−1/J.

We have thus shown that

Gk(x, h; q, a) ≤ ck(q)h

ζ(k)+O

((h log log h)

32k+1

(log h)6k−32k+1

+h

zk−1 log z+

W

log h

)for any J ≥ 2 and it is easy to verify that (2.8.1) is true and is in fact an improvement

of Theorem 2.1 in the stated range.

Theorem 2.4. Assume that Conjecture 2.2 holds. Then for any pair on integers (a, q)

satisfying (2.1.1) and any xed ε > 0 we have

gk(h; q, a) ≤ ck(q)h

ζ(k)+O(h

32k+1

+ε).

Proof. We apply the identity (2.6.3) with z := h2

2k+1 and set w := h1

2k+1 and s :=

blog(z/w)/ log 2c. Then we write

z∑′

p=2

zplog zp

≤∑

2≤p≤w

zplog zp

+s∑j=0

∑2jw<p≤2j+1w

zplog zp

:=∑

2≤p≤w

z

log z+

s∑j=0

∑2jw<p≤2j+1w

z/2j

log z/2j

85

wz

(log h)2+

wz

log h h

32k+1

log h.

The last term on the RHS of (2.6.3) is treated dierently. Using the Cauchy-Schwarz

inequality, we have

(2.8.2)x+h∑

n=x+1

ν(n) ≤

(x+h∑

n=x+1

ρ(n)

)1/2( x+h∑n=x+1

ν2(n)

)1/2

and it is easy to see that

(2.8.3)x+h∑

n=x+1

ν2(n) z2

(log z)2 h

42k+1

(log h)2

since each pair (p1, p2) with p1 6= p2 is counted at most once. Indeed, we have pk1pk2 | n

for some n ∈ [x + 1, x + h] and pk1pk2 > h because our choice implies pzp > h

1k for each

prime p. Also, by setting

M1,j := p ∈ [2jw, 2j+1w] : p - q and M2,j := p ∈ [z/2j, z] : p - q,

we can write

x+h∑n=x+1

ρ(n) ≤s∑j=0

|S(M1,j,M2,j)|(2.8.4)

s∑j=0

(2j+1w

log 2j+1w+

z

log z+

h

(wz)kwz

logw log z

)hε

(z +

h

(wz)kwz

log h

)hε h

22k+1

+ε.

In light of (2.8.3) and (2.8.4), (2.8.2) becomes

x+h∑n=x+1

ν(n) ≤ (h2

2k+1+ε)1/2

(h

42k+1

(log h)2

)1/2

=h

32k+1

+ε/2

log h

and the theorem is proved.

Remark 2.6. One can see that

(2.8.5)x+h∑

n=x+1

ν(n) = o

(z2

log2 z

)(x→∞),

for z close to (h log h)1k+1 (and zp just slightly smaller than z), is equivalent to an

improvement of Theorem 2.1. Also, we have seen that (2.8.5) is equivalent to the

estimate |S(M1,M2)| = o(M1M2) for some sets (M1,M2).

86

We deduce that if we can improve Theorem 2.1 for only one small value of q, it is

because we can succeed in proving some estimates for |S(M1,M2)| and therefore, using

Remark 2.5, we obtain a stronger inequality for all small values of q as well.

For a general result better than Theorem 2.1, we use a theorem of Konyagin [6].

Theorem 2.5. Dene θ implicitly by h = xθ. Then for θ ≤ k+12k

,

(2.8.6) Gk(x, h; q)− ck(q)h

ζ(k) 1

θ

(h

log x

)2/(k+1)(h

x

) k−1

(k+1)(2r2+r−1)

with

(2.8.7) r :=

[k + 1

2θ

]≤ 1− ε√

3

√log x

log log x

for any xed ε > 0.

Proof. Using Theorem 2 in [6] withW = 1 and f(u) := x(N+u)k

, we obtain the inequality

(2.8.8) S(N) N

(√r

(Nλr

Nk+r

) 12r−1

+1

r

(hN r

xλr

) 1r−1

+

(λ

N

) 12r

)+ rλ,

where S(N) is the number of integers u ∈ [N, 2N ] for which ‖f(u)‖ < δ. We have to

choose λ ≥ 1 and r ≥ 2. It is easy to see that if the above inequality is non-trivial,

that is S(N) N , the third term of (2.8.8) is larger than the fourth. Choosing λ such

that the second term is equal to the third we nd

(2.8.9) S(N) N

(h

x

) 12r2+r−1

+N

x

Nk

(h

x

) 2r2

2r2+r−1

12r−1

provided that

(2.8.10)N

r

(h

x

) 2r2r2+r−1

1.

We then make use of Remark 2.3. For w and z to be chosen later, we write the fourth

term of (2.6.3) as

∑′

2≤p≤z

zplog zp

≤∑

2≤p≤w

zplog zp

+s∑j=0

∑2jw<p≤2j+1w

zplog zp

87

≤∑

2≤p≤w

z

log z+

s∑j=0

∑2jw<p≤2j+1w

z/2j

log z/2j

wz

logw

where s = blog(z/w)/ log(2)c. We verify that the choice (2.8.7) implies that the second

term on the RHS of (2.8.9) has the exponent of N greater than 0 for each k ≥ 2 with

θ ≤ k+12k

. For the last term in (2.6.3) we write

∑n∈Qc

ν(n) 1

θ

∑n∈Qc

ρ(n) 1

θ

s∑j=0

S(2jwz) 1

θS(z2).

We are ready to choose w in order to have wzlogw≈ S(z2)

θ, that is we choose w so that

wz

logw=z2

θ

(h

x

) 12r2+r−1

which yields w :=z log z

θ

(h

x

) 12r2+r−1

and we see that with this choice of w we have w < z for all values of θ satisfying (2.8.7).

We are now ready to choose the value of z. To do so, we write

h

zk−1 log z=z2

θ

(h

x

) 12r2+r−1

which yields z :=

(θh

log h

) 1k+1 (x

h

) 1(k+1)(2r2+r−1)

and this gives the result (2.8.6). To complete the proof, we have to verify that, with

our choice of r given by (2.8.7) and the hypothesis θ ≤ k+12k

, (2.8.10) is satised, that is

the rst term on the RHS of (2.8.9) is the largest and that the second term in (2.6.3) is

smaller than the RHS of (2.8.6). The inequality in (2.8.7) corresponds to the domain

where the result is better than that of Theorem 2.1 and is used in the verication that

w < z and that (2.8.10) holds with our choice of r.

We can see that Theorem 2.5 yields an upper bound where the exponent of h satises

2

k + 1− 2(k − 1)θ(1− θ)

(k + 1)((k + 1)2 + θk + θ − 2θ2)≈ 2

k + 1− 2θ

k2

when (1+ε)√

3(k+1)2

√log log x

log x≤ θ ≤ k+1

2k. It can be compared with Theorem 2.3 which

provides the estimate

2

k + 1− k + 1− 2J

Jk(k + 1)≈ 2

k + 1− (1− 2α)

θ

k

when J ≈ 1θ≈ αk and provides no new estimate in other regions. In particular, we

cannot take θ less than 2k+1

and hope to have a good result.

88

2.9 Final remarks

For large values of θ, with h ≤ x, we use a completely dierent method. We proceed

directly with

Gk(x, h) =x+h∑

n=x+1

µk(n) =x+h∑

n=x+1

∑dk|n

µ(d)

=∑

d<2x1/k

µ(d)

([x+ h

dk

]−[x

dk

])= h

∑d≤h1/k

µ(d)

dk+O(h1/k) +

∑h1/k<d<2x1/k

µ(d)χ[x+1,x+h](dk)

=h

ζ(k)+O

( ∑h1/k<d<2x1/k

χ[x+1,x+h](dk)

)+O(h1/k),

where

χ[a,b](s) :=

1 if there is an n ∈ [a, b] such that s|n,0 otherwise.

Using the results from [5], we obtain for example

Gk(x, h) =h

ζ(k)+O(h1/k + x1/(k+2)).

To our knowledge, there are no results in the literature allowing one to get the same

result with an exponent of x of the form 1−ck

for some constant c > 0. Also, using

results from [4] and [5], we can establish that

G2(x, h) =h

ζ(2)+O(h1/2)

holds uniformly in h in the region h x3/7(log x)1/14 and that

G3(x, h) =h

ζ(3)+O(h1/3)

holds uniformly in h in the region h x1/2(log x)3. These are the best estimates that

we have obtained so far.

89

Bibliography

[1] P. Erd®s, Some problems and results in elementary number theory, Pub. Math.

Univ. Debrecen 2 (1951), 103− 109.

[2] M. Filaseta and O. Trifonov, The distribution of fractional parts with ap-

plications to gap results in number theory, Proc. Lond. Math. Soc. 73 (1996),

241− 278.

[3] J. Friedlander andH. Iwaniec, Opera de Cribro, American Mathematical Soci-

ety Colloquium Publication, Vol. 57, American Mathematical Society, Providence,

2010.

[4] M. N. Huxley, The integer points close to a curve. III. Number theory in progress,

Vol. 2 (Zakopane-Ko±cielisko, 1997), 911− 940, de Gruyter, Berlin, 1999.


de classe Cn, II, Functiones et Approximatio XXXV (2006), 91− 115.

[6] S. V. Konyagin, Estimates of the least prime factor of a binomial coecient,

Mathematika 46 (1999), no. 1, 41− 55.

[7] H. L. Montgomery and R. C. Vaughan, The large sieve, Mathematika 20

(1973), no. 40, 119− 134.

[8] D. I. Tolev, On the distribution of r-tuples of squarefree numbers in short inter-

vals, Int. J. Number Theory 2 (2006), no. 2, 225− 234.

[9] K.-M. Tsang, The distribution of r-tuples of square-free numbers, Mathematika

32 (1985), 265− 275.

90

Chapter 3

New upper bounds for the number of

divisors function

Jean-Marie De Koninck and Patrick Letendre

Résumé

Soit τ(n) le nombre de diviseurs d'un nombre entier positif n. Nous obtenons

des majorations de τ(n) en termes de log n et du nombre de facteurs premiers

distincts de n.

Abstract

Let τ(n) stand for the number of divisors of the positive integer n. We obtain

upper bounds for τ(n) in terms of log n and the number of distinct prime factors

of n.

91


Let τ(n) stand for the number of divisors of the positive integer n and ω(n) stand

for the number of prime factors of the positive integer n. We shall also be using the

functions

γ(n) :=∏p|n

p, a(n) :=∑p|n

1

log p, b(n) :=

∏p|n

1

log p.

In 1915, Ramanujan [8] obtained the inequality

(3.1.1) τ(n) ≤(

log(nγ(n))

ω(n)

)ω(n)

b(n) (n ≥ 2).

Recently, in [2], such an inequality has found an application in the problem of estimating

the maximal order of the function τ(τ(n)) for n ≤ x as x goes to innity.

In this paper, we compute explicitly some interesting limit cases of (3.1.1) and show

that for k = ω(n) ≥ 74,

τ(n) <

(1 +

log n

k log k

)k.

We also provide a short proof of (3.1.1) in Corollary 3.1.

From here on, for each integer k ≥ 0, we let

nk := p1p2 · · · pk, the product of the rst k primes (with n0 = 1).

Also, when we write log+ x, we mean log max(2, x).

Finally, given the factorization of an integer n = qα11 · · · q

αkk with q1 < · · · < qk, we call

the vector (α1, . . . , αk) the exponent vector of n.

3.2 Background results

It is well known that

(3.2.1) 2ω(n) ≤ τ(n) ≤ 2Ω(n) (n ≥ 1),

where Ω(n) stands for the number of prime divisors of n counting their multiplicity.

While the rst inequality in (3.2.1) is really good, the second one is rather poor. We can

92

already do better by using the arithmetic geometric mean inequality. In fact, writing

n = qα11 · · · q

αkk , where q1 < · · · < qk are primes and the αi's positive integers, we have

τ(n)1/k =k∏i=1

(1 + αi)1/k ≤ 1 + α1 + · · ·+ 1 + αk

k=k + Ω(n)

k.

This shows that

(3.2.2) τ(n) ≤(

1 +Ω(n)

ω(n)

)ω(n)

(n ≥ 2),

which is already better than the second inequality of (3.2.1).

The inequality (3.2.2) is of great interest. For instance, it is known that the quotientΩ(n)

ω(n)is near 1 for almost all integers n, as was shown for instance by the rst author

in [3]. In fact, one can use (3.2.2) and the estimate

|n ≤ x : Ω(n) ≥ αω(n)| x(log log x)(log x)21−α−1

valid for all α ≥ 1 and x ≥ 3 (see Corollary 3.6 in Tenenbaum [10] or for an even

sharper estimate, Balazard [1]) to show that for every xed ε > 0,

τ(n) ≤ (2 + ε)ω(n) for almost all n.

Finally, observe that there is no hope to nd an upper bound of the type τ(n) ≤ Cω(n)

for some xed positive integer C ≥ 2. Indeed, the inequality τ(n) ≤ Cω(n) would not

be true for n = nCk for every integer k ≥ 1.

Another natural approach is to use Rankin's method as follows.

τ(n) =∑d|n

1 ≤∑d|n

(nd

)σ= nσ

∑d|n

1

dσ

= nσ∏pα‖n

(1 +

1

pσ+ · · ·+ 1

pασ

)≤ nσ

∏p|n

(1 +

1

pσ − 1

)

= nσ∏p|n

(1 +

1

eσ log p − 1

)≤ nσ

∏p|n

(1 +

1

σ log p

)

≤ exp

σ log n+1

σ

∑p|n

1

log p

= exp

(σ log n+

1

σa(n)

).

Choosing σ =

√a(n)

log n, we then obtain the inequality

(3.2.3) τ(n) ≤ exp(

2√a(n) log n

).

93

This inequality gives an easy way to estimate τ(n) given that ω(n) ≤ k say, but it falls

short of (3.1.1).

Observe that a(n) ≤ ω(n) for all integers n not of the form 2α, 2α3β or 2α5β, in which

case it follows from (3.2.3) that

(3.2.4) τ(n) ≤ exp(

2√ω(n) log n

)(n ≥ 1).

The fact that (3.2.4) holds for all n follows after checking directly that it holds for the

exceptional cases n = 2α, 2α3β and 2α5β.

Observe that by using (3.4.23) (in Lemma 3.5 below), one can obtain from (3.2.3) that

(3.2.5) τ(n) ≤ exp

(η1

√ω(n) log n

log+ ω(n)

)(n ≥ 1),

where

η1 := 2

√a(n8) log 8

8= 2.277227 . . .

However, the problem with (3.2.5) is that for small values of ω(n), we must cope with

the huge factor√

log n appearing in the exponent.

We are motivated by the fact that, since Wigert [11], we know that

log τ(n) ≤ (log 2)(log n)

log log n+O

(log n

(log log n)2

),

and by the fact that it has been proved by Nicolas and Robin [6] that the maximum

value of the function

n 7→ log(τ(n)) log log n

log 2 log n(n ≥ 3)

is attained at n = 6 983 776 800 = 25 ·33 ·52 ·7 ·11 ·13 ·17 ·19 = 720n8 and that its value

is approximately 1.5379. But, meanwhile, those large values are almost never attained

since it has been proved by Erd®s and Nicolas [5] that, given any real θ ∈ (0, 1), the

cardinality of the set of those n ≤ x for which

ω(n) ≥ θlog x

log log x

is x1−θ+o(1) as x→∞. Furthermore, this set corresponds exactly to the set of values

where τ(n) is large as can still be seen from (3.3.4).

Remark 3.1. For each ε > 0, there is a constant C(ε) such that

τ(n) ≤ C(ε)nε.

94

In fact,

C(ε) =∏p<21/ε

maxα≥0

α + 1

pαε.

Robin [7] also designed an algorithm that allows one to easily obtain the list of all

highly composite numbers with less than k prime factors, which yields the absolute best

estimate for τ(n) for every n ≤ x for any given x.

Before stating our main results, we introduce the function λ(n) dened implicitly by

τ(n) =

(1 +

λ(n) log n

k log k

)k,

where k = ω(n) ≥ 2. Therefore, for each integer n ≥ 2 with ω(n) = k ≥ 2, we set

(3.2.6) λ(n) :=(τ(n)1/k − 1)k log k

log n.

3.3 Main results

Theorem 3.1. For every integer n ≥ 2,

(3.3.1) τ(n) ≤(

η2 log n

ω(n) log+ ω(n)

)ω(n)

,

where

η2 := exp

(1

6log 96− log

(log 60060

6 log 6

))= 2.0907132 . . .

Theorem 3.2. For every integer n > 24n16 = 782139803452561073520,

(3.3.2) τ(n) ≤(

2 log n

ω(n) log+ ω(n)

)ω(n)

.

Moreover, the inequality remains true for all n ≥ 2 with ω(n) ≤ 3.

Theorem 3.3. For every integer n ≥ 2,

(3.3.3) τ(n) ≤(

1 + η3log n

ω(n) log+ ω(n)

)ω(n)

where

η3 := λ(720n7) =(11521/7 − 1)7 log 7

log 367567200= 1.1999953 . . .

95

Theorem 3.4. For every positive integer n with k = ω(n) ≥ 74,

(3.3.4) τ(n) <

(1 +

log n

k log k

)k.

Remark 3.2. The number

n+ := 213 · 38 · 55 · 74 · 113 · 133 · 173 · 192 · · · 532 · 59 · · · 367,

whose prime factors are the rst 73 prime numbers, shows that Theorem 3.4 is best

possible since λ(n+) = 1.0008832 . . . In fact, one can nd similar examples n (that

is, with λ(n) > 1) for each ω(n) = k ∈ [3, 73]. Also, the methods used in the proof

of Theorem 3.4 allow one to show that the largest value of λ(n), with ω(n) = 74, is

attained only by the number

n∗ := 213 · 38 · 55 · 74 · 113 · 133 · 173 · 192 · · · 532 · 59 · · · 373

and for which λ(n∗) = 0.99991077 . . . (Observe that the number n+ realizes the unique

maximum of the function λ among the integers n with exactly 73 distinct prime factors.)

By comparing the lower bound in (3.2.1) with (3.3.4) and after some computation, one

can show that the inequality

n ≥ ω(n)ω(n) (n ≥ 2)

holds for each n satisfying ω(n) /∈ [4, 12] or n > 43n11. This helps to understand why

Theorem 3.4 is more powerful than Theorem 3.2.

Theorem 3.5. The largest integer n with k = ω(n) ≥ 44 for which λ(n) ≥ 1 is the

integer n made up of the rst 44 primes and has the exponent vector

(3.3.5)

(354, 223, 152, 125, 102, 95, 86, 83, 77, 72, 71, 67, 65, 64, 63, 61, 59, 59, 57, 57, 56,

55, 55, 54, 53, 52, 52, 52, 51, 51, 50, 49, 49, 49, 48, 48, 48, 47, 47, 47, 46, 46, 46, 46).

3.4 Preliminary lemmas

Let xi, with i ∈ 1, . . . , k, be xed real numbers that satisfy 0 < x1 ≤ · · · ≤ xk. Let

µ :=x1 + · · ·+ xk

k

96

and

$ :=k∑i=1

|xi − µ|.

Assume also that x1 ≤ · · · ≤ xm ≤ µ ≤ xm+1 ≤ · · · ≤ xk for a xed m ∈ 1, . . . , k− 1where k ≥ 2. Further set

µ1 :=x1 + · · ·+ xm

m= µ− $

2m,

µ2 :=xm+1 + · · ·+ xk

k −m= µ+

$

2(k −m)

and also

$1 :=m∑i=1

|xi − µ1|

and

$2 :=k∑

i=m+1

|xi − µ2|.

Lemma 3.1. Assume the above notation. (i) For k ≥ 1 we have

(3.4.1) x1 · · ·xk ≤ µk.

(ii) For k ≥ 2 we have

(3.4.2) x1 · · ·xk ≤ µm1 µk−m2 .

(iii) For k ≥ 4, let m ∈ 2, . . . , k − 2, m1 ∈ 1, . . . ,m− 1, m2 ∈ 1, . . . , k −m− 1and assume that

(3.4.3) 0<x1≤···≤xm1≤µ1≤xm1+1≤···≤xm≤µ≤xm+1≤···≤xm+m2≤µ2≤xm+m2+1≤···≤xk.

Then,

(3.4.4) x1···xk≤(µ1− $1

2m1

)m1(µ1+

$12(m−m1)

)m−m1(µ2− $2

2m2

)m2(µ2+

$22(k−m−m2)

)k−m−m2.

Proof. In each case, we simply use the arithmetic-geometric inequality for the corre-

sponding sub-product of variables for which we know the average.

From now on, we write n = qα11 · · · q

αkk . For each i ∈ 1, 2, . . . , k, we dene θi implicitly

by nθi = qαii , so that θ1 + · · ·+ θk = 1. We write

(3.4.5) xi :=(αi + 1) log qi

log n

97

and assume that the primes qi are ordered in such a way that (3.4.3) holds. In this case

we have

(3.4.6) µ =1

k

(1 +

log γ(n)

log n

),

$ =1

k

k∑i=1

∣∣∣∣(αi + 1)k log qi − log γ(n)

log n− 1

∣∣∣∣ =:$′

k,

µ1 =1

k

(1 +

log γ(n)

log n

)− $′

2km, µ2 =

1

k

(1 +

log γ(n)

log n

)+

$′

2k(k −m),

(3.4.7) $1 =1

k

m∑i=1


log n− 1 +

$′

2m

∣∣∣∣ =:$′1k

and

(3.4.8) $2 =1

k

k∑i=m+1


log n− 1− $′

2(k −m)

∣∣∣∣ =:$′2k.

Lemma 3.2. Let k ≥ 1 be an integer, zi > 0 and xi ≥ −1zi

be real numbers for i =

1, . . . , k, and assume that

x1 + · · ·+ xk = 1.

Then,

(3.4.9)k∏i=1

(1 + xizi) ≤k∏i=1

(zik

)(1 +

k∑i=1

1

zi

)k

,

with equality if and only if

xi =1

k

(1 +

k∑j=1

1

zj

)− 1

zi(i = 1, . . . , k).

Proof. Using the arithmetic geometric mean inequality, the hypothesis zi > 0 and the

fact that for each i we have 1 + xizi ≥ 0, we can write

k∏i=1

(1 + xizi) =k∏i=1

zi

k∏j=1

(xj +

1

zj

)

≤k∏i=1

(zik

)( k∑j=1

(xj +

1

zj

))k

98

=k∏i=1

(zik

)(1 +

k∑j=1

1

zj

)k

.

We have equality if and only if

xi +1

zi=

1

k

(1 +

k∑j=1

1

zj

)(i = 1, . . . , k),

thus completing the proof.

Corollary 3.1. Assume the above notation. Then, for every integer n ≥ 2,

(3.4.10) τ(n) ≤(

log n

ω(n)

)ω(n)(1 +

log γ(n)

log n

)ω(n)

b(n)

and

(3.4.11) τ(n) ≤(

2 log n

ω(n)

)ω(n)

b(n).

Proof of Corollary 3.1. Using inequality (3.4.1) we have

τ(n) =k∏i=1

(1 + αi) =k∏i=1

(1 +

θi log n

log qi

)≤(

log n

k

)k (1 +

log γ(n)

log n

)k∏p|n

1

log p,

which proves (3.4.10). Since log γ(n) ≤ log n, inequality (3.4.11) follows immediately

from (3.4.10).

In any event, observe that it follows from Corollary 3.1 that

(3.4.12) λ(n) ≤

∏p|n

log k

log p

1/k

+log γ(n)

log n

∏p|n

log k

log p

1/k

− k log k

log n.

Lemma 3.3. (i) Assume that µ > 0, $ ≥ 0, k − m ≥ 1, m ≥ 1 and µ − $2m

> 0.

Then, the function

(3.4.13)(µ− $

2m

)m(µ+

$

2(m− k)

)m−kdecreases when $ increases.

(ii) Assume that µ > 0, $1 ≥ 0, $2 ≥ 0, m ≥ 1, k − m ≥ 1, m1 ≥ 1, m2 ≥ 1,

k −m−m2 ≥ 1, µ− $2m− $1

2m1> 0 and µ+ $

2(k−m)− $2

2m2> 0. Then, the function

f($) :=

(µ− $

2m− $1

2m1

)m1(µ− $

2m+

$1

2(m−m1)

)m−m1

99

×(µ+

$

2(k −m)− $2

2m2

)m2(µ+

$

2(k −m)+

$2

2(k −m−m2)

)k−m−m2

(3.4.14)

has the property that if f ′($0) < 0 for some $0 > 0, then f($) < f($0) for each

$ > $0.

(iii) Assume that A > 0, B > 0, C > 0, z > 0, γ1 ≥ 0, γ2 ≥ 0, β1 ≥ 0, β2 ≥ 0,

β1 + β2 = 1 and C < AB. Then, the expression

(3.4.15) B

(γ1 +

A

z

)β1(γ2 +

A

z

)β2

− C

z

decreases when z increases.

(iv) Assume that A > 0, B > 0, C > 0, z > 0, γ1 ≥ 0, γ2 ≥ 0, γ3 ≥ 0, γ4 ≥ 0, β1 ≥ 0,

β2 ≥ 0, β3 ≥ 0, β4 ≥ 0, β1 + β2 + β3 + β4 = 1 and C < AB. Then, the expression

(3.4.16) B

(γ1 +

A

z

)β1(γ2 +

A

z

)β2(γ3 +

A

z

)β3(γ4 +

A

z

)β4

− C

z

decreases when z increases.

Proof. (i) Since the function (3.4.13) is assume to be positive, it follows that its deriva-

tive with respect to $ has the same sign than its logarithmic derivative with respect

to $. Then, since the logarithmic derivative is

−1

2µ− $m

+1

2µ+ $k−m

,

it is clearly strictly negative when $ > 0.

(ii) Again, the function f is assumed to be positive in which case its derivative with

respect to $ has the same sign than its logarithmic derivative with respect to $. Also,

we have(f ′($)

f($)

)′=

−m1

m2(2µ− $

m− $1

m1

)2 −m−m1

m2(2µ− $

m+ $1

m−m1

)2

− m2

(k −m)2(2µ+ $

k−m −$2

m2

)2 −k −m−m2

(k −m)2(2µ+ $

k−m −$2

k−m−m2

)2

which is clearly negative. We deduce that if f ′($0) < 0 for some$0 > 0, then f ′($) < 0

for each $ > $0 which in turn implies

f($)− f($0) =

∫ $

$0

f ′(t)dt < 0

100

for each $ > $0, thus establishing our claim.

(iii) We take the derivative of (3.4.15) with respect to z and multiply by z2. We then

see that the wanted property is equivalent to

(3.4.17) C < AB

(γ1 +

A

z

)β1(γ2 +

A

z

)β2(

β1

γ1 + Az

+β2

γ2 + Az

).

Now, from Jensen's inequality for the exponential function, we have

1

zβ1

1 zβ2

2

≤ β1

z1

+β2

z2

(z1, z2 > 0).

We deduce that the hypothesis C < AB implies (3.4.17). (iv) is done in the same

manner and the proof is complete.

Lemma 3.4. Let A and B be xed positive real constants. Consider the function

ψ := Z× R∗ × R→ R≥0 dened by

(3.4.18) ψ(α, x, ϕ) =

∣∣∣∣(α + 1)B − Ax

− ϕ∣∣∣∣ .

(i) Assume that x1, ϕ1 > 0. The minimum of the function ψ(α, x, ϕ) for α ∈ Z,x ∈ [x1, x2] and ϕ ∈ [ϕ1, ϕ2] is either 0 or is given by the minimum over the eight

possibilities provided by

α ∈⌊

x2ϕ2 + A

B

⌋− 1,

⌈x1ϕ1 + A

B

⌉− 1

, x ∈ x1, x2 and ϕ ∈ ϕ1, ϕ2.

The minimum is 0 if and only if

(3.4.19)

⌈x1ϕ1 + A

B

⌉≤⌊x2ϕ2 + A

B

⌋.

(ii) Let δ > 0 be a xed real number and assume that x1, ϕ1 > 0. The minimum of the

function ψ(α, x, 1) for x ∈ [x1, x2] and α ∈ Z\⌈

(1−δ)x1+AB

⌉− 1, . . . ,

⌊(1+δ)x2+A

B

⌋− 1

is given by the minimum over the four possibilities provided by

α ∈⌈

(1− δ)x1 + A

B

⌉− 2,

⌊(1 + δ)x2 + A

B

⌋and x ∈ x1, x2.

Proof. (i) First, assume that the minimum is 0. Choose (α, x, ϕ) that realizes 0. We

deduce that α + 1 = xϕ+AB

and then it is equivalent to having (3.4.19). Now, assume

101

that the minimum is not zero. In this case,⌊x2ϕ2+A

B

⌋<⌈x1ϕ1+A

B

⌉=⌊x2ϕ2+A

B

⌋+ 1. Also,

if (α, x, ϕ) realizes the minimum then there are two cases. We either have (α+1)B−Ax

−ϕ ≥ 0, in which case α + 1 ≥

⌈x1ϕ1+A

B

⌉, or we have (α+1)B−A

x− ϕ ≤ 0 , in which

case we have α + 1 ≤⌊x2ϕ2+A

B

⌋. It is then clear that the minimum is attained for

α ∈⌊

x2ϕ2+AB

⌋− 1,

⌈x1ϕ1+A

B

⌉− 1. To conclude, we remark that, once α is xed,

(α+1)B−Ax

− ϕ attained its extremum at the edges of the intervals since it is a sum of

independent monotone functions.

(ii) The choice for α is clear. Also, if we assume that the minimum is not 0 then the

choice for x is also clear. Now, assume the contrary, that the minimum is 0 and that

it is attained at (α, x) with α =⌈

(1−δ)x1+AB

⌉− 2 = (1−δ)x1+A

B− 2 + ξ for some ξ ∈ [0, 1].

In this case,

(α + 1)B − Ax

=

((1−δ)x1+A

B− 2 + ξ + 1

)B − A

x=

(1− δ)x1 + (ξ − 1)B

x≤ 1− δ < 1

and similarly for the other choice of α. This shows that the minimum is not 0 and the

proof is complete.

Lemma 3.5. We have

(3.4.20)k∑i=1

log pi ≤ k(log k + log log k − 1/2) for k ≥ 5,

(3.4.21)k∑i=1

log log pi ≥ k

(log log k +

log log k − 3/2

log k

)for k ≥ 6,

(3.4.22) b(nk) ≤ (log k)−k for k ≥ 44

and

(3.4.23) a(nk) ≤k

log kfor k ≥ 186.

Proof. We rst prove inequality (3.4.20) using induction. First observe that the in-

equality holds for k = 5. Assuming that the inequality holds for some k ≥ 5, we will

show that it must then hold for k+ 1. Since pj < 32j log j for each j ≥ 4, it follows that

k+1∑i=1

log pi ≤ k(log k + log log k − 1

2) + log pk+1

102

< (k + 1)(log(k + 1) + log log(k + 1)− 1

2) +

1

2+ log

3

2− k log(1 + 1/k)

< (k + 1)(log(k + 1) + log log(k + 1)− 1

2),

where this last inequality holds because of the fact that

log

(1 +

1

k

)≥ 1

k− 1

2k2for all k ≥ 1

implies that

1/2 + log3

2− k log(1 + 1/k) ≤ 1/2 + log

3

2− 1 +

1

2k< 0 for k ≥ 6,

thus completing the proof of (3.4.20).

To prove inequality (3.4.21), we rst verify using a computer that it holds for each k ∈

[6, 200 000]. For k ≥ 200 001, we proceed by induction. Since the functionπ(x) log x

xattains its maximum at x = 113 with the value 1/c := 1.255 . . . (see for instance Rosser

and Schoenfeld [9]), it follows that pj ≥ cj log j. It follows that

k∑i=1

log log pi

> k

(log log k +

log log k − 3/2

log k

)+ log log(k + 1) +

log log(k + 1) + log c

log k + 1

−1

2

(log log(k + 1) + log c

log(k + 1)

)2

= (k + 1)

(log log(k + 1) +

log log(k + 1)− 3/2

log(k + 1)

)+

3/2 + log c

log(k + 1)− 1

2


log(k + 1)

)2

+k

(log log k − log log(k + 1) +

log log k − 3/2

log k− log log(k + 1)− 3/2

log(k + 1)

).

It remains to show that the sum of these last three terms is positive. Using the mean

value theorem and the fact that log log ξ > 5/2 for ξ ≥ 200 000, we nd that it suces

to show that3/2 + log c

log(k + 1)≥ 1

2


log(k + 1)

)2

+k

ξ log ξ,

for some ξ ∈ (k, k + 1). It is therefore enough to show that

(3.4.24) 3/2 + log c− 1 ≥ 1

2

(log log(k + 1) + log c)2

log(k + 1)+

1

k log k.

103

Since each of the two terms on the right of (3.4.24) decreases as a function of k for

k ≥ 200 000 and since (3.4.24) is true for k = 200 000, it follows that (3.4.24) holds for

all k ≥ 200 000, thereby completing the proof of (3.4.21).

The proofs of (3.4.22) and (3.4.23) follow along the same lines as those of (3.4.20) and

(3.4.21).

Let us further introduce the function

(3.4.25) t(n) :=τ(n)1/k

log n(n ≥ 2).

Lemma 3.6. Let n ≥ 2 be an integer, 2 ≤ k = ω(n) and p be a prime number. If pα‖nwith α ≥ 2, then

(3.4.26)λ(n)

λ(n/p)≤(

1 +2

kα

)(1− log p

log n

)and

(3.4.27)t(n)

t(n/p)≤(

1 +1

kα

)(1− log p

log n

).

Also, for ` ∈ 1, 2, we have

(3.4.28)

(1 +

`

kα

)(1− log p

log n

)< 1⇐⇒ p > n

`αk+`

and

(3.4.29) α = max

(2,

⌈`

k

(log n

log p− 1

)⌉)=⇒

(1 +

`

kα

)(1− log p

log n

)< 1.

Proof. We write n = pαm, so that (p,m) = 1 and therefore,

λ(n)

λ(n/p)=

τ(n)1/k − 1

τ(n/p)1/k − 1

log n/p

log n

=

(1 +

τ(n)1/k − τ(n/p)1/k

τ(n/p)1/k − 1

)(1− log p

log n

)=

(1 +

τ(m)1/k

τ(n/p)1/k − 1((α + 1)1/k − α1/k)

)(1− log p

log n

)≤

(1 +

τ(n/p)1/k

τ(n/p)1/k − 1

1

kα

)(1− log p

log n

),

where the last inequality follows from the fact that

(α + 1)1/k − α1/k ≤ supξ∈[α,α+1]

ξ1/k

kξ=α1/k

kα.

104

Since the function z → zz−1

is strictly decreasing for z > 1, the result then follows from

the fact that τ(n/p)1/k ≥ 2. The proof of inequality (3.4.27) is similar and the proofs

of (3.4.28) and (3.4.29) follow from an easy computation.

Lemma 3.7. For any real z > 1 and integer n = qα11 · · · q

αkk ≥ 2, let

(3.4.30) υ(n, z) := log k

(1 +

log γ(n)

log z

)b(n)1/k − k log k

log z.

Then,

(3.4.31)d

dzυ(n, z) ≤ 0 (n ≥ 2)

with strict inequality if ω(n) ≥ 2. Also,

(3.4.32) υ(nk, nk) < 1 (k ≥ 95).

Proof. To prove (3.4.31), we rst observe that

d

dzυ(n, z) = − log k

log γ(n)

z log2 zb(n)1/k +

k log k

z log2 z,

which implies that (3.4.31) is equivalent to

k ≤ b(n)1/k log γ(n),

which itself is an immediate consequence of the arithmetic geometric mean inequality.

To prove (3.4.32), we must show that

(3.4.33) 2 <

(k

log nk+

1

log k

)b(nk)

−1/k.

Using inequalities (3.4.21) and then (3.4.20) we see that the right hand side of (3.4.33)

is

>

(exp

(log log k +

log log k − 3/2

log k

))·(

1

log k+

1

log k + log log k − 1/2

)=

(exp

(log log k − 3/2

log k

))·(

1 +log k


)>

(1 +

log log k − 3/2

log k

)·(

1 +log k


)= 2 +

log log k − 3/2

log k− 1

log k + log log k − 1/2= 2 + ξk,

105

say. Since ξk > 0 for all n ≥ 35 807, inequality (3.4.33) is proved for k ≥ 35 807. On the

other hand, using a computer, one can easily check that (3.4.32) holds for each integer

k ∈ [95, 35806], thus completing the proof of (3.4.32).

Lemma 3.8. Let α ∈ (0, 1), c1, c2 ∈ R with c1 > 0, c2 < 0 and I := (c−1/α1 − c2,∞).

Consider the function g : I → R dened by

g(z) :=c1(z + c2)α − 1

z.

Then, g attains its unique maximum at some point z0 > c−1/α1 − c2.

Proof. Consider the function h : I → R given by

(3.4.34) h(z) := z2(z + c2)1−αg′(z) = c1αz − c1(z + c2) + (z + c2)1−α.

It follows from this that h and g′ have the same sign and the same zeros in I. Moreover,

h(∞) = −∞. On the other hand,

h′(z) = c1(α− 1) +1− α

(z + c2)α,

in which case,

h′(z) = 0 ⇐⇒ 1 = c1(z + c2)α,

which is impossible for z ∈ I. Now, because h′(∞) < 0, this means that h′(z) < 0 for

z ∈ I. Our second claim then follows from the fact that the maximum is in I.


It is easy to verify that (3.3.1) holds when ω(n) = 1. For any n with ω(n) ≥ 2, we

introduce the function r(n) dened implicitly by

τ(n) =

(er(n) log n

ω(n) logω(n)

)ω(n)

.

Hence, for any n with ω(n) ≥ 2, we have

(3.5.1) r(n) :=1

ω(n)

(log τ(n)− ω(n) log

(log n

ω(n) logω(n)

)).

Observe that for n∗ := 60060 = 22 ·3 ·5 ·7 ·11 ·13 we have r(n∗) = 0.737505 . . . = log η2.

We claim that n∗ is the only integer n with ω(n) ≥ 2 that maximizes the function r

106

(this function is clearly bounded). To prove it, we proceed by contradiction. Assume

that, for some k ≥ 2, there exists an integer n′ 6= n∗ with ω(n′) = k for which (3.3.1) is

false and moreover that r(n′) is maximal. It is clear that the factorization of n′ takes

the form

(3.5.2) n =k∏i=1

pαii with α1 ≥ α2 ≥ · · · ≥ αk,

where the pi's are the primes in ascending order.

Using (3.4.11) (from Corollary 3.1) and (3.4.22) (from Lemma 3.5), one easily see that

r(n′) < log 2 = 0.693 . . . if k ≥ 44 which is non sense since r(n∗) = 0.737 . . . Thus we

must have k ≤ 43. Now, it follows from (3.4.11) that

(3.5.3) τ(n′) ≤(

2 log n′

k

)kb(n′) ≤

(2 log n′

k

)kb(nk).

Inserting (3.5.3) in (3.5.1), we then get

r(n′) ≤ 1

k

(log b(nk) + k log

(2 log n′

k

)− k log

(log n′

k log k

))= log 2 + log log k +

log b(nk)

k,

a quantity which depends only on k. On the other hand, using a computer reveals that

r(n′) < log η2 for each k ∈ [2, 3] ∪ [25, 43]. This contradicts the choice of n′. Therefore

we only need to consider the cases when k ∈ 4, . . . , 24.

Now, inserting (3.4.10) in (3.5.1), we have that

r(n′) ≤ 1

k

(log b(nk) + k log

(log n′

k

)+ k log

(1 +

log nklog n′

)− k log

(log n′

k log k

))=

log b(nk)

k+ log log k + log

(1 +

log nklog n′

)= r1(n′, k),

where

(3.5.4) r1(z, k) :=log b(nk)

k+ log log k + log

(1 +

log nklog z

).

We observe that the function r1(z, k) decreases when z increases. Thus, dening zk as

the unique solution in z of r1(z, k) = log η2, we obtain that n′ ≤ zk given that ω(n′) = k.

We now consider the function

(3.5.5) u(x) := max`≥0` : n` ≤ x.

107

Observe that, since n′ is of the form (3.5.2), u(zk/nk) is an upper bound for the rank j

of the largest prime pj such that p2j | n′. One may verify that for each k ∈ 4, . . . , 24

we have u(zk/nk) ≤ 3 implying that j ≤ 3. Now, recalling the denition of t(n) given

in (3.4.25), we may write

r(n) = log t(n) + log(ω(n) logω(n)).

Hence, for a xed value of k = ω(n), it follows that r(n) increases or decreases along

with t(n). Therefore, our hypothesis implies that t(n′) is maximal. Thus for each

j ∈ 1, 2, 3, using inequality (3.4.27) and the maximality of t(n′), we can write

1 ≤ t(n′)

t(n′/pj)≤(

1 +1

kα

)(1− log pj

log n′

)≤(

1 +1

kα

)(1− log pj

log zk

),

and we obtain the desired contradiction if this last expression is less than 1, which will

happen if the integer α ≥ 2 satisfying pαj ‖n′ is large enough. Using (3.4.29) we get an

upper bound for each of the rst three components in the exponent vector of n′. In

fact, one may verify that, for each k ∈ 4, . . . , 24,

(4, 2, 2, 1, . . . , 1︸︷︷︸k−3

)

is an upper bound (in each of its coordinates) for the exponent vector of n′, implying

that there are just a small number of cases to verify. After all the computations are

done, we obtain a nite set of pairs (n, r(n)) including (n∗, r(n∗)) and nd that all the

other pairs in this set satisfy r(n) < r(n∗). This contradicts the existence of n′ and

completes the proof of Theorem 3.1.


We rst verify that (3.3.2) does not hold for the integer n∗ := 782139803452561073520 =

24n16. If n is any integer such that ω(n) ≥ 44, then it follows from Corollary 3.1 and

Lemma 3.5 that inequality (3.3.2) is satised. Since it is clear that (3.3.2) holds when

ω(n) = 1, it remains only to consider the set of integers n such that 2 ≤ ω(n) ≤ 43.

For any such k, let zk be the unique solution in z to

r1(z, k) = log 2,

where r1(z, k) is the function dened in (3.5.4).

We proceed by contradiction by assuming that there exists an integer n′ such that

ω(n′) ∈ 17, . . . , 43 and for which (3.3.2) is false. We may also assume that n′ realizes

108

the maximum of the function r and moreover that n′ is of the form (3.5.2). As in

Theorem 3.1, we have n′ ≤ zk and one can verify that u(zk/nk) ≤ 5. Thus, the exact

same method that we used in the proof of Theorem 3.1 leads to an upper bound for

the exponent vector of n′ given by

(5, 3, 2, 2, 1, . . . , 1︸︷︷︸k−4

).

One can then verify, using a computer, that neither of these nite number of possibilities

leads to a number n that does not satisfy (3.3.2), thus contradicting the existence of

n′.

We can therefore assume that 2 ≤ k ≤ 16. Since z2 = 3.25 . . . , z3 = 36.12 . . . and

r(30) < log 2, we deduce that in the particular cases k = 2 and k = 3, there is no

counterexample in integers n of the form (3.5.2) to inequality (3.3.2). Thus, there

is no counterexample in integers n ≥ 2 with ω(n) ≤ 3. For 4 ≤ k ≤ 16 there are

counterexamples to (3.3.2) and thus we need to focus our attention on getting a good

upper bound for every such integer in terms of k only. In order to do this, we rst

exhibit the values of uk := u(zk/nk) (easily obtained using a computer) in Table 1.

k 4 5 6 7 8 9 10 11 12 13 14 15 16

uk 1 2 3 3 3 4 4 4 4 4 5 5 5

Table 1

We can use this information to obtain an upper bound for τ(n) for any such coun-

terexample n of (3.3.2). Indeed, by using the multiplicativity of the function τ and

inequality (3.4.10), we get that for any such n with ω(n) = k,

τ(n) ≤ dk :=2k−ukb(nuk)

uukk

(log

zkn2uk

nk

)uk.

A priori this inequality is valid only for integers n of the form (3.5.2), but it is then

clearly also true for any counterexample to (3.3.2) since any general counterexample to

(3.3.2) has an associated counterexample of the type (3.5.2) with the same exponent

vector once the prime factors are properly ordered. We use this inequality in (3.5.1)

and introduce the function

r2(z, k) :=1

k

(log dk − k log

(log z

k log k

)).

109

Now, let z′k be the unique solution in z to

r2(z, k) = log 2.

Since ddzr2(z, k) < 0, we deduce that z′k is an upper bound for the largest possible

counterexample n to (3.3.2) with an hypothetic value of τ(n) equal to dk; clearly this

is the largest among those we nd with any smaller value of τ(n). We then nd, using

a computer, that z′k is smaller than 24n16 for each k ∈ 4, . . . , 15.

For k = 16, the situation is somewhat dierent. Instead, we verify by using z′16 that

there are only three possible exponent vectors, namely

(3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)

(3, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)(3.6.1)

(4, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)

that yield a counterexample to (3.3.2) in integers n of the type (3.5.2). For each of these,

the smallest number strictly larger than the basic form is obtained by replacing the

largest prime factor p16 = 53 by 59. We then obtain numbers n which give r(n) < log 2.

We deduce that 24n16 (which corresponds to the last exponent vector in (3.6.1)) is the

largest of these. The proof of Theorem 3.2 is then complete.


We rst verify that for n∗ := 720n7 we have λ(n∗) = 1.1999953 . . . := η3. We will show

that n∗ is the only integer that maximizes λ. In order to reach a contradiction, we will

assume that there exists n′ 6= n∗ for which λ(n′) ≥ λ(n∗). Again, it is clear that the

maximal value of λ exists and is attained by an integer of the form (3.5.2). Therefore

we will assume that n′ is of this form with ω(n′) = k. From (3.4.12) and (3.4.22), it

follows that the inequality

λ(n′) ≤ 1 +

∑ki=1 log pi − k log k

log n′

is valid for each k ≥ 44. On the other hand, we cannot have

(3.7.1)

∑ki=1 log pi − k log k

log n′> η3 − 1

if k ≥ 44, the reason being that since n′ has k prime factors, it must satisfy log n′ ≥log nk =

∑ki=1 log pi in which case (3.7.1) would imply

(3.7.2) (2− η3)k∑i=1

log pi > k log k.

110

But, using (3.4.20), it is easy to verify that (3.7.2) is impossible when k ≥ 44. This

proves that we must have k ≤ 43. Considering (3.4.12), we let zk be the unique solution

in z of

υ(nk, z) =

∏p|n

log k

log p

1/k

+log nklog z

∏p|n

log k

log p

1/k

− k log k

log z= η3,

where υ(n, z) is the function dened in (3.4.30). Since ddzυ(nk, z) < 0 by (3.4.31),

we deduce that n′ ≤ zk. We nd that the only possibilities for n′ are those with

k ∈ 5, . . . , 13, since otherwise we would have n′ ≤ zk < nk which is impossible since

by hypothesis we have nk | n′.

Now, for 5 ≤ k ≤ 13 and from the fact that n′ is of the form (3.5.2) with ω(n′) = k, we

deduce that n′ = snk ≤ zk for some integer s which satises nj|s with j ≤ k. One can

calculate that the largest ratio zk/nk (for 5 ≤ k ≤ 13) is less than 264 507. This forces

j ≤ 6. Now, consider the set

U := s ≤ 264 507 : P (s) ≤ 13,

where P (s) stands for the largest prime factor of s, and the set V := (snk, λ(snk)) :

s ∈ U and 5 ≤ k ≤ 13. We observe that V contains the element (n∗, r(n∗)) and that

for any other n we have r(n) < r(n∗). This contradicts the existence of n′ and the proof

of Theorem 3.3 is then complete.


In order to reach a contradiction, let us assume that there exists an integer n′ with

ω(n′) = k, for some k ≥ 74, for which (3.3.4) does not hold. For xed values of ω(n)

and τ(n), we see by denition (3.2.6) that the function λ(n) decreases as n increases.

For this reason, we will assume that n′ is of the form (3.5.2). We will also assume that

λ(n′) is maximal.

For k ≥ 95, we deduce from (3.4.12), (3.4.30), (3.4.31) and (3.4.32) that

λ(n′) ≤ υ(n′, n′) ≤ υ(nk, nk) < 1.

This means that, inequality (3.3.4) holds for k ≥ 95.

For each integer k ∈ 74, . . . , 94, we cannot conclude since υ(nk, nk) > 1. However,

since by Lemma 3.7 we have ddzυ(nk, z) ≤ 0 and υ(nk,∞) < 1, we can dene zk

111

implicitly by υ(nk, zk) = 1, in which case n′ ≤ zk. Also, observe that

log zk/ log nk < 2

for each k. This last inequality implies that the largest prime factor of n′ has its

corresponding exponent equal to 1.

As we have already seen, uk := u(zk/nk) provides an upper bound for the rank j of the

largest prime pj such that p2j | n′ since n′ is of the form (3.5.2). Our goal from now on

is to verify all the remaining possibilities. To do so, we proceed in four steps. In the

rst step, we introduce a variable j1 that will take the values 0, 1, . . . , uk and a variable

j2 that will take the values 0, 1, . . . ,min(j1, uk). Then, we assume that

(3.8.1) n′ = pα11 · · · p

αj2j2· p2

j2+1 · · · p2j1· pj1+1 · · · pk

for some integers αi ≥ 3 and that n′ is of the form (3.5.2). Now, if 0 < j2 ≤ j1, by using

the multiplicativity of the function τ together with (3.4.12), recalling the denition of

λ in (3.2.6), we are lead to consider the function

f1(j2, j1, k, z) :=(c1(j2, j1, k)(log z + c2(j2, j1, k))j2/k − 1)k log k

log z

where

c1 = c1(j2, j1, k) := 2(k−j1)/k3(j1−j2)/k 1

jj2/k2

b(j2)1/k

and

c2 = c2(j2, j1, k) := log(nj2)− log(nknj1/(nj2)2).

Assume for now that each constant c2 that will be considered through this proof satises

(3.8.2) c2 < 0.

Then, using Lemma 3.8, we have

λ(n′) ≤ f1(j2, j1, k, n′) ≤ max

z>−c2(j2,j1,k)f1(j2, j1, k, z).

Therefore, we will get the desired contradiction if n′ is of the type (3.8.1) and the unique

maximum of f1(j2, j1, k, z) is proven to be less than 1 (see Remark 3.3 for more details).

The cases with j2 = 0 or j1 = 0 must be veried directly.

The values of uk are recorded in Table 2.

112

k 74 75 76 77 78 79 80 81 82 83 84

uk 45 43 41 39 37 35 33 30 29 26 25

k 85 86 87 88 89 90 91 92 93 94

uk 23 21 19 17 15 13 11 8 6 2

Table 2

All the computations being done, one is left with a reduced set of possibilities for the

form of n′. In fact, we now have that k ∈ 74, 75, 76, 77 and also that the number of

values that j1 can take is signicantly reduced. The nal result is given in Table 3.

k 74 75 76 77

j1 ∈ 14, . . . ,28 16, . . . ,26 18, . . . ,25 20, . . . ,23

Table 3

What we mean here is that a xed pair (k, j1) is not in Table 3 if for all j2 ≤ min(j1, uk)

we have maxz>−c2(j2,j1,k)

f1(j2, j1, k, z) < 1.

This is where the second step of verications starts. We now assume that

n′ = pα11 · · · p

αj3j3· p3

j3+1 · · · p3j2· p2

j2+1 · · · p2j1· pj1+1 · · · pk

for some integers αi ≥ 4 and we use the same argument as before to dene the well

suited function

f2(j3, j2, j1, k, z) :=(c1(j3, j2, j1, k)(log z + c2(j3, j2, j1, k))j3/k − 1)k log k

log z,

where

c1(j3, j2, j1, k) := 2(k−j1)/k3(j1−j2)/k4(j2−j3)/k 1

jj3/k3

b(j3)1/k

and

c2(j3, j2, j1, k) := log(nj3)− log(nknj1nj2/(nj3)3).

We still have the chain of inequalities

λ(n′) ≤ f2(j3, j2, j1, k, n′) ≤ max

z>−c2(j3,j2,j1,k)f2(j3, j2, j1, k, z).

This time, we run this over the remaining values of j1, and for

j2 ∈ 1, . . . ,min(j1, u(zk/(nj1nk)))

113

and

j3 ∈ 1, . . . ,min(j2, u(zk/(nj2nj1nk))).

The cases with j3 = 0 must be treated separately. Once again, these computations

lead to further progress. We record in Table 4 the remaining values which need to be

examined.

k 74 75 76

j1 ∈ 14, . . . ,23 16, . . . ,21 18,19

Table 4

We are now ready to begin the third step of verications. We assume that

n′ = pα11 · · · p

αj4j4· p4

j4+1 · · · p4j3· p3

j3+1 · · · p3j2· p2

j2+1 · · · p2j1· pj1+1 · · · pk

for some integers αi ≥ 5 and dene the function f3(j4, j3, j2, j1, k, z) in a similar manner

by using the same ideas. However, we do introduce a new idea in the way of reducing

the number of values that the variables js (s ≥ 3) can take. We rst assume that pα‖n′

for a xed α ≥ 2, then we use (3.4.26), the fact that n′ ≤ zk and the maximality of

λ(n′) in order to write

1 ≤ λ(n′)

λ(n′/p)≤(

1 +2

kα

)(1− log p

log n′

)≤(

1 +2

kα

)(1− log p

log zk

).

We nd a contradiction if p is large enough to force the last expression to be less than

1. In particular, we get an upper bound for the rank j of such a prime pj. Since this

upper bound decreases when α increases, we obtain an upper bound for the rank j of

any prime pj for which pαj | n′. Thus, by using (3.4.28), we obtain Table 5.

αk 74 75 76

4 11 11 10

5 7 6 6

6 4 4 4

Table 5

Using this, we let j1 take the values in Table 4, and we let

j2 ∈ 1, . . . ,min(j1, u(zk/(nj1nk))),

114

j3 ∈ 1, . . . ,min(j2, u(zk/(nj2nj1nk)), 11 or 10)

and

j4 ∈ 1, . . . ,min(j3, u(zk/(nj3nj2nj1nk)), 7 or 6).

Again, we treat the cases with j4 = 0 independently. The computations lead to the

result that we must have k = 74 and j1 ∈ 16, 17, 18. We rule out these cases by

dening f4(j5, j4, j3, j2, j1, k, z) and by using Table 5 to limit the range of the variables

j3, j4 and j5. This completes the verications.

It remains to prove (3.8.2). To do so, we use the fact that js+1 ≤ js and that

njs ≤zk

nknj1 · · ·njs−1

and c2(js, . . . , j1, k) = log njs + lognsjs


,

from which we deduce that

c2(js, . . . , j1, k) ≤ logzk


+ lognsjs


≤ logzk

nknj1+ log

njsnk

≤ logzkn2k

< −159.6

by direct computation, which proves (3.8.2). We also observe that

c2(js, . . . , j1, k) = logns+1js


> log1


> log1

nsk> −5 log n94 > −2342.

This completes the proof of Theorem 3.4.

Remark 3.3. We now provide some key details concerning the computations used in

the proof of Theorem 3.4. The information provided through the previous proof may

dier with other information obtained with another strategy. We used 50 decimals of

precision for all computations. We used the criterion fs(. . . ) < 0.999999 (for s = 1, 2, 3

or 4) for each comparison in the four steps of the computation and we kept a pair (k, j1)

if we found fs(. . . ) ≥ 0.999999 somewhere in the process. By considering the function

h dened in (3.4.34), we approximated c1 and c2 with 50 decimals and we called c′1 and

c′2 these approximations. Then we solved for z1 in h(z1) = 0. From

0 = c′1αz1 − c′1(z1 + c′2) + (z1 + c′2)1−α > c′1αz1 − c′1(z1 + c′2),

115

α ≥ 1/94 and c′2 < −159, we deduced that z1 + c′2 > 1.61. The same is true also for the

solution z of

0 = c1αz − c1(z + c2) + (z + c2)1−α.

It is easy to see that we always have c1 < 6!/ log 2 < 1039. With this information at

hand and using the mean value theorem, one nds that∣∣∣∣c′1(z1 + c′2)α − 1

z1

− c1(z1 + c2)α − 1

z1

∣∣∣∣ < 10−43,

thus concluding that the two functions are of about the same size for all values of z or

z1 such that z1 + c′2 > 1.6 or z + c2 > 1.6. From the fact that 94 log 94 < 428 and that

an error of about z1 · 10−50 on z1 cost less than 10−45 in the evaluation ofc′1(z1+c′2)α−1

z1,

we end up with an error of at most 10−40. This is small enough for the criterion we

used.


First, we verify that the integer n in (3.3.5) satises λ(n) > 1 and that it is of size

exp(10640.8428 . . . ). Then, we claim that n is the largest integer n with ω(n) ≥ 44

and λ(n) ≥ 1. To do so, we proceed by contradiction and assume that there exists an

integer n′ such that n′ > n with λ(n′) ≥ 1 and ω(n′) ≥ 44. The argument is done in

several steps.

3.9.1 Preliminary steps

The rst step consists in showing that we must have ω(n′) = 44. For this, we use

(3.4.12), (3.4.30) and (3.4.31) to deduce that if we dene zk by υ(nk, zk) = 1 then we

must have n′ ≤ zk. We verify that zk ≤ exp(4569.68) < n for each k ∈ 45, . . . , 73and then we conclude using Theorem 3.4.

We then want to show that γ(n′) = n44. This is done in two steps. We rst assume

that n′ is made of a choice of a set S of 44 distinct primes in p1, . . . , p45 and that this

choice is not S ′ := p1, . . . , p44. There are 44 possibilities and if we write nS :=∏p∈S

p,

then using again the same argument as previously, we dene zS by υ(nS, zS) = 1 and

verify that zS ≤ exp(9927.67) < n for each S.

Now, we assume that n′ has a general set of prime factors which has not been previously

considered (and is not S ′), implying that there exists an integer n′′ < n′ such that

116

τ(n′′) = τ(n′) and such that the set of prime divisors of n′′ is S for some S 6= S ′. We

then have

λ(n′) < λ(n′′) < 1

if n′′ > zS. This proves that the set of prime factors of n′ must be S ′.

We solve for z in the equation υ(n44, z) = 1 to nd that n′ < exp(10758.21). We have

thus proved that

10640.8 < log n′ < 10758.8.

Consider the intervals Ij := [10639.8 + j, 10640.8 + j] for each j = 1, . . . , 118. From

now on, we want to show that log n′ cannot be in any of these Ij.

3.9.2 A rst argumentation

Recall the notation in (3.4.5) and (3.4.6), that is xi (i = 1, . . . , 44), µ, µ1, µ2, $ and $′.

The rst argument that we use to eliminate some intervals Ij relies on the inequality

(3.4.2) and on the proof of Corollary 3.1. For a value of m ∈ 1, . . . , 43, we have

τ(n′) ≤ log44 n′b(n44)µm1 µ44−m2

= b(n44)

(log n′

44

)44(1 +

log n44

log n′− $′

2m

)m(1 +

log n44

log n′+

$′

2(44−m)

)44−m

so that if we write

υm(z, w) := b(n44)144 log 44

(1 +

log n44

z− w

2m

)m44(1 +

log n44

z+

w

2(44−m)

)1−m44

−44 log 44

z(3.9.1)

then we have

λ(n′) ≤ maxm∈1,...,43

υm(log n′, $′).

Thus, we dene zm,$ by υm(zm,$, $) = 1. We have seen that zm,0 = 10758.2 . . . From

(i) and (iii) of Lemma 3.3, we know that zm,w decreases when w increases. We record

in Table 6 a value of w := w(j) such that

maxm∈1,...,43

zm,w(j) < 10639.8 + j

for the rst 39 values of j.

j 1 2 3 4 5 6 7 8 9

w 0.2137 0.2128 0.2119 0.2109 0.2100 0.2091 0.2081 0.2072 0.2062

j 10 11 12 13 14 15 16 17 18

w 0.2053 0.2043 0.2034 0.2024 0.2014 0.2004 0.1995 0.1985 0.1975

117

j 19 20 21 22 23 24 25 26 27

w 0.1965 0.1955 0.1945 0.1935 0.1925 0.1914 0.1904 0.1894 0.1884

j 28 29 30 31 32 33 34 35 36

w 0.1873 0.1863 0.1852 0.1841 0.1831 0.1820 0.1809 0.1799 0.1788

j 37 38 39

w 0.1777 0.1766 0.1755

Table 6

In the opposite direction, a lower bound for $′(= $′(n′)) can be computed for n′

assuming that log n′ is in Ij. To do so, we split the interval Ij in 210 subintervals of

length 1210

that we call Ij,j1 where 1 ≤ j1 ≤ 210. We use Lemma 3.4 (i) with ϕ = 1

term by term to compute

minz∈Ij,j1

44∑i=1

minαi∈Z

∣∣∣∣(αi + 1)44 log pi − log n44

z− 1

∣∣∣∣and take the minimum over the variable j1 to get the lower bound for $′(n′) for log n′

in Ij. We record the result in Table 7.

j 1 2 3 4 5 6 7 8 9

$′ 0.1814 0.1812 0.1810 0.1808 0.1804 0.1802 0.1800 0.1798 0.1797

j 10 11 12 13 14 15 16 17 18

$′ 0.1797 0.1798 0.1800 0.1800 0.1800 0.1798 0.1797 0.1797 0.1798

j 19 20 21 22 23 24 25 26 27

$′ 0.1800 0.1800 0.1800 0.1798 0.1796 0.1792 0.1788 0.1784 0.1781

j 28 29 30 31 32 33 34 35 36

$′ 0.1779 0.1777 0.1775 0.1773 0.1771 0.1769 0.1765 0.1763 0.1761

j 37 38 39

$′ 0.1755 0.1751 0.1748

Table 7

Also, we verify that for each j ∈ 40, . . . , 118 we have $′(j) > w(j), thereby implying

that there exist no n′ with log n′ in Ij.

All of this gives rise to a new concept that will be crucial for the remaining of the proof.

This is the dierence between the upper and lower bounds for $′. Just saying that here

118

the dierence when j = 1, that is 0.2137− 0.1814 = 0.0323, is too large for us. In fact,

it will be convenient to work with a slightly dierent concept. Consider the function

dened on primes p by

(3.9.2) εj(p) := minz∈Ij

∣∣∣∣(α + 1)44 log p− log n44

z− 1

∣∣∣∣ .The value of εj(p) is computed by using Lemma 3.4 (i). For each j ∈ 1, . . . , 39 wesum the εj(pi) for i ∈ 1, . . . , 44 and subtract the answer from the upper bound w(j).

We call these value δ′(= δ′(j)) and record them in Table 8.

j 1 2 3 4 5 6 7 8

δ′ 0.03422 0.03353 0.03283 0.03203 0.03142 0.03083 0.03005 0.02936

j 9 10 11 12 13 14 15 16

δ′ 0.02848 0.02753 0.02638 0.02536 0.02436 0.02340 0.02253 0.02178

j 17 18 19 20 21 22 23 24

δ′ 0.02075 0.01958 0.01846 0.01748 0.01650 0.01561 0.01481 0.01398

j 25 26 27 28 29 30 31 32

δ′ 0.01340 0.01279 0.01216 0.01133 0.01053 0.00964 0.00874 0.00795

j 33 34 35 36 37 38 39

δ′ 0.00705 0.00618 0.00547 0.00458 0.00392 0.00331 0.00258

Table 8

The value δ′ is to be interpreted as an upper bound to the extra error that can produce

n′.

3.9.3 A rst verication

We want to make some direct verications to prove that λ(n′) ≥ 1 is impossible if the

exponent vector of n′ is of a certain type. Consider the sets

(3.9.3) Jδ(p, j) :=⌈

(1−δ)(10639.8+j)+logn44

44 log p

⌉− 1, . . . ,

⌊(1+δ)(10640.8+j)+logn44

44 log p

⌋− 1.

The set Jδ(p, j) has the property that if∣∣∣∣(α + 1)44 log p− log n44

log n− 1

∣∣∣∣ ≤ δ with log n ∈ Ij

then α ∈ Jδ(p, j).

119

We divide the verications into two distinct types. Type 1 concerns the sets

Sj(δ) := Jδ(p1, j)× · · · × Jδ(p44, j).

We take δ = 0.011 for j ∈ 1, . . . , 4 and δ = 0.01 for j ∈ 5, . . . , 14. Also, to speed

up the process, we consider the union term-by-term of S1(0.011), . . . , S4(0.011) to get

a new set S1 say, so that S1 = J0.011(2, 1) ∪ · · · ∪ J0.011(2, 4) × . . . We do the same

with S5(0.01), . . . , S14(0.01) to get S2. These sets have respectively 92160 and 53760

elements. For each vector v = (α1, . . . , α44) in each of these two sets, we take one of the

946 possible choices of two elements in a set of 44 elements, say (i1, i2), and construct

the new set

α1×···×αi1−1×Jεj(pi1)+δ′(j)(pi1 ,j)×αi1+1×···×αi2−1×Jεj(pi2

)+δ′(j)(pi2 ,j)×αi2+1×···×α44.

We verify that all these exponent vectors v give rise to an integer n such that λ(n) < 1,

log n < 10640.8 or is itself n.

Type 2 concerns the sets

S ′j(δ) := Jεj(p1)+δ(p1, j)× · · · × Jεj(p44)+δ(p44, j).

This time, we take δ = 0.0055 for j ∈ 1, . . . , 6, δ = 0.0054 for j ∈ 7, . . . , 9 andj ∈ 10, . . . , 13, δ = 0.005 and j ∈ 14, . . . , 19, δ = 0.0044 for j ∈ 20, . . . , 23,δ = 0.004 for j ∈ 24, 25, 26, δ = 0.0035 for j ∈ 27, 28, 29 and δ = 0.003 for

j ∈ 30, . . . , 39. Again, to speed up the process, we consider the unions term-by-term

the same way, so that we have S ′1 = Jε1(2)+0.0055(2, 1) ∪ · · · ∪ Jε6(2)+0.0055(2, 6)× . . . andthe same for S ′2, . . . , S

′8. These sets have respectively 98304, 73728, 49152, 49152, 32768,

32768, 32768 and 24576 elements. For each vector v, we do the exact same process as

for the type 1.

At the end of these verications, we know that there are at least three entries in the

exponent vector that produce a large error and this occurs in both type 1 and 2.

3.9.4 Reducing the upper bound for δ′

Our strategy begins with a lower bound for $′1 and $′2. For each j, there are four cases

to consider depending on the position of the xi (3.4.5) compared to µ (3.4.6). Indeed,

we have seen in the previous section, with the type 1 verication, that there are at

least three xi that are far from µ but this does not tell us where they are. So any lower

bound for $′1 and $′2 will come in pair ($′1, $

′2) with the total number m of xi that are

120

less than or equal to µ and with a position signature s in 0, 1, 2, 3 that tells us thatthe number of xi that are less than µ in these three we assume to have. This number

m can be shown to take the values we recorded in Table 9.

j ∈ 1,. . . ,6 7 8,. . . ,12 13 14

m ∈ 11,. . . ,33 12,. . . ,33 12,. . . ,32 13,. . . ,32 13,. . . ,31

Table 9

To do so, we use Table 7 and verify that zm,$′(j) < 10639.8 + j (see section 9.2) for all

the values of m not listed in Table 9.

Also, the denition of $′1 and $′2 in (3.4.7) and (3.4.8) includes the exact value of $′,

something that we cannot know precisely. So we assume an interval containing the value

of $′, and look for a contradiction. More precisely, we will assume, for j ∈ 1, . . . , 14,that $′ belongs to W (j) := [w(j)− 0.01, w(j)].

To get these lower bounds, we x j, m and a signature s. Then, we split the interval Ijin 30 subintervals Ij,r1 (r1 = 1, . . . , 30) of equal length and we split the interval W (j)

in 80 subintervals W (j, r2) (r2 = 1, . . . , 80) of equal length.

We x Ij,r1 and W (j, r2) and begin with $′1. At rst, we focus on the s points xi that

are less than µ. We will show that in this case the minimum of

(3.9.4)

∣∣∣∣(α + 1)44 log p− log n44

x− 1 +

$′

2m

∣∣∣∣is attained with α :=

⌈(1−δ)(10639.8+j)+logn44

44 log p

⌉− 2 and, as in Lemma 3.4 (ii), at the

extremity of the intervals Ij,r1 and W (j, r2). In fact, from the denition of Jδ, it isenough to show that (α+1)44 log p−logn44

x− 1 + $′

2m< 0 and this follows from

(α+ 1)44 log p− logn44

x− 1 +

$′

2m=

(⌈(1−δ)(10639.8+j)+logn44

44 log p

⌉− 1)44 log p− logn44

x− 1 +

$′

2m

=

((1−δ)(10639.8+j)+logn44

44 log p+ ξ − 1

)44 log p− logn44

x− 1 +

$′

2m

=(1− δ)(10639.8 + j)

x− (1− ξ)44 log p

x− 1 +

$′

2m

≤ −δ + $′

2m< 0

since 2mδ > 2 > $′ from our choices, where 0 ≤ ξ < 1 and both $′ and x are seen as

xed. We keep the s smallest such values among the 44 prime numbers.

121

Then, we compute the minimum value of (3.9.4), without any constraint on α, using

Lemma 3.4 (i). We keep the m−s smallest ones among the 44 prime numbers. We sum

the m values we have kept so far and we take the minimum among the 30 · 80 = 2400

possible values of (r1, r2) and this is the wanted lower bound for $′1 in I(j) with this

value of m and s. We do the same for $′2 with 3− s values of xi greater than µ along

with the choice α :=⌊

(1+δ)(10640.8+j)+logn44

44 log p

⌋instead and the function

(3.9.5)

∣∣∣∣(α + 1)44 log p− log n44

x− 1− $′

2(44−m)

∣∣∣∣as in the denition of $′2 in (3.4.8). The proof is similar. We obtain the value for

$′2 in Ij with parameters 3 − s and 44 − m instead. We keep the pair ($′1, $′2)(=

($′1(j,m, s), $′2(j,m, s))). For example, here is the output we get as lower bound with

j = 1 and m = 11:

(0.010296421544, 0.093154520284), (0.010296421544, 0.089438737225),

(0.011179764497, 0.087104430865), (0.012637967643, 0.085223479629)

for ($′1, $′2) when s = 0, 1, 2, 3 respectively. Note that these values are just stated as

an example, they are sensitive to the way the program is done. Now, using the same

reasoning as we did to get to (3.9.1), we are lead to consider

υm,m1,m2(z, w,w1, w2) := b(n44)144 log 44

(1 +

log n44

z− w

2m− w1

2m1

)m144

·(1 +

log n44

z− w

2m+

w1

2(m−m1)

)m−m144

·(1 +

log n44

z+

w

2(44−m)− w2

2m2

)m244

·(1 +

log n44

z+

w

2(44−m)+

w2

2(44−m−m2)

)1−m+m244

−44 log 44

z(3.9.6)

for xed values of m1 ∈ 1, . . . ,m − 1 and m2 ∈ 1, . . . , 43 − m. We also dene

zj,m,m1,m2,s implicitly by

υj,m,m1,m2,s(zj,m,m1,m2,s, w(j)− 0.01, $′1(j,m, s), $′2(j,m, s)) = 1.

and verify that

maxm

maxs∈0,...,3

maxm1∈1,...,m−1

maxm2∈1,...,43−m

zj,m,m1,m2,s < 10639.8 + j,

122

where the maximum is taken over the values of m appearing in Table 9. We justify

that zj,m,m1,m2,s is the appropriate choice by using Lemma 3.3 part (i), (ii) and (iv)

(there is a condition to verify in part (ii)). We nd that υm,m1,m2 would be smaller

with larger values of the variables. This is the contradiction we were looking for. So

we have in fact that $′ /∈ W (j) and a new upper bound for δ′ recorded in Table 10.

j 1 2 3 4 5 6 7

δ′ 0.02422 0.02353 0.02283 0.02203 0.02142 0.02083 0.02005

j 8 9 10 11 12 13 14

δ′ 0.01936 0.01848 0.01753 0.01638 0.01536 0.01436 0.01340

Table 10

3.9.5 The last verication

Our strategy of verication begins with a preliminary computation. We use the type 2

computations that we did previously to prove that at least three points xi dened in

(3.4.5) are far from µ dened in (3.4.6). We rst want to show that, among the 13244

possibilities of triplets of primes, at most a few hundreds can produce these three values

of xi.

To do so, we x j and split the interval Ij into 25 subintervals Ij,r of equal length. We

also x a triplet (q1, q2, q3). Now, the type 2 computations reveal that the exponent of

a prime p ∈ q1, q2, q3 that divide exactly n′ is not in Jδ+εj(p)(p, j) where δ = δ(j) can

be found in section 9.3. We are thus in the exact situation of Lemma 3.4 (ii). So that

we compute and sum the three minimal errors, we take the minimum over r = 1, . . . , 25

and call this minimum ζ(= ζ(q1, q2, q3)). If

ζ − εj(q1)− εj(q2)− εj(q3) > δ′(j)

then the triplet (q1, q2, q3) is rejected. Otherwise, we keep

(q1, q2, q3, ζ(q1, q2, q3)− εj(q1)− εj(q2)− εj(q3))

to the last verication in a set T (j), say. The value of δ′ is picked from in Table 10 if

j ≤ 14 and from Table 8 if 15 ≤ j ≤ 39.

Now, for the very last verication, after all the T (j) have been computed, we use a new

idea. We assume that j is xed. For a prime p in a xed vector (q1, q2, q3, ρ) ∈ T (j), we

123

observe that it is enough to check the integers n with the exponent in Jδ′(j)+εj(p)(p, j).

Then, for the remaining 41 primes p, it is enough to verify with the exponent in the set

Jδ′(j)/2−ρ/2+εj(p)(p, j)

for all but one prime p for which it can be in Jδ′(j)−ρ+εj(p)(p, j).

With these observations in mind, we design an algorithm. We compute the largest

fourth component in any of the vectors in T (j) and call it t. Then we consider only the

vectors such that the fourth component is in [t− u/1000, t− (u− 1)/1000] for a xed

u ∈ 1, . . . , 7, which we denote by Tu(j). With u xed, we store in memory all the

vectors in

Jδ′(j)/2−(t−u/1000)/2+εj(2)(2, j)× · · · × Jδ′(j)/2−(t−u/1000)/2+εj(193)(193, j)

to which we add two dimensions: one of which is the value of τ(n)1/44 of the integer n

with this exponent vector whereas the other is its logarithm. Then, for all such vectors,

only four exponents have to be modied at each verication and thus the last two

informations need only a small adjustment to be used to compute the value of λ in each

case. So, for each vector of 46 dimensions, for each exponent in Jδ′(j)+εj(p)(p, j) of each

prime p in each triplet (in a vector) in Tu(j) and for each exponent in Jδ′(j)−ρ+εj(p)(p, j)

of each other 41 prime p we compute the corresponding value of λ. We try each value

of u and then all the values of j.

After all these verications, no value of n′ have been found. This is the contradiction

we were searching for and thus n is the largest number n such that λ(n) > 1 and

ω(n) ≥ 44. The proof is complete.

3.10 Final remarks

One can show that∑n≤x

∣∣∣∣λ(n)− log log x log log log x

log x

∣∣∣∣2 x log log x(log log log x)2

log2 x,

from which we may conclude that for almost all n ≤ x,

λ(n) = (1 + o(1))log log x log log log x

log x(x→∞).

On the other hand, we can show that there are innitely many n for which λ(n) > 1.

Indeed, to any set S of primes satisfying(∏p∈S

log k

log p

)> 1 and #S = k,

124

we can associate a sequence of integers m1, m2, . . . such that their exponent at each

prime factor, and then the associated θi as dened in Corollary 3.1, is as close as

possible to the optimal value as dened in Lemma 3.2. Precisely, for each p′ ∈ S,wecan choose mj to be an integer for which the exponent of p′, αp′ , is the closest integer

to 1k

(log zjlog p′

+∑

p∈Slog plog p′

)− 1 for a xed large zj. One veries that

λ(mj)→

(∏p∈S

log k

log p

)1/k

(zj →∞).

Finally, we can also show that the set of limit points of λ(n) is the interval [0, b(6)1/6 log 6]

= [0, 1.145206 . . . ] and that there exists a positive constant η such that

#n ≤ x| λ(n) ≥ 1 = (η + o(1)) log43 x (x→∞).

Moreover, we have

supω(n)=k

λ(n) = 1− log log k − 1

log k+

(log log k)2 − 3 log log k

log2 k+O

(1

log2 k

)(k →∞).

125

Bibliography

[1] M. Balazard, Sur la moyenne des exposants dans la décomposition en facteurs

premiers, Acta Arith. 52 (1989), no. 1, 1123.

[2] Y. Buttkewitz, C. Elsholtz, K. Ford and J.-C. Schlage-Puchta, A prob-

lem of Ramanujan, Erd®s and Kátai on the iterated divisor function, Inter. Math.

Research Notices, 2012, 40514061.

[3] J. M. De Koninck, Sums of quotients of additive functions, Proc. Amer. Math.

Soc. 44 (1974), 35− 38.

[4] J. L. Duras, J.-L. Nicolas and G. Robin, Large values of the function dk,

Number theory in progress, Vol. 2 (Zakopane-Ko±cielisko, 1997), 743 − 770, de

Gruyter, Berlin, 1999.

[5] P. Erd®s and J.-L. Nicolas, Sur la fonction nombre de facteurs premiers de n,

Séminaire Delange-Pisot-Poitou. Théorie des nombres, tome 20, no. 2 (1978-1979),

exp. no. 32, 119.

[6] J. L. Nicolas and G. Robin, Majorations explicites pour le nombre de diviseurs

de N , Canad. Math. Bull. Vol. 26 (1983), no. 4, 485492.

[7] G. Robin, Méthodes d'optimisation pour un problème de théorie des nombres,

RAIRO-Informatique théorique, tome 17, no. 3 (1983), 239247.

[8] S. Ramanujan, Highly composite numbers. Annotated and with a foreword by

Jean-Louis Nicolas and Guy Robin. Ramanujan J. 1 (1997), no. 2, 119153.

[9] J. B. Rosser and L. Schoenfeld, Approximate formulas for some functions of

prime numbers, Illinois J. Math. 6 (1962), 64− 94.

[10] G. Tenenbaum, Introduction à la théorie analytique et probabiliste des nombres,

3e édition, Belin, 2008.

126

[11] S. Wigert, Sur l'ordre grandeur du nombre de diviseurs d'un entier. Ark. Mat.

3, no. 18 (1907), 19.

127

Chapter 4

Lattice points close to a

three-dimensional smooth curve

Patrick Letendre

Résumé

Nous obtenons des majorations du nombre de points entiers près d'une courbe

lisse dans R3.

Abstract

We obtain various upper bounds of the number of lattice points close to a smooth

curve in R3.

128


Let X ⊆ [M, 2M ] be a xed interval, with M ≥ 2. We write f(x) := (x, f1(x), f2(x)),

where f1 and f2 are xed functions in Cn(X ) for some n ≥ 1. Much work has been

done to count the number of integer points close or on a given curve in R2. Techniques

greatly dier to t a given family of curve and a precise question. In this paper, we

mainly focus on a generalization to the third dimension of some of the ideas that can

be found in [3], [4], [5], [7] and [8] for example. More precisely, we see the main result

as the three-dimensional analog to the Corollary in [4], which is also the main result of

[2].

This problem in three dimension may also be seen as the number of approximate integer

solutions to a system of two equations in an integer, that is the rst coordinate of f

being an integer and the two other coordinates being close to an integer. Thus one

asks for three integers in this problem. As a very rst idea, it is clear that one gets

a nontrivial estimate by dropping one of the three conditions so that the cited results

can be applied. This paper is our rst attempt to fully use the three informations at

the same time to get a nontrivial estimate.

Throughout the paper, we will assume that there are constants C1 and C2, both greater

than 1, such that

(4.1.1)λi,j

Ci+1j

≤ |f (i)j (x)| ≤ Ci+1

j λi,j, (x ∈ X , j ∈ 1, 2, i = 1, 2) ,

Also, the main result of this paper is stated with the hypothesis

(4.1.2)λi1,j

Ci1+1j

≤ |f (i1)j (x)| and |f (i2)

j (x)| ≤ Ci2+1j λi2,j

for x ∈ X , j ∈ 1, 2, i1 = 1, 2, 3 and i2 = 1, 2, 3, 4. Here and throughout the paper we

assume that 0 < λi,j = Mλi+1,j for i ∈ 0, 1, 2, 3 and j ∈ 1, 2. Often, we will use

the notation C3 := min(C1, C2), C4 := max(C1, C2). We further set

(4.1.3) ∆i(x) := f(i)1 (x)f

(i+1)2 (x)− f (i+1)

1 (x)f(i)2 (x) (x ∈ X , i = 1, 2).

We will assume throughout the paper that

(4.1.4)∆i

2Ci+13 Ci+2

4

≤ |∆i(x)| ≤ 2Ci+13 Ci+2

4 ∆i, (x ∈ X , i = 1, 2)

129

where ∆i := λi,1λi+1,2 = λi+1,1λi,2. Given numbers

(4.1.5) 0 < δ1, δ2 ≤1

4,

we are interested in the set

(4.1.6) S := n ∈ X ∩ N| ‖f1(n)‖ < δ1 and ‖f2(n)‖ < δ2,

where ‖ · ‖ measures the distance to the nearest integer, and in the quantity S := #S.We also consider the set

(4.1.7)

Λ := (v1, v2, v3) ∈ R3| ∃x ∈ X s.t. v1 = x, |v2 − f1(x)| < δ1 and |v3 − f2(x)| < δ2

as the region in R3 that contains the lattice points corresponding to those included in

S. In particular S = #Λ ∩ Z3. We write f(x) = (x, f1(x), f2(x)) where fj(x) is the

closest integer to fj(x).

Throughout the paper, we use the following notation, often freely:

α0 := δ1 + δ2, α1 := δ1C22λ1,2 + δ2C

21λ1,1, α2 := δ1C

32λ2,2 + δ2C

31λ2,1,

X := |X |, X := ρM, δ0 := max(δ1, δ2), δ := min(δ1, δ2),(4.1.8)

λi,0 := max(λi,1, λi,2), λi := min(λi,1, λi,2)

and also the convention

(4.1.9)δ1

λ0,1

≤ δ2

λ0,2

.

Also, we write C := Cj where j is the value that corresponds to λi = λi,j.

For a general lattice Γ, we write |Γ| for its determinant. We adopt some of the notation

in [5], that is

(4.1.10) D(g1(x), . . . , gn(x);x1, . . . , xn) := det(gj(xi))n×n

where gj(x) are some functions for j = 1, . . . , n. In particular, we write

(4.1.11) V (x1, . . . , xn) := D(1, x, . . . , xn−1;x1, . . . , xn)

for the Vandermonde determinant.

In what follows, θ, with or without subscript, is a real number with |θ| ≤ 1 and not

necessarily the same at each occurrence. Similarly, ξ, η, ν, ζ are real numbers that come

either from Lemma 4.1, from a mean value theorem or from Taylor's theorem and thus

belong to some obvious interval at each occurrence.

130

4.2 Statement of the Theorems

Theorem 4.1. Let X ∈ [M, 2M ] be an interval satisfying

(4.2.1) X ≤ M

16C73C

164

and assume that δ1, δ2 ∈ R satisfy (4.1.5). Consider two functions f1, f2 ∈ C4(X )

satisfying (4.1.2) and (4.1.4). Assume the general notation that has been established in

the introduction along with

(4.2.2) α := α0 + α1, β0 :=δ1δ

λ2,1

, β1 :=δ2

1

λ2,1

, β2 :=δ2

2

λ2,2

, Ω := λ1,2 + 1 +λ2,2

λ2,1

,

Y := max

(λ1,1

λ1,2

,λ1,2

λ1,1

, λ1,1,1

λ1,1

, λ1,2,1

λ1,2

)which is specic to the statement of this theorem. Assume also that (4.1.9) holds and

that

(4.2.3)C2

4δ2

λ1,2

≤ 1.

Let S be dened in (4.1.6). Then for a xed i ∈ 0, 1, 2 we have

S ∆1/62 X + (α0∆1)1/3X + (α0λ2,1)1/3X + (α0λ2,2)1/3X + 1

+α1/20 αX log2M

+ min

(α

1/30 αX,

(δ2λ0,2

)1/5

α1/50 αρ3/5M+min

((δ2λ0,2

)2/15

α1/50 αρ3/5M,α

1/30 α X

Y 2/3

))log2 M

+ min

α 130 (δ1δΩ)

12X,

(δ2λ0,2

) 120

(δ1δΩ)12 α

3100 ρ

910M+min

( δ2λ0,2

) 130

(δ1δΩ)12 α

3100 ρ

910M,α

130 (δ1δΩ)

12 X

Y16

logM

+(λ2,1Ω)1/3

(δ2

λ0,2

)1/2

M +

(λ2,1Ωδ2

Y λ0,2

)1/3

M + (δ1λ2,1)1/2M

+δ1Ω

(δ2

λ0,2

)1/2

M + α

(δ2

λ0,2

)1/2

M log2M

+α

Y

(δ2

λ0,2

)1/3

M log2M +

(δ1δ2δ0Ω

λ0,2

)1/2

M logM +

(δ1δ0Ω

Y

)1/2(δ2

λ0,2

)1/3

M logM.

δ1/41 δ

1/22 λ

1/42,1X logM + λ

1/32,i

(δ1

λ2,1

)1/2

∆1/62 ρ2/3M + ΞiX,

131

where

(4.2.4)

Ξi :=∑′

1≤Q≤L(

δ1λ2,1

)1/2

QL2/βi

min

(βi max

(∆

1/42

Q1/2

L,α

1/22

Q

L3/2,δ1δ2

Q2

L3

),max

(∆

1/52

L

Q4/5,α

1/32

L2/3

Q2/3

),λ2,iβi

Q2

ρL

)

where the apostrophe ′ indicates that the values taken by Q and L are only powers of 2.

We also have

S ∆1/62 X + (α0∆1)1/3X + (α0λ2,1)1/3X + (α0λ2,2)1/3X + 1

+α1/20 αX log2X + min

((δ2

λ0,2

)2/15

α1/50 αX,α

1/30 α

X

Y 2/3

)log2X

min

((δ2

λ0,2

) 130

(δ1δΩ)12α

3100 X,α

130 (δ1δΩ)

12X

Y16

)logX

+α

Y

(δ2

λ0,2

)1/3

X log2X +

(δ1δ0Ω

Y

)1/2(δ2

λ0,2

)1/3

X logX.

δ1/41 δ

1/22 λ

1/42,1X logX + λ

1/32,i

(δ1

λ2,1

)1/2

∆1/62 X + ΞiX,

provided that ρ 1.

Theorem 4.2. Let X be an interval satisfying (4.2.1) and δ1, δ2 ∈ R satisfying (4.1.5).

Consider two functions f1, f2 ∈ C4(X ) satisfying (4.1.2) and (4.1.4). Assume the

general notation that has been establish in the introduction along with

(4.2.5) α := α0 + α1, Y := max

(λ1,1

λ1,2

,λ1,2

λ1,1

, λ1,1,1

λ1,1

, λ1,2,1

λ1,2

)which is specic to the statement of this theorem. Assume also that (4.1.9) holds and

that q0 ≥ 1 is the integer dened in the proof of Lemma 4.16 below. Let S be dened

in (4.1.6). Then for i = 1 and 2 we have

S ∆1/62 X + (α0∆1)1/3X + (α0λ2,1)1/3X + (α0λ2,2)1/3X + α

1/20 αX + 1

+ min

(α

1/30 αX,

(δ2

λ0,2

)1/5

α1/50 αρ3/5M + min

((δ2

λ0,2

)2/15

α1/50 αρ3/5M,

α1/30 αX

Y 2/3

))

+(∆1/31 + λ

1/32,1 + λ

1/32,2 + α)

((δ2

λ2,2

)1/2

+1

Y

(δ2

λ3,2

)1/3)

+ δiX +1

q0

(δiλ2,i

)1/2

.

132

Theorem 4.3. Let f1, f2 ∈ C2(X ) be two functions satisfying (4.1.1), (4.1.5), (4.1.9)

and the hypothesis

(4.2.6)C2

4δ1

λ1,1

≤ 1.

Then,

S λ1,2X + λ1/32,1 δ2X +

λ1/32,1 δ2

λ1,2

+δ1δ2

λ1,2

+ 1 + (Mλ1,2δ1)1/2 +Mδ1δ2L+1

q0

(δ1

λ2,1

)1/2

.

Theorem 4.4. Let f1, f2 ∈ C1(X ) be two functions satisfying

(4.2.7)λ1,j

C2j

≤ |f (1)j (x)| ≤ C2

j λ1,j, (x ∈ X , j ∈ 1, 2) .

Assume also that (4.1.5) and (4.1.9) holds. Then,

S λ1X + δ1λ1

λ1,1

X + δ2λ1λ1,1

λ1,2

X + δ1δ2λ1

λ1,2

X +δ1

λ1,1

+δ2λ1,1

λ1,2

+δ1δ2

λ1,2

+ 1.

Proof. We follow the idea that the structure of solutions in S is a sequence of intervals

that are intersection of intervals in which f1(x) and f2(x) are close to an integer. From

Lemma 4.10 below, with n = 1, and (4.2.7) we know that an interval in the variable x

in which |f2(x)| ≤ δ2 is of length at most 2C22δ2λ1,2

. Also the distance in x between the

beginning of two consecutive intervals such that |f1(x)| ≤ δ1 is at least 1C2

1λ1,1from the

mean value theorem and (4.2.7). Thus there are at most

2C21C

22δ2λ1,1

λ1,2

+ 2

intervals in which f1(x) is constant for any given interval in which f2(x) is constant.

Each such interval yields at most 2C21δ1

λ1,1+ 1 solutions. There are at most C2λ1X + 2

such intervals, from which we deduce that

S ≤ (C2λ1X + 2)

(2C2

1C22δ2λ1,1

λ1,2

+ 2

)(2C2

1δ1

λ1,1

+ 1

).

4.3 Preparatory lemmas

Lemma 4.1. Let I be an interval. Assume that g1(x) = 1 and that g2(x), . . . , gn(x) in

C1(I). Then, given a1 < a2 < · · · < ar distinct points in I, there are points b1, . . . , br−1

with a1 < b1 < b2 < · · · < br−1 < ar for which

D(g1(x), . . . , gr(x); a1, . . . , ar)

V (a1, . . . , ar)=D(g′2(x), . . . , g′r(x); b1, . . . , br−1)

(r − 1)!V (b1, . . . , br−1).

133

Proof. This is Lemma A2 in [5].

Lemma 4.2. Let h(x) ∈ C2(X ) stand for f1(x) or f2(x) and assume that (4.1.1).

Then, for any two xed real numbers x1, x2 in X with x1 < x2, the real ξ for which

(4.3.1) h(x2)− h(x1) = (x2 − x1)h′(ξ)

satises

ξ =x1 + x2

2+ θ

C7|x2 − x1|2

2Mfor some |θ| ≤ 1 where C stands for C1 or C2 depending on which function is h (only

in this lemma).

Proof. From equation

h(x2)− h(x1) = (x2 − x1)h′(x1) + (x2 − x1)2h′′(η)

2

and (4.3.1), we deduce that

(x2 − x1)h′′(η)

2= h′(ξ)− h′(x1) = (ξ − x1)h′′(ν),

so that2(ξ − x1)

x2 − x1

=h′′(η)

h′′(ν)= 1 +

(η − ν)h′′′(ζ)

h′′(ν)= 1 + θ

C7|x2 − x1|M

.

The result then follows immediately.

Lemma 4.3. Assume that x1 ≤ x2 ≤ x3 ≤ x4 are real numbers. Then,

(4.3.2) V (x1, x2, x3) ≤ (x3 − x1)3

4,

(4.3.3) V (x1, x2, x3, x4) ≤ (x4 − x1)6

55/2

and

(4.3.4) V (x1, x2, x3) + V (x1, x2, x4) + V (x1, x3, x4) + V (x2, x3, x4) ≤ 16(x4 − x1)3

27.

Proof. We use the well-known formula

V (x1, . . . , xn) =∏

1≤i<j≤n

(xj − xi).

To prove (4.3.3), we write x1 := x, x2 := x+tl, x3 := x+ul, x4 := x+l where l := x4−x1

and 0 ≤ t ≤ u ≤ 1. The result then follows from the easily established inequality

tu(u− t)(1− t)(1− u) ≤ 1

55/2.

Inequalities (4.3.2) and (4.3.4) are proved in a similar manner.

134

4.3.1 Minor arcs

Lemma 4.4. Assume that f1, f2 ∈ C3(X ) satisfy (4.1.1) and also that (4.1.4) holds.

Assume also that x1 < x2 < x3 < x4 are in S and that D(x1, x2, x3, x4) 6= 0, where

D(x1, x2, x3, x4) :=

∣∣∣∣∣∣∣∣∣∣1 x1 f1(x1) f2(x1)

1 x2 f1(x2) f2(x2)

1 x3 f1(x3) f2(x3)

1 x4 f1(x4) f2(x4)

∣∣∣∣∣∣∣∣∣∣.

Then,

(4.3.5) |x4 − x1| ≥ min

((2 · 55/2

C153 C

44∆2

)1/6

,

(9

8α2

)1/3

,1

24δ1δ2

).

Proof. Without any lost in generality, we assume that C3 = C1 (otherwise we permute

the last two columns in D). We write fi(xj) = fi(xj) + δi,j for each i = 1, 2 and

j = 1, 2, 3, 4, so that we have D(x1, x2, x3, x4) = D1 +D2 +D3 +D4 where

D1 :=

∣∣∣∣∣∣∣∣∣∣1 x1 f1(x1) f2(x1)

1 x2 f1(x2) f2(x2)

1 x3 f1(x3) f2(x3)

1 x4 f1(x4) f2(x4)

∣∣∣∣∣∣∣∣∣∣, D2 :=

∣∣∣∣∣∣∣∣∣∣1 x1 δ1,1 f2(x1)

1 x2 δ1,2 f2(x2)

1 x3 δ1,3 f2(x3)

1 x4 δ1,4 f2(x4)

∣∣∣∣∣∣∣∣∣∣,

D3 :=

∣∣∣∣∣∣∣∣∣∣1 x1 f1(x1) δ2,1

1 x2 f1(x2) δ2,2

1 x3 f1(x3) δ2,3

1 x4 f1(x4) δ2,4

∣∣∣∣∣∣∣∣∣∣, D4 :=

∣∣∣∣∣∣∣∣∣∣1 x1 δ1,1 δ2,1

1 x2 δ1,2 δ2,2

1 x3 δ1,3 δ2,3

1 x4 δ1,4 δ2,4

∣∣∣∣∣∣∣∣∣∣.

We need an upper bound for each of these determinants. By using Lemma 4.1, we have

D1

V (x1, x2, x3, x4)=

1

6V (ξ1, ξ2, ξ3)

∣∣∣∣∣∣∣1 f ′1(ξ1) f ′2(ξ1)

1 f ′1(ξ2) f ′2(ξ2)

1 f ′1(ξ3) f ′2(ξ3)

∣∣∣∣∣∣∣=

1

12V (η1, η2)

∣∣∣∣∣f ′′1 (η1) f ′′2 (η1)

f ′′1 (η2) f ′′2 (η2)

∣∣∣∣∣(4.3.6)

=f ′′1 (η1)f ′′1 (η2)∆2(ν)

12f ′′1 (ν)2,

so that using (4.1.1), (4.1.4) and (4.3.3), we get |D1| ≤ C121 C3

3C44∆2

6·55/2 |x4 − x1|6. For

D2, we expand the determinant with the third column and use Lemma 4.1 for each

135

of the four determinants that arise and then we use (4.3.4). In the end, we obtain

|D2| ≤ 827δ1C

32λ2,2|x4 − x1|3 and similarly we get |D3| ≤ 8

27δ2C

31λ2,1|x4 − x1|3. For D4,

we subtract the third row from the fourth one, the second from the third and the rst

from the second and then we expand from the rst column. The computation leads to

|D4| ≤ 8δ1δ2|x4 − x1|.

Now, by hypothesis, we have 1 ≤ |D(x1, x2, x3, x4)| ≤ |D1| + |D2| + |D3| + |D4|. This

implies that max(|D1|, |D2|+ |D3|, |D4|) ≥ 13and the conclusion follows.

An ordered set of four points x1, x2, x3, x4 of S for which D(x1, x2, x3, x4) 6= 0 is called

minor arc.

Lemma 4.5. Assume that f1, f2 ∈ C3(X ) satisfy (4.1.1) and also that (4.1.4) holds.

Assume also that x1, x2, x3, x4 is a set for which D(x1, x2, x3, x4) 6= 0. Let s1 + 1,

resp. s2 + 1, be the number of lattice points of Z3 on the line `1 between f(x1) and

f(x2), resp. on the line `2 between f(x3) and f(x4). Write l1 := x2 − x1, l2 := x4 − x3

and K := x4 − x1. Then, the inequality

(4.3.7) s1s2 ≤ max

(C15

3 C44

2∆2K

4l1l2, 3α2K2(l1 + l2), 24δ1δ2K

)holds.

Proof. The proof follows along the lines of that of Lemma 4.4. We expand D(x1, x2, x3, x4)

in a similar way. For D1 we use (4.3.6) along with the upper bound

|V (x1, x2, x3, x4)| ≤ K4l1l2.

For D2 and D3 we proceed in the same manner, but we use K to estimate xj − xi, aslong as it is not l1 or l2, in the Vandermonde determinants that one needs when using

Lemma 4.1. For D4, we do the same. To conclude, we use the fact that the hypothesis

implies that |D(x1, x2, x3, x4)| ≥ s1s2.

4.3.2 Relationship between f and a general plane

Lemma 4.6. Assume that f1, f2 ∈ C3(X ) satisfy (4.1.1) and that (4.1.4) holds. Then,

each set of values x1 < x2 < x3 < x4 are such that the points f(x1), f(x2), f(x3), f(x4)

are not in the same plane. Also, for a xed plane π, there are at most 7 intervals I

such that for all x ∈ I we have f(x) ∈ Λ ∩ π.

136

Proof. Let D(x1, x2, x3, x4) be the analogue of D(x1, x2, x3, x4) formed with f instead

of f . Then, as in the proof of Lemma 4.4, we have from (4.3.6)

D(x1, x2, x3, x4)

V (x1, x2, x3, x4)=f ′′1 (η1)f ′′1 (η2)∆2(ν)

12f ′′1 (ν)2.

From (4.1.1) and (4.1.4), it follows that D(x1, x2, x3, x4) 6= 0 and the four lattice points

f(x1), f(x2), f(x3), f(x4) are not in the same plane.

Now, for the second part, we assume for a contradiction that there are 8 connected

intervals I in x such that f intersects Λ ∩ π. Consider the four functions f(x) +

ε1(0, δ1, 0) + ε2(0, 0, δ2) with ε1, ε2 ∈ −1, 1. It is clear that these four functions sharethe property with f of not having four points in the same plane. Also, these four

functions are the four corners of the region Λ (dened in (4.1.7)). But each such I thatdoes not contain the point f(X1) or f(X2), where X =: [X1, X2], has an entry point

in the region Λ and an exit point. This implies that the plane crosses one of the four

functions each time. So, with 8 we nd at least 14 cross points. Then there must be a

function with at least 4 points in the same plane. This is a contradiction.

Lemma 4.7. Assume that f1, f2 ∈ C3(X ) satisfy (4.1.1) and that (4.1.4) holds. Let

a, b, c, d be real numbers and consider the function g : X → R dened by

(4.3.8) g(x) = ax+ bf1(x) + cf2(x)− d

and set

(4.3.9) ε := |b|δ1 + |c|δ2.

Then, there are at most 3 intervals I such that for all x ∈ I the inequality |g(x)| ≤ ε

holds.

Proof. If b = c = 0, then the result is easily established since g is a line. If b or c = 0,

then hypothesis (4.1.1) implies that g′′(x) = 0 has no solution. If bc 6= 0, then

g′′(z) = 0 ⇐⇒ f ′′1 (z)

f ′′2 (z)=−cb

and since the function f ′′1 (x)

f ′′2 (x)is monotone by (4.1.1) and (4.1.4), we deduce that there

is at most one such solution z. It follows that g′(x) = 0 has at most two solutions.

Now, since between any two intervals such that |g(x)| ≤ ε there is at least one zero of

the derivative of g, we conclude that there are at most three such intervals.

137

Lemma 4.8. Assume that f1, f2 ∈ C4(X ) satisfy (4.1.2) and (4.1.4). Consider the

function g dened in (4.3.8). Assume that (4.2.1) holds. Then, for all x ∈ X , we have

(4.3.10)

|g′′(x)| ≥ 1

16max

(|b|λ2,1

C31C

73C

84

,|c|λ2,2

C32C

73C

84

)or |g′′′(x)| ≥ 1

16max

(|b|λ3,1

C31C

73C

84

,|c|λ3,2

C32C

73C

84

).

Proof. Let x ∈ X be xed and consider the matrix

H(x) :=

(f ′′1 (x) f ′′2 (x)

f ′′′1 (x) f ′′′2 (x)

).

We have

H(x)−1 =1

∆2(x)

(f ′′′2 (x) −f ′′2 (x)

−f ′′′1 (x) f ′′1 (x)

)

and since

(g′′(x)

g′′′(x)

)= H(x)

(b

c

), we deduce that

(b

c

)=

1

∆2(x)

(f ′′′2 (x)g′′(x)− f ′′2 (x)g′′′(x)

−f ′′′1 (x)g′′(x) + f ′′1 (x)g′′′(x)

).

Then we must havemax

(∣∣∣∣f ′′′2 (x)g′′(x)

∆2(x)

∣∣∣∣ , ∣∣∣∣f ′′2 (x)g′′′(x)

∆2(x)

∣∣∣∣) ≥ |b|2max

(∣∣∣∣f ′′′1 (x)g′′(x)

∆2(x)

∣∣∣∣ , ∣∣∣∣f ′′1 (x)g′′′(x)

∆2(x)

∣∣∣∣) ≥ |c|2 ,so that, using (4.1.2) and (4.1.4), we deduce that

(4.3.11)

|g′′(x)| ≥ 1

4min

(|b|λ2,1

C42C

33C

44

,|c|λ2,2

C41C

33C

44

)or |g′′′(x)| ≥ 1

4min

(|b|λ3,1

C32C

33C

44

,|c|λ3,2

C31C

33C

44

).

From this, we prove that

(4.3.12)

|g′′(x)| ≥ 1

8max

(|b|λ2,1

C31C

73C

84

,|c|λ2,2

C32C

73C

84

)or |g′′′(x)| ≥ 1

8max

(|b|λ3,1

C31C

73C

84

,|c|λ3,2

C32C

73C

84

).

Indeed, assume that we are in the rst case of (4.3.11) and that we want to show the

last part of the rst inequality in (4.3.12). First assume that

|c|λ2,2

4C41C

33C

44

<|b|λ2,1

4C42C

33C

44

138

in which case the corresponding inequality in (4.3.11) has already been proved to be

true and the latter in (4.3.12) is also true since C4 ≥ C1. Otherwise we have

(4.3.13)|b|λ2,1

4C42C

33C

44

<|c|λ2,2

4C41C

33C

44

.

Assume, as a rst possibility, that the inequality

|g′′(x)| ≥ 1

2max (|bf ′′1 (x)|, |cf ′′2 (x)|) =

1

2|cf ′′2 (x)|

does not hold (if it holds, the result follows from (4.1.2)). Then,

|g′′(x)| = |cf ′′2 (x)| − |bf ′′1 (x)|,

so that |bf ′′1 (x)| > 12|cf ′′2 (x)| and then, again by (4.1.2), |b|C3

1λ2,1 > |c|λ2,2

2C32

so that

|b|λ2,1 >|c|λ2,2

2C31C

32and the result follows by inserting this last inequality in (4.3.13). The

second possibility is that |bf ′′1 (x)| ≥ |cf ′′2 (x)| ≥ 12|cf ′′2 (x)| which is the same as considered

previously. The other cases are treated similarly.

Now, we let x be the middle point in X . It xes the case, but since X ≤ M16C7

3C164

from

(4.2.1), we have for example

|g′′(x+ h)− g′′(x)| ≤ |b||f ′′1 (x+ h)− f ′′1 (x)|+ |c||f ′′2 (x+ h)− f ′′2 (x)|

= |b||hf ′′′1 (ξ1)|+ |c||hf ′′′2 (ξ2)|

≤ 2 max

(|b|λ2,1

32C73C

124

,|c|λ2,2

32C73C

124

)≤ 1

16max

(|b|λ2,1

C31C

73C

84

,|c|λ2,2

C32C

72C

84

),

for each x+h ∈ X , in the rst case. The other case is similar. Using this with (4.3.12)

proves that (4.3.10) holds for all x ∈ X .

4.4 Linear major arcs

In what follows, by a linear major arc we mean an ordered set x1, . . . , xj of j ≥ 3

consecutive points of S for which the elements of the set f(x1), . . . , f(xj) are all onthe same line.

Lemma 4.9. Assume that x1 < x2 < x3 are on a linear major arc. Then (x1, fi(x1)),

(x2, fi(x2)), (x3, fi(x3)) are on a linear equation for both i = 1 and 2.

139

Proof. By hypothesis, we have that f(x1) =: (x1,1, x1,2, x1,3), f(x2) =: (x2,1, x2,2, x2,3)

and f(x3) are on the line

f(x1) + t(f(x2)− f(x1)) = (x1,1 + t(x2,1−x1,1), x1,2 + t(x2,2−x1,2), x1,3 + t(x2,3−x1,3)).

Now, x2,1 − x1,1 = x2 − x1 > 0 and then by setting x = x1,1 + t(x2,1 − x1,1) we obtain(x,

(x2,2 − x1,2)x+ x1,2x2,1 − x1,1x2,2

x2,1 − x1,1

,(x2,3 − x1,3)x+ x1,3x2,1 − x1,1x2,3

x2,1 − x1,1

)and the result follows.

Lemma 4.9 tells us that on a linear major arc each function fi are approximated linear

function as well. This is exactly a situation that has been studied in the papers [4], [7]

and [8]. In our case, we have a linear major arc A of equation y =a′1x+b′1q1

= a1x+b1q

with

(a′1, q1) = 1 for f1 and of equation y =a′2x+b′2q2

= a2x+b2q

with (a′2, q2) = 1 for f2 where

q = [q1, q2]. The denominator of A is then q. We also says that a′1q1

is the gradient of Afor f1 and a′2

q2the gradient of A for f2. The length, noted l, is l = xj − x1 and we also

write A := [x1, xj].

Remark 4.1. We know from Lemma 3 of [7] that a single function is approximated

in at most 2 by the same linear equation and we deduce from Lemma 4.9 that a line

` intersects Λ in at most three intervals. Indeed, there are at most two separations

between intervals that contain a linear major arc.

A linear major arc x1, . . . , xj, on the line `, that is entirely included in one of the (at

most) three intervals, as explained in Remark 4.1, is called a proper linear major arc.

Lemma 4.10. Let h ∈ Cn(X ) and assume that the inequality |h(x)| ≤ ε holds on an

interval I ⊆ X of length l and that h(n)(x) 6= 0 for x ∈ I. Then, we have

l ≤ 2n

(ε

|h(n)(ξ)|

)1/n

for some ξ in I.

Proof. This is simply a restatement of Lemma 4 of [7].

Remark 4.2. It follows from Lemma 4.10 (with n = 2) and (4.1.9) that the length of

any proper linear major arc is bounded by L0, where

(4.4.1) L0 :=

(16C3

1δ1

λ2,1

)1/2

.

140

Lemma 4.11. Let h ∈ Cn(X ) and assume that the inequality |h(x)| ≤ ε holds on an

interval I ⊆ X of length l. Let x0 be such that |h(x0)| ≥ σ for some σ > ε. Le K be

the distance from x0 to the farthest point in I. Then, if n ≥ 2, we have

(4.4.2) K ≥ min

((σn!

2(2n)n

)1/(n−1)l

ε1/(n−1),

(σn!

2

)1/n1

|h(n)(ξ)|1/n

)

for some ξ ∈ I. For n = 1 we have

(4.4.3) K ≥ σ − ε|h(1)(ξ)|

for some ξ ∈ I.

Proof. This is a restatement of the last part of Lemma 19 of [4]. We will provide a

proof for this particular generalization. Starting with the case n ≥ 2, we will assume

that x0 < x whenever x ∈ I, the other case being similar. Let x′ be the point in I

which is the closest of x0. We consider the n values xi := x′ + iln(i = 1, . . . , n). Then,

using Lemma 1 of [7], there exists a point ξ such that

h(n)(ξ)

n!=

n∑i=0

h(xi)∏j 6=i(xi − xj)

.

Hence, using the inequality |x0 − xj| ≥ jKn, we have∣∣∣∣h(n)(ξ)

n!

∣∣∣∣ ≥ σ∏j 6=0 |x0 − xj|

− εn∑i=1

1∏j 6=i |xi − xj|

≥ σ

Kn− εnn

Kln−1

n∑i=1

1∏j 6=i |i− j|

≥ σ

Kn− ε(2n)n

n!Kln−1.

From this, we deduce that we either have∣∣∣∣h(n)(ξ)

n!

∣∣∣∣ ≥ σ

2Knor

ε(2n)n

n!Kln−1≥ σ

2Kn,

and (4.4.2) follows. The inequality (4.4.3) follows from

|Kh(1)(ξ)| ≥ |(x′ − x0)h(1)(ξ)| = |h(x′)− h(x0)| ≥ σ − ε,

where we used the mean value theorem.

141

Lemma 4.12. Assume that f1, f2 ∈ C2(X ) satisfy (4.1.1) and that (4.1.9) holds. Con-

sider a sequence of proper linear major arcs Aj for j = 1, . . . , J , all with the same

gradient for both f1 and f2. Assume that the sequence Aj is in order of appearance on

the real line (in the variable x) and that their denominator q and their length lj satisfy

q ∈ [Q, 2Q) and lj ∈ [L, 2L) for each j = 1, . . . , J . Assume also that each consecutive

pairs (b1,j, b2,j), (b1,j+1, b2,j+1) (j = 1, . . . , J − 1) are distinct in both entries. If

(4.4.4) Q ≤ 1

34δ,

then we either have

(4.4.5) J ≤ 5

4C6

4 or J ≤ 272C3

1δ1δQ

λ2,1L2.

If

(4.4.6) Q ≥ 1

34δ,

then we either have

(4.4.7) J ≤ 5

4C6

4 or J ≤ 8C3

1δ1

λ2,1L2.

Proof. We begin with the proof of the rst assertion under the hypothesis (4.4.4). We

use Lemma 4.11 with n = 2 and h = gi,j where

gi,j(x) := aix− qfi(x) + bi,j.

Let Ki,j be the quantity dened in Lemma 4.11 for j = 1, . . . , J − 1 where the x0 is the

rst point in Aj+1, εi := 2Qδi and σi := 1− εi. We have

(4.4.8) Ki,j ≥ min

(σiL

32Qδi,

σ1/2i

(2QC3i λ2,i)1/2

).

There are two possibilities for each xed i: we either have the rst or the second lower

bound valid for each j in (4.4.8). Call this lower bound Ki and set K := max(K1, K2).

Now, by using twice the Cauchy mean value Theorem and the fact that in a proper

linear major arc we have |gi,j(x)| ≤ qδi (i = 1, 2), we deduce that

(4.4.9) |f ′′i (η)| =∣∣∣∣fi(x1 + L′ +D)− fi(x1 +D)− fi(x1 + L′) + fi(x1)

L′D

∣∣∣∣ ≤ 4δiLD

142

where x1 is the rst point in A1, D is the distance from x1 to the rst point in AJ and

L′ = min(L1, Lj). We deduce from (4.4.9) the inequality

(4.4.10) J ≤ 4C3i δi

λ2,iLK+ 1

for both values of i.

We are now ready for the argument. We rst assume that K is the second term in

(4.4.8) for i0, in which case, we deduce that

L ≥ 32Qδi0(2σC3

i0Qλ2,i0)1/2

by comparing both terms in (4.4.8). By using these inequalities in (4.4.10), we nd

J ≤4C3

i0δi0(2C3

i0Qλ2,i0)1/2

λ2i0Lσ1/2

+ 1 ≤4C3

i0δi0(2C3

i0Qλ2,i0)

λ2,i032Qδi0+ 1

=C6i0

4+ 1 ≤ 5C6

4

4.

This gives the rst possibility in (4.4.5). We now assume that the estimate 5C64

4does

not hold, so that J ≥ 2 and K is the rst term in (4.4.8) for both values of i. In this

case, by choosing i1 so that δi1 = δ, we have directly

J ≤ 4C31δ132Qδi1λ2,1L2σi1

+ 1 ≤ 136C31δ1δQ

λ2,1L2+ 1,

where we have used the inequality σi1 ≥ 1617

that follows from (4.4.4). The result follows

from the fact that the rst term has to be larger that 1. Observe that we motivated

our choices by (4.1.9) and (4.1.8).

Now we proceed to the proof of the second assertion under the hypothesis (4.4.6).

We rst observe that the inequality for the space between the two rst elements of

consecutive linear major arcs cannot be less than L. Using this in (4.4.10), we obtain

that

J ≤ 4C31δ1

λ2,1L2+ 1.

Now, if J ≤ 5C64

4does not hold, then we have J ≥ 2. Thus the result follows.

Lemma 4.13. Assume that f1, f2 ∈ C2(X ) satisfy (4.1.1) and that (4.1.9) holds. Con-

sider a sequence of proper linear major arcs Aj for j = 1, . . . , J , all with the same

gradient for both f1 and f2. Assume that the sequence Aj is in order of appearance on

143

the real line (in the variable x) and that their denominator q and their length lj satisfy

q ∈ [Q, 2Q) and lj ∈ [L, 2L) for each j = 1, . . . , J . Assume also that each consecutive

pairs (b1,j, b2,j), (b1,j+1, b2,j+1) (j = 1, . . . , J − 1) are distinct, not necessarily in both

entries. If

(4.4.11) Q ≤ 1

34δ0

,

then we either have

(4.4.12) J ≤ 5

4C6

4 or J ≤ 272C3

1δ1δ0Q

λ2,1L2.

If

(4.4.13) Q ≥ 1

34δ0

,

then we either have

(4.4.14) J ≤ 5

4C6

4 or J ≤ 8C3

1δ1

λ2,1L2.

Proof. The proof is similar to the proof of Lemma 4.12. The only dierence is that

since two consecutive proper linear major arcs may have only one entry for which the

equation diers, we deduce that we can use only one of the two estimates in (4.4.8) in

general. The simplest way to determine which one is by saying that

(4.4.15) K ≥ min

(σL

32Qδ0

,σ1/2

(2QC31λ2,1)1/2

,σ1/2

(2QC32λ2,2)1/2

)for each Aj (j = 1, . . . , J). From there, if the minimum is achieved by one of the last

two terms, then we get the rst result in (4.4.12) by working with the corresponding

function from (4.4.9). Also, if the minimum is achieved by the rst term, then we get

the last term of (4.4.12).

Remark 4.3. Suppose that we have a sequence of parallel lines `j that are all in the

same plane π. We verify that Lemma 4.12 can be used as long as the equation of

the plane π has bc 6= 0 (see (4.3.8) above and (4.5.1) below). If it is not the case,

then at least we know that the corresponding function g (see (4.3.8)) satises the rst

possibility in Lemma 4.8. Also, we know that we cannot have b = c = 0 since otherwise

the equation of π would x x which is a nonsense.

Lemma 4.14. Let fi ∈ C2(X ) be one of the two functions of section 4.1 that satises

(4.1.1). Assume that we have a sequence of proper linear major arcs Aj for j = 1, . . . , J ,

144

all with the same gradient. Assume that the sequence Aj is in order of appearance on

the real line (in the variable x) and that their denominator q (as dened in [7]) and

their length lj satisfy q ∈ [Q, 2Q) and lj ∈ [L, 2L) for each j = 1, . . . , J . Assume also

that for each j = 1, . . . , J − 1, there exists a point of S between Aj and Aj+1 that is

not on the equation of Aj for fi. If

(4.4.16) Q ≤ 1

34δi,

then we either have

(4.4.17) J ≤ 5

4C6i or J ≤ 272

C3i δ

2iQ

λ2,iL2.

If

(4.4.18) Q ≥ 1

34δi,

then we either have

(4.4.19) J ≤ 5

4C6i or J ≤ 8

C3i δi

λ2,iL2.

Proof. First assume that (4.4.16) holds. For a xed Aj, we consider the function

gj(x) := ax− qfi(x) + bj.

By Lemma 4.11 with n = 2 we get

(4.4.20) K ≥ min

(σL

32Qδi,

σ1/2

(2QC3i λ2,i)1/2

).

The remainder of the proof is similar to the proof of Lemma 4.12, but only the infor-

mation on the function fi is used at each stage.

Lemma 4.15. Let fi ∈ C1(X ) be one of the two functions of section 4.1 that satises

(4.2.7). We assume that we have a family of proper linear major arcs, all of length l in

[L, 2L) and of denominator q in [Q, 2Q). We assume that their gradients are pairwise

distinct and nonzero. Then the total contribution of these arcs is at most

≤ 12C2iMλ2,iLQ.

Proof. Using Lemma 1 of [7] with n = 1, we deduce that for a xed linear major arc Aof gradient a

qwe have

(4.4.21)

∣∣∣∣f ′i(ξ)− a

q

∣∣∣∣ ≤ 2δiq

145

for some ξ in A. We deduce that, for q ∈ [Q, 2Q) and using (4.1.5) and (4.1.1), we

have 1 ≤ |a| ≤ 4C2i λ1,iQ with the sign determined by the sign of f ′i on X . So, since the

integer points on A are in arithmetic progression, we deduce that the largest possible

contribution is given by∑a/q

3L

Q≤ 12C2

i λ1,iL∑

q∈[Q,2Q)

1 = 12C2i λ1,iLQ

and the result follows.

Remark 4.4. The hypothesis (4.2.3) implies that there are no linear major arcs with

gradient 0 in one of the coordinates. Indeed, in such an interval we would have

fi(x+ 2)− fi(x) < 2δi

for i = 1 or 2. This is in contradiction with (4.2.3) from hypothesis (4.1.9). In other

words, Lemma 4.15 is not restrictive for Theorem 4.1.

Lemma 4.16. Let fi ∈ C2(X ) be one of the two functions f1 and f2 introduced in

section 4.1 that satises (4.1.1). Then, there exists a subset S ′i ⊆ S such that any three

consecutive points in S ′i are not aligned, and there exists another subset R ⊆ S such

that

(4.4.22) #S ≤ 3 ·#S ′i + #R

and

#R ≤ 91C12i δiX +

9C6i

2maxA

s(4.4.23)

≤ 91C12i δiX +

18C6i

q0

(C3i δiλ2,i

)1/2

holds. We have dened q0 as the smallest denominator of any proper linear major arc

in S (see the proof for details).

Proof. We follow the proof of Lemma 2 in [8]. Consider the sets

Si := n ∈ X ∩ N| ‖fi(n)‖ < δi

and

(4.4.24) Λi := (v1, v2) ∈ R2| ∃x ∈ X s.t. v1 = x, |v2 − fi(x)| < δi.

146

Clearly, S ⊆ Si. For a xed line `, we know from Lemma 3 of [7] that ` ∩ Λi is made

of at most two intervals (in the variable x) that may contain points of Si. In this proof

only, by proper major arcs we mean a sequence of consecutive points x1 < · · · < xj

(j ≥ 3) of Si for which, if we write f(x) := (x, fi(x)) and f(x) := (x, fi(x)), the set of

points f(x1), . . . , f(xj) is included in a line ` and [x1, xj] is completely included in

one of the at most two previously dened intervals. Our aim is to transform S into a

set S ′i that does not contain three consecutive points on a line and such that

(4.4.25) |S| ≤ 3|S ′i|+ |U∗|,

where U is the set of points of S that are in some proper major arcs and where U∗ :=

U\S ′i. The basic idea is simple. We assume that S is composed of x1 < · · · < xJ .

Just by considering the points S as member of Si, there is a line `1 generated by

f(xr0) := f(x1) and f(x2). This line may contain f(x3) and other points, but there is

an indice r1 such that f(xr1) is on `1 but not f(xr1+1). Then, there exists another line

`2 generated by f(xr1) and f(xr1+1). This new line contains consecutive points up to

f(xr2), and so on. We have thus constructed a sequence xr0 < xr1 < · · · < xj =: xrw

and this is the set S ′i. We see from Lemma 4.9 that S ′i does not have three consecutivepoints on a line. Now, inspired by Lemma 3 of [7] and by the desire to work with proper

major arcs, we multiply |S ′i| by 3. This operation would count, for each j = 0, . . . , w−1,

two points between xrj and xrj+1, xrj+1 and xrj+2 if they exist. These two points are

enough to estimate the contribution of the (at most) two segments that miss a point

to be a proper linear major arc. Also, for a given proper major arc of length l and

denominator q (see [7] here and throughout the proof) it will be enough to count the

contribution of s = lqinstead of s + 1 (by construction). The formula (4.4.25) is thus

proved.

For a xed proper major arc of length l and denominator q, we consider the function

g(x) := ax− qfi(x) + b.

By applying Lemma 4.11 with n = 2, ε = qδi, σ := 1− qδi and (4.1.1), we get

(4.4.26) K ≥ min

(σl

16qδi,

σ1/2

(C3i qλ2,i)1/2

).

We are thus lead to examine 3 cases. These are

(1) q ≥ 1

17δi;

(2) q <1

17δiand

σl

16qδi<

σ1/2

(C3i qλ2,i)1/2

;

147

(3) q <1

17δiand

σl

16qδi≥ σ1/2

(C3i qλ2,i)1/2

.

In case (1), the contribution is clearly s = lq≤ 17δil and by summing over all the

corresponding proper major arcs, we get a total contribution of at most 17δiX. In

this case only, there is no need to multiply by 2. In the two last cases, we count the

contribution of one proper major arc per line ` and multiply by 2 (we keep the one

with the largest contribution). This is done because if q does not satisfy (1) then a

point xi not on ` is such that |g(xi)| ≥ 1617, so that we know that the line exits Λ once.

That is, it either happens between xrj and xrj+1or it happens anyway between xrj+1

and xrj+1+1, if we encounter ` for the rst time. Also, we may have added some points

of Si in the argument that were not in S in the cases (2) and (3) because we assumed

that (a, q) = 1.

In case (2), we deduce that K ≥ l17qδi

so that we have

l

K≤ 17δiq ⇒

s

K≤ 17δi

and thus the density is 17δi and by adding up all these proper major arcs, we get an

upper bound of 17δiX + maxA s (the last term comes from the last proper major arc).

The case (3) requires a little bit more work. By using the exact same argument as

in Lemma 4.12, we get that the number of dierent proper major arcs of the same

gradient is bounded by 5C6i

4. Also, the contribution of an arc is given by l

q≤ Li

q, where

Li ≤ 4(C3i δiλ2,i

)1/2

from Lemma 4.10 with n = 2. Thus, the idea is to sum over all the

gradients with their multiplicities. For that, we consider (4.4.21), from which we nd

that all the gradients aqare in an interval J of length ≤ C3

iXλ2,i + 4δiq1

where q1 is

the smallest denominator that is greater than or equal to q0 that is itself the smallest

denominator that appears in the list of all the denominators of such proper major arcs.

Now, we can write

(4.4.27)∑aq∈J

(a.q)=1q≤Q0q 6=q0

1

q≤ |J |Q0

(see Lemma 1.2.3 in [6]). We deduce from the argument that leads to (4.4.27) that

q1 ≥ max(

1|J |1/2 , 1

). We also deduce that |J | ≤ 2C3

iXλ2,i. Indeed, if C3iXλ2,i ≥ 4δi

q1

then the result follows, otherwise we would have q1 ≥(q1

8δi

)1/2

which is in contradiction

148

with the rst hypothesis in case (3). We conclude that the total contribution is bounded

by5C6

i

4

(2C3

iXλ2,iQ0Li +l

q0

)≤ 5C6

i

4

(2C3

iXλ2,iQ0Li + maxA

s).

Now we use the inequality Q0 ≤ Li2on the rst term to conclude.

Lemma 4.17. Let fi ∈ C2(X ) be one of the two functions f1 and f2 introduced in

section 4.1 that satises (4.1.1). Assume the notation used in Lemma 4.16. Then,

#Si ≤ 6

(C3i λ2,i

4

)1/3

X + 115C12i δiX + 6 +

9C6i

2maxA

s.

Proof. The idea is to apply the argument in the proof of Lemma 4.16 to the set Sidirectly instead of S. It does not make a notable dierence. The result is that (4.4.22)

holds with Si instead of S. So it remains to estimate #S ′i. In order to do that, for any

x1 < x2 < x3, we consider the determinant

D(x1, x2, x3) :=

∣∣∣∣∣∣∣1 x1 fi(x1)

1 x2 fi(x2)

1 x3 fi(x3)

∣∣∣∣∣∣∣ .Then, we expand it as D(x1, x2, x3) = D1 +D2 where

D1 :=

∣∣∣∣∣∣∣1 x1 fi(x1)

1 x2 fi(x2)

1 x3 fi(x3)

∣∣∣∣∣∣∣ and D2 :=

∣∣∣∣∣∣∣1 x1 δi,1

1 x2 δi,2

1 x3 δi,3

∣∣∣∣∣∣∣ .From the construction of S ′i, we know that three consecutive points f(x1), f(x2) and

f(x3) are not on a line. So that we have |D(x1, x2, x3)| ≥ 1. Now, proceeding as in

Lemma 4.4, we get that

(4.4.28)1

x3 − x1

≤ max

((C3i λ2,i

4

)1/3

, 4δi

).

By writing z0 < z1 < · · · < zt for the sequence of points in S ′i, we deduce that

j

z2j − x0

≤ max

((C3i λ2,i

4

)1/3

, 4δi

).

Using z2j − z0 ≤ X and the fact that #S ′i = t+ 1 ≤ 2j + 2 (t = 2j + 0 or t = 2j + 1),

we obtain that

#S ′i ≤ 2

(C3i λ2,i

4

)1/3

X + 8δiX + 2.

The result follows by inserting this inequality in (4.4.22) and by using (4.4.23).

149

Now, we prove a result very similar to Lemma 12 of [8].

Lemma 4.18. Let f ∈ C2(X ) be any one of the two functions f1 and f2 introduced in

section 4.1 that satises (4.1.1) and (4.1.5) (in this lemma, we write C, δ and λr for

the corresponding Ci, δi and λr,i). Suppose that

(4.4.29)C2δ

λ1

< 1.

Assume that we have a sequence of proper linear major arcs Aj, j = 1, . . . , K, all of

length at most L3. Assume that any two of them are distant by at least d (in the variable

x). Then,

(4.4.30)K∑i=1

maxAi

s ≤ (512C5KMδ)1/2 +16C5Mδ

dQ0L,

where L is dened in (4.6.1) and Q0 = min

(L0

2,(

8KC2λ1

)1/2)

(L0 is dened in (4.4.1)).

Proof. The basic idea is to choose theK gradients aqthat would give the highest possible

densities for these linear major arcs. Hypothesis (4.4.29) ensures us that each equation

are of degree one, see Remark 4.4. In the proof of Lemma 4.15 we have seen the

inequality 1 ≤ |a| ≤ 2C2λ1q, which is valid for the gradient aqof any linear major arc of

degree one. We will be looking at this inequality as an upper bound for the multiplicity

of a given denominator.

We will distinguish three cases depending on the size of K. The rst one is when

4C2Kλ1 ≤ 1. In this case, we choose |q0| ≥ 1 the smallest integer for which 2C2λ1|q0| ≥1. It is then clear from Remark 4.2 that we have

K∑i=1

maxAi

s ≤ KL0

q0

≤ 2C2λ1KL0

≤ (C2λ1K)1/2L0

= (16C5KMδ)1/2.

In the second case, we assume that 2K ≤ C2λ1. We simply write

K∑i=1

maxAi

s ≤ KL0 ≤ KL0

(C2λ1

2K

)1/2

150

= (8C5KMδ)1/2.

In the third case, we assume that we are not in one of the above situations. We choose

q0, . . . , Q where |q0| ≥ 1 is minimal with 2C2λ1q0 ≥ 1 and where Q is chosen so that

K ≤∑aq

1 ≤Q∑

q=q0

2C2λ1q = C2λ1(Q(Q+ 1)− q0(q0 − 1)).

Now, by minimality of q0 and the fact that Q ≥ q0 − 1, it is enough to have

Q2 ≥ 2K

C2λ1

≥ K

C2λ1

+1

4C4λ21

where we have used the inequality 4C2Kλ1 ≤ 1. We deduce, from 2K ≥ C2λ1, that it

will be enough to take Q as the largest integer of size at most

(8K

C2λ1

)1/2

. First of all,

we will evaluate the contribution of all these gradients, assuming that they are taken

only once. We get

K∑i=1

maxAi

s ≤∑aq

L0

q

≤Q∑

q=q0

2C2λ1L0 ≤ 2C2λ1QL0

= (512C5KMδ)1/2.

Now, we still have to take into account the possibility that the same gradient occurs

more than once. By hypothesis, there is a distance of at least d between two proper

linear major arcs. Consider only the contribution of the proper linear major arcs of

length l ∈ [L, 2L) for a xed L. A similar argument that has led to inequality (4.4.10),

and the hypothesis that we have more than one such proper linear major arcs show

that we have at most 4C3δλ2Ld

extra such proper linear major arcs. So that we have

K∑′

i=1

maxAi

s ≤ 4C3δ

λ2Ld

∑aq

2L

q

≤ 8C3δ

λ2d

Q∑q=q0

2C2λ1

≤ 16C5Mδ

dQ0,

151

where ′ means that the sum is taken over the linear major arcs of length l ∈ [L, 2L) and

where Q0 is any upper bound for Q. It can be L3/2, L0/2 or

(8K

C2λ1

)1/2

for example.

The result follows by taking the sum over L.

Remark 4.5. It is not hard to remove the hypothesis about the size of the spacing

between proper linear major arcs in Lemma 4.18. Indeed, a slight modication of the

above argument gives the upper bound (KMδ)1/2 logM instead of (4.4.30). To do

that, we observe that only the last stage needs d. So we use U0 := 4C3δλ2L2 as an upper

bound for the extra multiplicity (with L xed). Then we choose all the proper linear

major arcs of length l ∈ [L, 2L) for which the extra multiplicity is u ∈ [U, 2U) and

remark that there are at most KU

of them. The same ideas and a nal sum over U

and then L completes the proof.

4.5 Major arcs

An ordered set of j ≥ 3 consecutive points x1, . . . , xj of S for which the set

f(x1), . . . , f(xj)

lies into a plane π, not all on the same line, is called a major arc for which we refer to

as A. Also, x1, xj are the edges of A, l := xj − x1 is the length of A and A := [x1, xj].

Each point in A satises the equation of π that we call the equation of A. We adopt

the convention that the equation of A is of the form

(4.5.1) ax+ by + cz = d where a, b, c ∈ Z and (a, b, c) = 1.

The function g dened in (4.3.8) is associated to A. The set of all lattice points (x, y, z)

that satisfy (4.5.1) forms a lattice Γ such that

(4.5.2) |Γ| = (a2 + b2 + c2)1/2.

Often, we will use the inequality

(4.5.3) max(|a|, |b|, |c|) ≤ |Γ| ≤√

3 max(|a|, |b|, |c|).

We also say that Γ is the lattice of A and that π is the plane of A.

We have seen in Lemma 4.6 that a xed plane intersects Λ in at most 7 intervals. A

major arc contained in one of these intervals is called a proper major arc.

152

Lemma 4.19. Assume that f1, f2 ∈ C2(X ) satisfy (4.1.1) and (4.1.5). Assume also

that (4.2.1) holds. Let x1 < x2 ≤ x3 < x4 be four points of S in a proper major arc of

plane π, with equation ax+ by + cz = d and lattice Γ. Set

a1 = x2 − x1, a2 = x4 − x3, b1 =x1 + x2

2, b2 =

x3 + x4

2and K := b2 − b1.

Let s1 + 1, resp. s2 + 1, be the number of lattice points of Γ on the line `1 between f(x1)

and f(x2), resp. on `2 between f(x3) and f(x4). Assume that the lines `1 and `2 are

not parallel. Then, the inequality

(4.5.4)

s1s2|Γ| ≤ a1a2K

(69633

16384C10

3 C34∆1 +

17

8C3

1λ2,1 +17

8C3

2λ2,2

)+ 4(a1 + a2)(α0 + α1)

holds.

Proof. First, we write

v1 := f(x2)− f(x1) = a1(1, f1(ξ1), f2(ξ2)) + (0, 2θ1,1δ1,1, 2θ2,1δ2,1)

v2 := f(x4)− f(x3) = a2(1, f1(η1), f2(η2)) + (0, 2θ1,2δ1,2, 2θ2,2δ2,2).

Now, its is clear that the area of the parallelogram formed by (0, 0, 0), v1, v2 and v1 + v2

is s1s2|Γ|. Also, since the vector (a, b, c) is normal to π, we deduce that

s1s2|Γ| =| det(vt1, v

t2, (a, b, c)

t)|(a2 + b2 + c2)1/2

so that s1s2|Γ|2 = | det(vt1, vt2, (a, b, c)

t)|

from (4.5.2). We expand det(vt1, vt2, (a, b, c)

t) as D1 +D2 +D3 +D4 where

D1 :=

∣∣∣∣∣∣∣a1 a2 a

a1f′1(ξ1) a2f

′1(η1) b

a1f′2(ξ2) a2f

′2(η2) c

∣∣∣∣∣∣∣ , D2 :=

∣∣∣∣∣∣∣0 a2 a

2θ1,1δ1,1 a2f′1(η1) b

2θ2,1δ2,1 a2f′2(η2) c

∣∣∣∣∣∣∣ ,

D3 :=

∣∣∣∣∣∣∣a1 0 a

a1f′1(ξ1) 2θ1,2δ1,2 b

a1f′2(ξ2) 2θ2,2δ2,2 c

∣∣∣∣∣∣∣ , D4 :=

∣∣∣∣∣∣∣0 0 a

2θ1,1δ1,1 2θ1,2δ1,2 b

2θ2,1δ2,1 2θ2,2δ2,2 c

∣∣∣∣∣∣∣ .By expanding D2 with the rst column and using (4.1.1), (4.1.8) and (4.5.3) we get

|D2| ≤ 2a2|Γ|(α0 + α1). Similarly we have |D3| ≤ 2a1|Γ|(α0 + α1) and clearly |D4| ≤8δ1δ2|Γ|. For D1, we use Lemma 4.2 and expand as D1 = a1a2(E1 + E2 + E3 + E4)

where

E1 :=

∣∣∣∣∣∣∣1 1 a

f ′1(b1) f ′1(b2) b

f ′2(b1) f ′2(b2) c

∣∣∣∣∣∣∣ , E2 :=

∣∣∣∣∣∣∣0 1 a

θ′1,1C10

1 λ3,1a21

2f ′1(b2) b

θ′2,1C10

2 λ3,2a21

2f ′2(b2) c

∣∣∣∣∣∣∣ ,

153

E3 :=

∣∣∣∣∣∣∣1 0 a

f ′1(b1) θ′1,2C10

1 λ3,1a22

2b

f ′2(b1) θ′2,2C10

2 λ3,2a22

2c

∣∣∣∣∣∣∣ , E4 :=

∣∣∣∣∣∣∣0 0 a

θ′1,1C10

1 λ3,1a21

2θ′1,2

C101 λ3,1a2

2

2b

θ′2,1C10

2 λ3,2a21

2θ′2,2

C102 λ3,2a2

2

2c

∣∣∣∣∣∣∣ .Now, we expand E1 with the last column and use Lemma 4.1 along with (4.1.1) and

(4.1.4) to get to

|E1| ≤ |Γ|K(2C103 C

34∆1 + C3

1λ2,1 + C32λ2,2).

We expand E2 with the rst column and E3 with the second column to get

|E2|+ |E3| ≤ |Γ|K(

∆1

8+λ2,1

16+λ2,2

16

)using the fact that a2

1+a22

2≤ KX since a1+a2

2≤ K and a1 +a2 ≤ X. We get |E4| ≤ |Γ|K∆1

323

by using a21a

22

2≤ KX3

8. We thus have shown that

s1s2|Γ| ≤ a1a2K

(69633

32768C10

3 C34∆1 +

17

16C3

1λ2,1 +17

16C3

2λ2,2

)+2(a1 + a2)(α0 + α1) + 8δ1δ2

and the result follows from 8δ1δ2 ≤ 12by using (4.1.5).

For any real number x ∈ I, we write f(x) for the closest point to f(x) on π with the

rst coordinate x.

Lemma 4.20. Assume that f1, f2 ∈ C2(X ) satisfy (4.2.7). Assume that we have a

proper major arc A of plane π with A = [x′1, x′j] and equation ax+ by+ cz = d. Assume

also that the points x1, x2, x3 are in A and that the points f(x1), f(x2), f(x3) are not on

a line. Then (a, b, c) = t(a′, b′, c′) for some t ∈ R\0 with

a′ :=f ′1(ξ1)f ′1(ξ2)∆1(η1)

2f ′1(η1)2V (x1, x2, x3) + 2θ1α1|x3 − x1|+ 33/2θ2δ1δ2,

−b′ := f ′′2 (η2)

2V (x1, x2, x3) + 2θ3δ2|x3 − x1|,

c′ :=f ′′1 (η3)

2V (x1, x2, x3) + 2θ4δ1|x3 − x1|.

Proof. A point (x, y, z) is on π if and only if

0 =

∣∣∣∣∣∣∣∣∣∣1 x y z

1 x1 f1(x1) + δ1,1 f2(x1) + δ2,1

1 x2 f1(x2) + δ1,2 f2(x2) + δ2,2

1 x3 f1(x3) + δ1,3 f2(x3) + δ2,3

∣∣∣∣∣∣∣∣∣∣,

154

the development of which yields the equation of A. By unicity, (a, b, c) = t(a′, b′, c′) for

some t ∈ R\0 where

a′ :=

∣∣∣∣∣∣∣1 f1(x1) + δ1,1 f2(x1) + δ2,1

1 f1(x2) + δ1,2 f2(x2) + δ2,2

1 f1(x3) + δ1,3 f2(x3) + δ2,3

∣∣∣∣∣∣∣ , −b′ :=

∣∣∣∣∣∣∣1 x1 f2(x1) + δ2,1

1 x2 f2(x2) + δ2,2

1 x3 f2(x3) + δ2,3

∣∣∣∣∣∣∣and

c′ :=

∣∣∣∣∣∣∣1 x1 f1(x1) + δ1,1

1 x2 f1(x2) + δ1,2

1 x3 f1(x3) + δ1,3

∣∣∣∣∣∣∣ .The result follows by expanding these determinants the same way we did with D in

Lemma 4.4. Note that we have used the inequality of Hadamard for the last term of

a′.

Remark 4.6. The relation found for a′ in Lemma 4.20 could have been stated with the

roles of f2 and f1 reversed.

We say that a proper major arc is short if its length l satises l ≤ L1, where

(4.5.5) L1 :=

(512C7

3C114

δ2

λ2,2

)1/2

.

A proper major arc that is not short is said to be long.

Lemma 4.21. Assume that f1, f2 ∈ C4(X ) satisfy (4.1.2) and (4.1.4). Assume that Ais a long proper major arc of plane π and length l. Then,

(4.5.6) l ≤(

6912C73C

114

δ2

λ3,2

)1/3

:= L2

and the corresponding lattice Γ satises

(4.5.7)

|Γ| ≥ 1

6C103 C

64

max

(2C7

3C34λ1,1

λ1,2

,2C7

3C34λ1,2

λ1,1

, λ1,1,1

λ1,1

, λ1,2,1

λ1,2

, 6C103 C

64

)=: Y.

Proof. We assume throughout that x1 and x3 are respectively the entrance and exit

points (in the variable x) of f in π ∩Λ of the interval that contains A. We assume also

that x2 = x1+x3

2. We rst suppose that the rst possibility in (4.3.10) holds. In this

case, we use Lemma 4.10 with n = 2 and h = g, as dened in (4.3.8), to get

l ≤ 4

(ε

|g′′(ξ)|

)1/2

≤ 4

(|b|δ116C3

1C73C

84

|b|λ2,1

+|c|δ216C3

2C73C

84

|c|λ2,2

)1/2

155

≤(

512C73C

114

δ2

λ2,2

)1/2

,

where we have used (4.1.9) for the last inequality. This is in contradiction with the

fact that A is assumed to be long. So the second possibility of (4.3.10) holds. The

same computation, with n = 3 in Lemma 4.10, leads to (4.5.6). In the eventuality that

(4.5.6) is still no longer than L1, it would simply mean that there are no such proper

major arcs.

To conclude the proof, it suces to show that each of the rst terms in the denitions

of a′, b′ and c′ in Lemma 4.20 are greater than twice the contribution of the other terms.

Also, before we apply Lemma 4.20, we have to show that f(x1), f(x2), f(x3) are not

on a line. This is done by using the structure of linear major arcs (as in Lemma 4.9),

the proof of Lemma 4.10 on the function f2 (which gives the same result as stated in

Lemma 4.10 with n = 2) and the size of L1. Assuming that, and using (4.5.3) and

Lemma 4.20, we can at least say that

|Γ| ≥ |b′||c′|≥ |f

′′2 (η2)|/4

3|f ′′1 (η3)|/4≥ λ1,2

3C31C

32λ1,1

where we have used (4.1.2) in the last inequality. The result (4.5.7) follows by verifying

the ve other possibilities.

Now, to conclude the proof, we will verify our claim with a′ since it is the hardest case.

We write a′ := a1 + θ1a2 + θ2a3 (see Lemma 4.20) and we will show that |a1| ≥ 4|a2|and |a2| ≥ |a3|, so that |a1| ≥ 4|a2| ≥ 2(|a2| + |a3|). Beginning with |a1| ≥ 4|a2|, wehave that

f ′1(ξ1)f ′1(ξ2)∆1(η1)

2f ′1(η1)2V (x1, x2, x3) ≥ ∆1

4C81C

23C

34

(x3 − x1)3

4≥ 8α1(x3 − x1)

since we have

(x3 − x1)2 > L21 = 512C7

3C114

δ2

λ2,2

≥ 128C81C

23C

34

(δ1C

22

λ2,1

+δ2C

21

λ2,2

)where we used (4.1.2), (4.1.4) and Remark 4.6. Then we have |a2| ≥ |a3| since otherwisewe would have

L1 < (x3 − x1) ≤ 33/2

2

δ1δ2

α1

≤ 33/2

2

δ2

C22λ2,2M

from which we deduce from L1 < (x3 − x1) ≤M that

L1 ≤ (L1M)1/2 ≤(

33/2δ2

2C22λ2,2

)1/2

which is false. The two other cases are simpler to verify and the result follows.

156

Lemma 4.22. Assume that f1, f2 ∈ C4(X ) satisfy (4.1.2). Assume the previous nota-

tion (4.3.9) and consider the functions

gt(x) := atx+ bf1(x) + cf2(x)− dt

for t = 1, . . . , T where it is assumed that (b, c) 6= (0, 0) (see Remark 4.3). For each t,

assume that the second possibility of (4.3.10) holds and that there is one interval of the

at most three (as dene in Lemma 4.7) that contains A(= A(t)), that is the interval

that corresponds to the proper major arc A(= A(t)) of equation atx+ by+ cz− dt = 0.

Assume also that only the second possibility of (4.4.2) holds with h = gt, n = 3 and

σ := 1− ε. If

(4.5.8) ε <1

2,

then

(4.5.9) T ≤ 182

27C7

3C154 .

Proof. We observe that the whole family of functions gt share the same second and

third derivative. Because of that, we drop the subscript t for these two functions. Also,

the choice of σ is explained by the fact that if the point x0 ∈ S is not in a major arc Aof equation atx+ by + cz − dt = 0, then we must have |gt(x0)| ≥ 1− ε > 1

2.

The rst step is to get some control on the size of the third derivative of these gtfunctions. We will assume, without losing in generality, that the inequality

(4.5.10) |g′′′(x)| ≥ |b|λ3,1

16C31C

73C

84

≥ |c|λ3,2

16C32C

73C

84

holds in Lemma 4.8. We now establish the inequality

(4.5.11) |g′′′(x)| ≤ 2|b|λ3,1 max

(C4

1 ,C7

2

C31

).

By setting

ϕ :=|c|λ3,2C

42

|b|λ3,1C41

,

we see from (4.5.10) that ϕ ≤ C72

C71and the inequality (4.5.11) follows.

The second step is to establish the lower bound l ≥ l1 where

l1 := ε1/2(

72

σ

)1/2 3σ

2|b|λ3,1 max(C4

1 ,C7

2

C31

)1/3

157

for the length of each proper major arc A that we are dealing with. This is done by

using the hypothesis that only the second inequality of (4.4.2) holds.

Now we are ready for the argument. Using the proof of Lemma 4 of [7], we deduce that

for each of the T proper major arcs A of length l := l(A), there exists ξ := ξ(A) ∈ Asuch that

(4.5.12) |g′′(ξ)| ≤ 16ε

l2≤ 16ε

l21.

From there, we label the ξj in ascending order on the real line with j = 1, . . . , T . If

T ≤ 2 then (4.5.9) holds, if not then the result follows from the inequalities

(T − 2)

(3σ

2|b|λ3,1 max

(C4

1 ,C7

2C3

1

))1/3

|b|λ3,1

16C31C

73C

84≤ (ξT − ξ1)|g′′′(ν)| = |g′′(ξT )− g′′(ξ1)| ≤ 32ε

l21

and the denition of l1. The rst inequality is self-explanatory; it is obtained by

considering each of the T − 2 intern proper major arcs together with the interval

between it and the previous one, which is exactly the same situation as in Lemma 4.11,

and by using (4.5.10). The other inequalities are respectively the mean value theorem

and (4.5.12) together with the triangular inequality. The result follows.

Lemma 4.23. Let π1 and π2 be two planes and let u1, u2 ∈ π1 and v1, v2 ∈ π2 be

four vectors pairwise not parallel. Assume that each of the four pairs (u1, v1), (u1, v2),

(u2, v1) and (u2, v2) generates its own plane (at most 4 dierent such planes). Then

π1 = π2.

Proof. Consider the four lines `1, `2, `3 and `4 that are generated by u1, u2, v1, v2

respectively. By hypothesis, no two such lines are parallel and, since `1 and `3 are on

the same plane there must exist a crossing point between them, say p1,3. Then, we also

have the points p1,4, p2,3 and p2,4 that can be dened in the same way. Now it is clear

that all these points pi,j are on both π1 and π2. The points p1,3 and p1,4 are on the line

`1 and the points p1,3 and p2,3 are on the line `3. We deduce that if the four points are

aligned, then we have `1 = `3 which is false. Thus these 4 points dene a plane and

this plane is π1 = π2.


Throughout the proof, we use a strategy that needs some explanation. We assume that

L0 ∈ [2ι0 , 2ι0+1) where L0 is dened in (4.4.1). Our argumentation depends on the size

158

of the length l of a proper linear major arc and the size of its denominator q. So we

assume also that l ∈ [L, 2L) and that q ∈ [Q, 2Q) where L and Q are some powers of 2

with Q ≤ L/2. We write

(4.6.1) L :=log(L0)

log 2,

where L0 has been dened in (4.4.1). We thus have ι0 ≤ L. In Lemmas 4.24 and 4.25,

we record without proof some estimations that will be used in the proof of Theorem

4.1.

Lemma 4.24. Let s ≥ 1 be an integer, ϑ > 0 a real number and ι0 as previously

dened. Then,

ι0∑j=1

j−1∑i=0

1 ≤ L(L+ 1)

2,(4.6.2)

s∑i=0

2ϑi ≤ 2ϑ

2ϑ − 12ϑs,(4.6.3)

∑i≥s

1

2ϑi≤ 2ϑ

2ϑ − 1

1

2ϑs.(4.6.4)

Lemma 4.25. Consider the function

h(z) := Azα +B

zβ(z ≥ 0)

where α, β,A and B are xed positive constants. Then,

minn∈Z

h(2n) ≤

(AβBα

)1/(α+β)

(1

2α

(β

α

)α/(α+β)

+ 2β(α

β

)β/(α+β))

if β ≤ α,

(AβBα

)1/(α+β)

(2α(β

α

)α/(α+β)

+1

2β

(α

β

)β/(α+β))

otherwise.

4.6.1 Setting a structure on S

The set S is formed of integers z that we put in order M ≤ z1 < · · · < zS ≤ 2M .

We will give some structure to the set S. We start with the rst element z1 and then

consider the three ones that follow, z1 < z2 < z3 < z4. If they form a minor arc, then

we move to z4 as the rst element and restart. In the case that four consecutive points

x1, . . . , x4 satisfy D(x1, x2, x3, x4) = 0, we know that x1, . . . , x4 is in some major arc

A. If the plane π of A is xed by x1, . . . , x4, we consider all the points xi that follow

159

x4 for which f(xi) is on π until the last one, xj. The major arc A is then x1, . . . , xj.If the plane is not xed by x1, . . . , x4, then we consider the following points in Suntil we get to x a plane π (if it does) and then proceed as we did in the rst case.

Then we restart the process with xj as the rst element. At most three points remain

as unclassied at the last stage.

Now, from the preceding structure, we want to deduce that the following inequality

(4.6.5) S ≤ 21S1 + 7S2 + 82S3 + 44

holds where S1 is the contribution arising from proper linear major arcs, S2 is the

contribution of the proper major arcs and S3 is the number of successive minor arcs.

Two consecutive proper linear major arcs counted by S1 are from distinct lines. Also,

two consecutive proper major arcs A counted by S2 come either from distinct planes

or there exists a point between them that is in S but not in π.

We remove these (at most) two unclassied points through all the argument (three were

claimed but it is only if there are actually exactly 3 points in S). If there are at most

42 points then the result is trivial. So we may assume that there are at least 43 points.

If S3 = 0 then the inequality is true since the points are in a line or in a plane. Since,

from Remark 4.1, a line has at most three components, and since, from Lemma 4.6, a

plane has at most seven components, we deduce that the most popular component has

at least 7 points and the inequality (4.6.5) follows. Now we assume that S3 6= 0.

Usually a minor arc should have a contribution of 4 points. But, for the sake of the

structure that we need, we will use minor arcs to eliminate the major arcs that have a

small number of points. First, if x1 < · · · < xj form a major arc, then there is an i with

1 < i < j such that D(x1, xi, xj, xj+1) 6= 0, if there is such a xj+1. From Lemma 4.6 and

since we need at least 3 points to do the counting argument (either in a proper linear

major arc and in a proper major arc), we will include the contribution of any major

arcs with less than 43 points into the number of minor arcs. To do so, we organize

the points into quadruples of successive minor arcs. We assume that we have chosen

the quadruples up to x1 < x2 < . . . If x1, . . . , x4 forms a minor arc then we dene

a new quadruple and we restart with x4 as the rst element. If x1 < · · · < xj is a

major arc and xj < · · · < xj1 is another major arc, then we take (x1, xi, xj, xj+1) as

a quadruple (with the i such that D(x1, xi, xj, xj+1) 6= 0) and take either xj1 or xj as

the rst element depending whether if there are less than or at least 43 points in the

second major arcs respectively. If x1 < · · · < xj is a major arc and xj < · · · < xj+3 is

a minor arc, then we take (x1, xi, xj, xj+1) as a quadruple and we take xj+3 as a rst

160

element. Now, an easy verication reveals that the worst possible case is when we are

in the second situation with two consecutive major arcs of exactly 42 points. In such

a situation we have to consider that the quadruple account for 82 points. The rst

situation being at most 3 and the third at most 44. An exception is when there are no

further minor arcs in x1 < . . . but this is the basic case that we did rst. It remains

from this last one a proper linear major arc, a proper major arc or at most 42 points.

We will refer to this set of points as A1.

Also, it is easy to give an estimate for S3. Let K be the interval described in (4.3.5).

There are at most XK

such intervals, so that

(4.6.6) S3 ≤(C15

3 C44∆2

2 · 55/2

)1/6

X +

(8α2

9

)1/3

X + 24δ1δ2X.

We remove from S any points that is not a complete major arc (including its edges) of

at least 43 points, including A1. Some edges may be counted twice in the process. We

still call the resulting set S. We now have a sequence of major arcs (maybe not with

the same number of points but with at least 43) and like previously we can nd the

most popular proper sequence of points. This component is either a line or a plane. We

now consider the resulting subsequence of proper linear major arcs and the subsequence

of proper major arcs separately. We may also add A1 at the end of the appropriate

sequence. Now, the sequence of major arcs satises the desired property. The sequence

of linear major arcs may contains at most three consecutive proper linear major arcs

with the same equation and the result follows by keeping the linear major arcs with

the most points. Indeed, since there are at least 7 points, at least one component has

at least 3 points. Gathering these observations, one nds (4.6.5).

4.6.2 The S2 term

The estimate for the contribution of the major arcs has two components. We rst

organize the density of points inside one proper major arc with the help of formula

(4.5.4). Then, we combine together these densities with another density by a use of the

spacing principle (4.4.2) on the functions g as dened in (4.3.8).

There are two possibilities in (4.3.10) and these greatly inuence our ability to estimate

the density of major arcs in X. Precisely, for a major arc A of length l, we are

distinguishing two cases depending on whether we can prove that the closest point of

S not in A is at a distance lε1/2

from A or at a distance lε1/3

. We now show that

161

case 1 includes each case for which the rst possibility of (4.3.10) holds. For this, we

use formula (4.4.2) with n = 2, h = g (g associated to A) and σ ≥ 1− ε to get

(4.6.7) K ≥ min

(σl

16ε,

σ1/2

|g(2)(ξ1)|1/2

).

If the rst possibility of (4.6.7) holds then we clearly are in the case 1. If instead the

second possibility holds, then we use Lemme 4.10 with n = 2 to get

(4.6.8) l ≤ 4

(ε

|g(2)(ξ2)|

)1/2

.

Hence, without any loss in generality, we can assume the hypothesis

|g(2)(x)| ≥ |b|λ2,1

16C31C

73C

84

≥ |c|λ2,2

16C32C

73C

84

.

Then, we deduce that |g(2)(x)| ≤ 2|b|λ2,1 max(C31 ,

C62

C31) by using essentially the argument

that we can nd in Lemma 4.22 following (4.5.10). Combining these leads to

K ≥ σ1/2

|g(2)(ξ1)|1/2≥ σ1/2l

4ε1/2|g(2)(ξ2)|1/2

|g(2)(ξ1)|1/2

≥ σ1/2l

4ε1/2

|b|λ2,1

16C31C

73C

84

1

2|b|λ2,1 max(C31 ,

C62

C31)

1/2

(4.6.9)

≥ σ1/2l

ε1/21

(512C73C

144 )1/2

,

as wanted. Now, if the second possibility of (4.3.10) holds, then this time we use (4.4.2)

with n = 3 to get to

(4.6.10) K ≥ min

(( σ72

)1/2 l

ε1/2,

(3σ)1/3

|g(3)(ξ3)|1/3

)from which it is clear that the rst possibility is also in the case 1. Also, if the second

possibility of (4.6.10) occurs, then we use Lemma 4.10 with n = 3 along with the in-

equalities on |g(3)(x)| that we found in Lemma 4.22 from (4.5.10) to prove an inequality

analogue to (4.6.9), namely

(4.6.11) K ≥ σ1/3l

ε1/31

(2304C73C

154 )1/3

.

This is the case 2.

Now, the case 1 has three possibilities and one can verify that in the region

(4.6.12) ε ≤ 1

513C73C

144

,

162

the worst of the three inequalities is given by (4.6.9) and thus we state the inequality

that will be used in case 1 when (4.6.12) holds, namely

(4.6.13)l

K≤ ε1/2(513C7

3C144 )1/2.

The case 2 has only one possibility and when

(4.6.14) ε ≤ 1

2305C73C

154

holds then we have

(4.6.15)l

K≤ ε1/3(2305C7

3C154 )1/3.

We now turn to the structure of points into a major arc. By using (4.5.4), we can state

that

(4.6.16) s1s2 ≤ max

(a1a2K

∆

|Γ|, (a1 + a2)

α

|Γ|

)where we have set

(4.6.17) ∆ :=69633

8192C10

3 C34∆1 +

17

4C3

1λ2,1 +17

4C3

2λ2,2

and

(4.6.18) α := 8(α0 + α1).

There is a huge dierence between the situation when si = 1 for an i and the one when

si ≥ 2. In the second case we know that the distance between the rst and the last

element on the line `i (Lemma 4.19) is bounded by (4.4.1) and in the rst case we do

not have any information.

4.6.2.1 The case s = 1

We assume that there are more than one proper major arc in both case 1 and 2. We

x a given proper major arc A that contains the points x1 < · · · < xJ , so that x1 and

xJ are the edges of A. This sequence can be seen as a sequence of lines starting with

the line generated by f(x1) and f(x2) that ends with f(xr1), say. Then another line

starts with f(xr1) and f(xr1+1) and ends at f(xr2) and so on until we get to f(xJ). The

sequence of points that we are counting in this section is

xr0 := x1 < xr1 < xr2 < · · · < xJ := xrt .

163

From Remark 4.1, we know that there are at most three intervals in the variable x

such that the line between any xed f(xri) and f(xri+1) intersect Λ. Each of them may

generate lattice points. So that we will include a possibility of 4 points between each

xed f(xri) and f(xri+1) by obtaining an estimate for 5t + 1. The reason for this is

that, in the next section, we want to deal with proper linear major arcs and for this we

need at least 3 points on the same component, so that we remove the contribution of

lines with less than 7 points in this section.

For a xed triple of consecutive such points xri < xri+1< xri+2

, we apply (4.6.16) with

a1 := xri+1− xri , a2 := xri+2

− xri+1and K1 := a1 + a2(= 2K) (with the notation of

Lemma 4.19). We thus have s1 = s2 = 1, so that

(4.6.19)1

xri+2− xri

=1

K1

≤ max

((∆

8|Γ|

)1/3

,α

|Γ|

)where we have used the inequality a1a2 ≤ K2. By applying this inequality to xr0 <

xr1 < xr2 , to xr2 < xr3 < xr4 and so on, we get that

(4.6.20)j

xr2j − xr0≤ max

((∆

8|Γ|

)1/3

,α

|Γ|

).

From this, we deduce that the worst possibility is when t = 3 (t = 2j or t = 2j + 1),

which yields the result

(4.6.21)5t+ 1

xJ − x1

≤ 16 max

((∆

8|Γ|

)1/3

,α

|Γ|

).

Now, we are ready for the estimate. We collect together all the edges o all such

proper major arcs. Assume that the complete list of edges is x1,1 < x1,2 < x2,1 <

x2,2 < · · · < xw,2, where the indice (i, j) means that we are talking about the i-th

proper major arcs and the j is for the rst or the last element. We will also write

x11,1 < x1

1,2 < x12,1 < x1

2,2 < · · · < x1w1,2

and x21,1 < x2

1,2 < x22,1 < x2

2,2 < · · · < x2w2,2

for the subsequences in case 1 and case 2 respectively. We denote by J ′i = 5ti + 1 the

number of points that we want to count in the i-th proper major arcs, and we also use

J ′1i and J ′2i when we need to specify which case. In both cases 1 and 2, we consider all

the proper major arcs for which the rst term in (4.6.20) holds. In the region (4.6.14)

we have (4.6.15), as can be veried, so that we deduce that we have for these proper

major arcs the estimate

(4.6.22)J ′i

xi+1,1 − xi,1≤ 16

(∆

8|Γ|

)1/3

ε1/3(2305C73C

154 )1/3 ≤ 8(2305C7

3C154 α0∆)1/3

164

sinceε

|Γ|≤ α0, which follows from (4.5.3). Also, since |Γ|2 ≥ b2 + c2, one can show

with calculus that

(4.6.23) ε ≥ β =⇒ |Γ| ≥ β

α0

.

We thus deduce that the inequality (4.6.22) still holds when (4.6.14) does not hold. By

applying (4.6.22) to i = 1, . . . , w − 1 and by summing we arrive at

(4.6.24)J ′1 + · · ·+ J ′w−1

X≤J ′1 + · · ·+ J ′w−1

xw,1 − x1,1

≤ 8(2305C73C

154 α0∆)1/3.

If the second case of (4.6.20), then one possibility is to use the exact same argument

to get to

(4.6.25)J ′1 + · · ·+ J ′w−1

X≤ 16(2305C7

3C154 )1/3α

1/30 α.

Another one is to examine case 1 and case 2 separately. For case 2 we use another kind

of strategy and it will be explained in section 4.6.2.3. For case 1, we do the same proof

but we use (4.6.13) in the region (4.6.12) to get

(4.6.26)J ′11 + · · ·+ J ′1w1−1

X≤ 16(513C7

3C144 )1/2α

1/20 α.

The possible remaining proper major arc in (4.6.24), (4.6.25) and (4.6.26) will be con-

sidered in section 4.6.4.

4.6.2.2 The case s ≥ 2

We consider a xed proper major arc A. We collect all the proper linear major arcs ` in

A that have length l in [L, 2L) and denominator q in [Q, 2Q) where both Q and L are

powers of 2. We have s = lqand the total contribution of the line ` is s+ 1. In section

4.6.2.1 all the lines with at most 6 points (as dened there) have been considered in the

sum. So now, since there are at most 3 proper linear major arcs with the same equation,

we consider only the one with the most points (which is at least 3) and multiply by 3

at the end. We can also only consider a contribution of s points on ` since 5 ≥ 3 and

at least one has been counted already on each. We have s ≥ 2 so that we assume that

Q ≤ L/2 and from Remark 4.2 we know that L ≤ L0 from (4.4.1).

We will use formula (4.6.16) in two dierent ways. First, assume that two lines `1 and

`2 are not parallel and satisfy the previous hypothesis. In this case one gets

(4.6.27)s1

K≤ max

(4LQ∆

|Γ|, 6Qα

L|Γ|

)

165

by using s2 = a2

q2along with the hypothesis. Here K still has the denition from Lemma

4.19. Now, assume that we have a line `1 that satises the previous hypothesis and a

line `2 that has s2 = 1, and thus an unknown length, that link `1 to the next line `3

that satises the hypothesis of the section. We then have

(4.6.28)s1

K≤ max

((8L2∆

Q|Γ|

)1/2

, 2α

|Γ|

)where the rst inequality is done by writing

s1 ≤ a1a2K∆

|Γ|≤ 2a1K

2 ∆

|Γ|⇒ s2

1 ≤ 2a21K

2 ∆

q1|Γ|.

We know from Lemma 4.13 that there is either 1 parallel lines to a given line or

some function described in (4.4.12) and (4.4.14). In view of this, we collect together all

the lines in A that satisfy the hypothesis in order of appearance in the variable x. We

rst consider the subsequence of lines that happen in less than 5C64

4consecutive parallel

lines. In this situation, we use only formula (4.6.27). Let `1, . . . , `µ be a complete

sequence of parallel lines. The corresponding contribution s1 + · · ·+ sµ thus satises

(4.6.29)s1 + · · ·+ sµ

K1

≤ max

(5C6

4

LQ∆

|Γ|,15

2C6

4

Qα

L|Γ|

)where K1 is the distance between the middle of `1 and the middle of the line `′1, that is

the rst line in the following set of parallel lines. The argument that leads to (4.6.29)

is as follows: we take the maximum value of si, we apply (4.6.27) to the line `i and `′1,

multiply by 5C64

4and extend the corresponding K to K1. Now, by summing up all the

contributions inside A together, we obtain

(4.6.30)JA(L,Q)

K2

≤ max

(5C6

4

LQ∆

|Γ|,15

2C6

4

Qα

L|Γ|

)where K2 is the distance between the rst and the last element in A. Of course the

possible last set of parallel lines is not included in this estimate and it will be treated

in section 4.6.3. Now, to extend this estimate to the whole interval eciently, we use

(4.6.14), (4.6.15) and (4.6.23) in the same way we did to get (4.6.24). We thus obtain

(4.6.31)J(L,Q)

X≤ (2305C7

3C154 α0)1/3C6

4 max

(5LQ∆,

15

2

Qα

L

).

Again, the possibility of only one such proper major arc A will be studied in section

4.6.4. Now, to get the complete contribution, it remains to sum over Q and then over

L using Lemma 4.24. In fact, we use∑Q≤L/2

Q ≤ L,∑

1≤n≤ι0

1 ≤ L and∑L≤L0

L2 ≤ 4

3L2

0

166

to get to the result

(4.6.32)J

X≤ (2305C7

3C154 α0)1/3C6

4 max

(320

3C3

1

δ1∆

λ2,1

,15

2αL)

with the help of (4.4.1).

As in section 4.6.2.1, there is another way to proceed by examining separately cases 1

and 2. The case 1 is done like we just did, but by using (4.6.12) and (4.6.13) instead

to go from the proper major arcs to the interval X , in which case we obtain

(4.6.33)J1

X≤ (513C7

3C144 α0)1/2C6

4 max

(320

3C3

1

δ1∆

λ2,1

,15

2αL).

The case 2 is handled in section 4.6.2.3.

We now turn to the evaluation of the contribution of the subsequence of lines that

occur in more than 5C64

4consecutive parallel lines. In this situation, we use inequality

(4.6.28). Indeed, since most consecutive lines are parallel, we cannot use inequality

(4.6.27). Thus, if two consecutive lines `1 and `2 are parallel, then the strategy is to

apply (4.6.28) to the line `1 along with the line joining the last point of `1 to the rst

point of the line `2. Proceeding as previously, we get

(4.6.34)J ′A(L,Q)

K1

≤ max

((8L2∆

Q|Γ|

)1/2

, 2α

|Γ|

)

where K1 is the distance between the rst and the last element of A. Now, we use thesame approach as when we went from (4.6.30) to (4.6.31), and obtain

(4.6.35)J ′(L,Q)

X≤ (2305C7

3C154 α0)1/3 max

((8L2∆

Q

)1/2

, 2α

).

Again, at most one line remains and it will be treated in section 4.6.3. To get the total

contribution, we run the sum over Q and then over L in the restricted domain where

(4.6.36) 272C3

1δ1δ0Q

λ2,1L2≥ 5C6

4

4, that is, Q ≥ 5C6

4

1088C31

λ2,1L2

δ1δ0

noticing that (4.4.12) remains true in the region (4.4.13). Again, using Lemma 4.24,

we get ∑Q

1

Q1/2≤ 21/2

21/2 − 1

(2176C3

1

5C64

δ1δ0

λ2,1L2

)1/2

167

so that the result is

(4.6.37)

J ′

X≤ (2305C7

3C154 α0)1/3 max

((34816C3

1

(21/2 − 1)25C64

)1/2(δ1δ0∆

λ2,1

)1/2

L, αL(L+ 1)

).

Finally, the second method, examining separately cases 1 and 2, the analog of (4.6.33),

leads to

(4.6.38)

J ′1X≤ (513C7

3C144 α0)1/2 max

((34816C3

1

(21/2 − 1)25C64

)1/2(δ1δ0∆

λ2,1

)1/2

L, αL(L+ 1)

).

for the contribution of case 1. The case 2 is examined in section 4.6.2.3. Similarly,

at most one line remains and it will be treated in section 4.6.3. Also, we postpone to

section 4.6.4 the estimate for the exceptional contribution in the case that there is only

one proper major arc.

4.6.2.3 Another method

Here we consider only case 2. The key observation is the subject of Lemma 4.22. We

deduce from it that a xed pair (b, c) xes at most 18227C7

3C154 proper major arcs. We

deduce from b2 + c2 ≤ |Γ|2 that the number of possibilities for z ≤ |Γ| < 2z is at most

(4.6.39)339

4C7

3C154 z

2.

Now, by looking back to sections 4.6.2.1 and 4.6.2.2, we see that there are ve situations

to handle, namely the second possibility in (4.6.21) and both possibilities of (4.6.30)

and of (4.6.34). We assume that condition (4.5.8) holds. First, we consider the second

possibility of (4.6.21) as a model for the method. We begin with the collection of all

the short proper major arcs in case 2. We assume that z ≤ |Γ| < 2z so that we have

the estimate

(4.6.40) J ′21i (z) ≤ 16α

z

(512C7

3C114

δ2

λ2,2

)1/2

=16α

z

(512C7

3C114

δ2

λ0,2

)1/2

M

for the number of extremities in the i-th such proper major arc (see section 4.6.2.1).

The total contribution is thus

(4.6.41) J ′21(z) ≤ 16α

(512C7

3C114

δ2

λ0,2

)1/2339

4C7

3C154 zM.

By summing up to Z on powers of 2, we get

(4.6.42) J ′211 ≤ 16α

(512C7

3C114

δ2

λ0,2

)1/2339

4C7

3C154 ZM.

168

This is the total contribution when |Γ| < Z, where Z is some power of two to be chosen

later. For the remaining values, we apply the methods of section 4.6.2.1 that leads to

(4.6.25), but we keep the |Γ| term that we can now estimate non trivially to get

(4.6.43) J ′212 ≤ 16(2305C7

3C154 )1/3α

1/30 αX

Z2/3.

By optimizing (4.6.42) with (4.6.43) and summing with the help of Lemma 4.25, we

obtain that

(4.6.44)

J ′21 ≤ 16ϕ1(2305C73C

154 )1/5

((512C7

3C114

δ2

λ0,2

)1/2339

4C7

3C154

)2/5

α1/50 αρ3/5M,

where

ϕ1 :=1

2

(2

3

)3/5

+ 22/3

(3

2

)2/5

.

We now turn to the collection of all the long proper major arcs in case 2. The line

(4.6.40) changes only by the use of the estimate (4.5.6) instead of (4.5.5) and it is the

only dierence up to line (4.6.42). The line (4.6.43) remains unchanged and we optimize

the result as we did for line (4.6.44). We get

(4.6.45)

J ′22 ≤ 16ϕ1(2305C73C

154 )1/5

((6912C7

3C114

δ2

λ0,2

)1/3339

4C7

3C154

)2/5

α1/50 αρ3/5M.

The interesting property of long proper major arcs is that second non trivial lower

bound that we have for |Γ| that is given by (4.5.7). So that we have the alternative

possibility for line (4.6.43) given by

(4.6.46) J ′22 ≤ 16(2305C73C

154 )1/3α

1/30 αX

Y 2/3.

Now, it is clear that the combined result of (4.6.45) and (4.6.46) is far stronger than

(4.6.25) by considering the sixth term in the denition of Y (given in(4.5.7)) along with

line (4.6.45) that contains λ0,2 = Mλ1,2. We can verify that (Mλ1,2)2/15 + 1

λ2/31,2

M1/9.

Also, usually the term λ0,2 is itself large so that the line (4.6.44) is essentially α2/50 αM

which is also better than (4.6.25).

Finally, we have to add the contribution in the cases where ε ≥ 1/2 for which we cannot

apply Lemma 4.22. In this situation we use inequality (4.6.23) to get a density of size

≤ 32α0α inside such a proper major arc, long or short, and thus the result

(4.6.47) J ′23 ≤ 32α0αX.

169

The method does not change for any of the 4 other possibilities, but there is a sum

running over Q and then over L to perform at the very end. Also, the sum is done in

the restricted domain

(4.6.48) 272C3

1δ1δQ

λ2,1L2≥ 5C6

4

4, that is, Q ≥ 5C6

4

1088C31

λ2,1L2

δ1δ

for the rst possibility of (4.6.34) (see Remark 4.3). Tracking back the computations,

we see that γ|Γ| becomes the upper bound

(4.6.49) ϕ1γ(2305C73C

154 )1/5α

1/50

((512C7

3C114

δ2

λ0,2

)1/2339

4C7

3C154

)2/5

ρ3/5M

in the short situation and

(4.6.50) ϕ1γ(2305C73C

154 )1/5α

1/50

((6912C7

3C114

δ2

λ0,2

)1/3339

4C7

3C154

)2/5

ρ3/5M

in the long one. The other possibility gives

(4.6.51) γ(2305C73C

154 )1/3α

1/30

X

Y 2/3.

So the rst term of (4.6.30) becomes respectively

(4.6.52)

320

3ϕ1C

31C

64

δ1∆

λ2,1

(2305C73C

154 )1/5α

1/50

((512C7

3C114

δ2

λ0,2

)1/2339

4C7

3C154

) 25

ρ35M,

(4.6.53)320

3ϕ1C

31C

64

δ1∆

λ2,1

(2305C73C

154 )1/5α

150

((6912C7

3C114

δ2

λ0,2

) 13 339

4C7

3C154

) 25

ρ35M

and

(4.6.54)320

3C3

1C64

δ1∆

λ2,1

(2305C73C

154 )1/3α

1/30

X

Y 2/3.

The second term of (4.6.30) becomes

(4.6.55)15

2ϕ1C

64αL(2305C7

3C154 )1/5α

1/50

((512C7

3C114

δ2

λ0,2

)1/2339

4C7

3C154

)2/5

ρ35M,

(4.6.56)15

2ϕ1C

64αL(2305C7

3C154 )1/5α

1/50

((6912C7

3C114

δ2

λ0,2

)1/3339

4C7

3C154

)2/5

ρ35M

170

and

(4.6.57)15

2C6

4αL(2305C73C

154 )1/3α

1/30

X

Y 2/3.

The second possibility of (4.6.34) becomes

(4.6.58) ϕ1L(L+ 1)α(2305C73C

154 )1/5α

1/50

((512C7

3C114

δ2

λ0,2

)1/2339

4C7

3C154

)2/5

ρ35M,

(4.6.59) ϕ1L(L+ 1)α(2305C73C

154 )1/5α

1/50

((6912C7

3C114

δ2

λ0,2

)1/3339

4C7

3C154

)2/5

ρ35M

and

(4.6.60) L(L+ 1)α(2305C73C

154 )1/3α

1/30

X

Y 2/3.

Then, the rst possibility of (4.6.34) becomes

(4.6.61) ϕ2

(34816C3

1

(21/2−1)25C64

) 12(δ1δ∆λ2,1

) 12L(2305C7

3C154 α0)

310

(512C73C

114

δ2λ0,2

) 12

3394C7

3C154

110

ρ910M,

(4.6.62) ϕ2

(34816C3

1

(21/2−1)25C64

) 12(δ1δ∆λ2,1

) 12L(2305C7

3C154 α0)

310

(6912C73C

114

δ2λ0,2

) 13

3394C7

3C154

110

ρ910M

and

(4.6.63)

(34816C3

1

(21/2 − 1)25C64

) 12(δ1δ∆

λ2,1

) 12

L(2305C73C

154 )1/3α

1/30

X

Y 1/6

where

ϕ2 :=1

3621/231/5 + 21/631/5.

The terms (4.6.62) and (4.6.63) look like the largest contributions obtained in this

section. The rst one has a strong delta component" and the mostly unavoidable ratio

of second derivative λ2,2/λ2,1 that one hopes to be small, at least it is less than δ2/δ1 by

(4.1.9) so essentially the term is stronger than the corresponding one in (4.6.37). The

other one would be similar, but again (Mλ1,2)1/30 + 1

λ1/61,2

M1/36.

Finally, we still have to add the contributions where ε ≥ 1/2 and we proceed as we did

for (4.6.47). We get respectively

(4.6.64) 6403C3

1C64δ1α0

∆λ2,1

X, 15C64α0αLX, 2L(L+1)α0αX and

(69632C3

1

(21/2−1)25C64

) 12(δ1δ∆λ2,1

) 12Lα1/2

0 X.

It remains to recall that every estimate obtained in section 4.6.2.2 has to be multiplied

by 3.

171

4.6.3 The S1 term

We consider the ordered list of all the remaining proper linear major arcs of length in

[L, 2L) and of denominator in [Q, 2Q) where L and Q are powers of 2 with Q ≤ L/2

and L ≤ L0. By this, we mean every proper linear major arcs in S1 and every proper

linear major arcs left in sections 4.6.2.2 and 4.6.2.3 that satisfy the hypothesis. Since

we have kept at most two linear major arcs from each proper major arc treated in

section 4.6.2, we will multiply the contribution of this section by two and work with at

most one proper linear major arc, the one with the largest contribution to S, for each

proper major arc.

Recalling the situation developed in the rst paragraph of section 4.6.1, a major arc Aof plane π that is made of the points x1 < · · · < xj contains a line ` on which there

is a proper linear major arc of length l and denominator q that is made of the points

xm < · · · < xm+s with s = lq. Since A is a major arc, it contains a point of S that is

not on the line `, say xi. We deduce that the set xm, xm+s, xi, xj+1 is a minor arc

(xj+1 exists for all but at most one proper major arc). We will use this structure in

two dierent ways in Lemma 4.5.

We remove all the lines with index of appearance congruent to i1 and i2 modulo 3, we

keep the congruence with the largest contribution. Now, between two remaining lines

`m and `m+3, there are four points on a minor arc. We deduce that both the equation

for f1 and for f2 (of `m) there is a point not on the equation before `m+3 (in the variable

x). Given any xed line ` in this list, the number of lines parallel to ` in the list is either

the rst or the second estimate that we nd in (4.4.5). The proof is similar, even if

the hypotheses are dierent. We also remark, as a second possibility, that the estimate

(4.4.17) holds for both values of i.

We now consider the subset of lines that occur at most 5C64

4times and we keep only one

of each. We have a sequence `j1 , . . . , `js of lines and we write xj1,1, xj1,2, xj2,1, . . . for the

corresponding sequence of endpoints. Now, the set

xj1,1, xj1,2, xj2,1, xj2,2

is either a minor arc or these points are on the same plane and dene that plane. Maybe

some of the following lines are on this plane, up to the line `jr1−1 , say. We restart this

with `jr1 and `jr1+1 and so on until there are no more lines. We thus have a sequence of

planes that contains lines, separated by a sequence of pairs made of two lines for which

the set of endpoints is a minor arc. We assume that this sequence contains m = 8j+w

172

elements and we remove the last w elements. There is also at most one line left at the

end. We extract the sequence of pairs that give minor arc, the sequence of planes with

at most 7 lines and the sequence of planes with at least 8 lines and we assume that they

contain respectively m1,m2,m3 elements, where m1 + m2 + m3 = 8j. We either have

m1 +m3 ≥ 2j or m2 ≥ 6j. In the rst case, we have a contribution of at least 4j lines

by the rst and the third sequences and a contribution of at most 42j by the second

sequence. Now since we have conditions on length and denominator, we always haves1s2≤ 4 where s1 +1, s2 +1 are the contributions of any given lines `1 and `2. We deduce

that 43 times the contributions from the rst and the third sequence will count all the

points (and we will multiply be 32as usual to add 1). Now, if the second possibility

occurs, that is m2 ≥ 6j, then we go back to the original sequence before the separation

into three sequences and consider the 8j rst elements. We form pairs of consecutive

elements, we thus have 4j of them, and since we have at most 2j elements from the rst

and the third sequence, we deduce that we have at least 2j pairs made only of elements

from the second sequence. Now, each of these pairs is made of two consecutive planes of

between 2 and 7 elements. They are distinct. From Lemma 4.23 we deduce that there

exists a line in the rst set and another in the second set such that their endpoints is

a minor arc. Again, by multiplying the contribution of each of these sets of two lines

by 25 we control the total contribution of the two sets. We then multiply by 2 to get

the (at most) 8j of them and also we still add in between the contribution from the

rst sequence and add to this the contribution of the plane. In the end, we see that

the worst possibility is a multiple of 150 times the contribution of the rst sequence

and of 129 the contribution of the planes and 24 lines. Also, since a plane contains at

most 7 components, the worst possible case is if each of the 6 components contains a

proper linear major arc and the other one contains a proper major arc. This implies

that the total contribution of one such plane is at most 13 times the contribution of

the proper major arc with the highest contribution. Since we left at most two lines by

proper major arc in section 4.6.2, one can avoid circular reasoning by multiplying the

whole contribution of section 4.6.2 by 65 = 5 · 13.

To evaluate the contribution of the sequence of lines that produce a minor arc, we use

Lemma 4.5. We get

(4.6.65)s1 + s2

K≤ max

(4

(C15

3 C44∆2

L4

Q2

)1/4

, 12α1/22 L1/2, 24

Qδ1δ2

L

).

To see this, we rst assume that the rst possibility of (4.3.7) occurs. We write

s1s2 ≤C15

3 C44∆2K

4l1l22

⇒ s1 ≤C15

3 C44∆2K

4l1q2

2

173

⇒ s41 ≤

C153 C

44∆2K

4l41q2

2q31

⇒ s41

K4≤ 16

C153 C

44∆2L

4

Q2

which provides the rst upper bound from (4.6.65) after a multiplication by 2. We have

used the notation s1 = l1q1

and s2 = l2q2. The two other results are done similarly. Now,

by adding together all of the contributions and multiplying by 32, since 3si

2≥ si + 1, we

get

(4.6.66)J1(L,Q)

X≤ max

(6

(C15

3 C44∆2

L4

Q2

)1/4

, 18α1/22 L1/2, 36

Qδ1δ2

L

).

Also, there is another estimate for this situation given by Lemma 4.15. We can write

(4.6.67) J1(L,Q) ≤ 24C24λ2,0QLM

in the case that we are using the estimate from Lemma 4.12, because in the construction

there is a function for which the gradient has only one occurrence, and

(4.6.68) J1(L,Q) ≤ 12C2i λ2,iQLM

in the case where we are using Lemma 4.14.

Assume that the rst possibility in (4.6.66) occur. We sum over Q the minimum

between the latter term and (4.6.67) or (4.6.68) with the help of Lemma 4.24. This

optimized sum is then summed over L ≤ L0 and we get respectively

(4.6.69) J11 ≤ ϕ3(48C2

4λ2,0)1/3

(6 · 21/2

21/2 − 1

(C15

3 C44∆2

)1/4)2/3(

16C31δ1

λ2,1

)1/2

ρ2/3M

and

(4.6.70) J11 ≤ ϕ3(24C2

i λ2,i)1/3

(6 · 21/2

21/2 − 1

(C15

3 C44∆2

)1/4)2/3(

16C31δ1

λ2,1

)1/2

ρ2/3M

where

ϕ3 := 25/6 +1

25/3.

Assume now that the second possibility of (4.6.66) occurs. We just sum over Q and

then over L with the help of Lemma 4.24. We get

(4.6.71) J21 ≤

18 · 21/2

21/2 − 1

(16C3

1δ1

λ2,1

)1/4

α1/22 LX.

174

We do the same with the third possibility to get

(4.6.72) 36δ1δ2LX.

We remind that every estimate has to be multiplied by 1505C6

4

4if we use (4.4.5) and by

1505C6

i

4if we use Lemma 4.14.

We are ready to start the second part of the proof. That is, the subsequence that

contains w parallel lines where

(4.6.73)5C6

4

4< w ≤ 272

C31δ1δQ

λ2,1L2

or

(4.6.74)5C6

i

4< w ≤ 272

C3i δ

2iQ

λ2,iL2.

We deduce from this inequality that the sum over Q can be performed respectively over

the domains

(4.6.75) Q >5C6

4

1088C31

λ2,1L2

δ1δor Q >

5C3i

1088

λ2,iL2

δ2i

.

There are three ways to proceed in this case. At rst, we can use the exact same

argument by using only the line with the most points for each parallel family, and then

multiply by one of the two upper bounds on w. This leads, by using the rst possibility

and multiplying by 150, to the result

(4.6.76)

J2(L,Q)

X≤ 40800C3

1

δ1δQ

λ2,1L2max

(6

(C15

3 C44∆2

L4

Q2

)1/4

, 18α1/22 L1/2, 36

Qδ1δ2

L

).

By summing on Q ≤ L/2 we get

(4.6.77)J2(L)

X≤ 40800C3

1

δ1δ

λ2,1

max

(6

21/2 − 1

(C153 C

44∆2)

1/4

L1/2, 18

α1/22

L1/2, 12

δ1δ2

L

).

Similarly, we can do that with the estimate in Lemma 4.15 and get

(4.6.78) J2(L,Q) ≤ 6528C24C

31M

λ2,0δ1δ

λ2,1

Q2

L

and by summing on Q and get

(4.6.79) J2(L) ≤ 2176C24C

31M

λ2,0δ1δ

λ2,1

L.

175

By summing and optimizing, we get to respectively

(4.6.80)

ϕ3

(4352C2

4C31

λ2,0δ1δ

λ2,1

)1/3(40800C3

1

δ1δ

λ2,1

6 · 21/2

(21/2 − 1)2

(C15

3 C44∆2

)1/4)2/3

ρ2/3M,

(4.6.81) ϕ3

(4352C2

4C31

λ2,0δ1δ

λ2,1

)1/3(40800C3

1

δ1δ

λ2,1

18 · 21/2

21/2 − 1α

1/22

)2/3

ρ2/3M

and

(4.6.82)5

2

(4352C2

4C31

λ2,0δ1δ

λ2,1

)1/2(40800C3

1

δ1δ

λ2,1

24δ1δ2

)1/2

ρ1/2M.

Using the exact same argument but with (4.6.74) and (4.6.68) instead, we get

(4.6.83) ϕ3

(2176C5

i δ2i

)1/3(

40800C3i

δ2i

λ2,i

6 · 21/2

(21/2 − 1)2

(C15

3 C44∆2

)1/4)2/3

ρ2/3M,

(4.6.84) ϕ3

(2176C5

i δ2i

)1/3(

40800C3i

δ2i

λ2,i

18 · 21/2

21/2 − 1α

1/22

)2/3

ρ2/3M

and

(4.6.85)5

2

(2176C5

i δ2i

)1/2(

40800C3i

δ2i

λ2,i

24δ1δ2

)1/2

ρ1/2M

for i = 1, 2.

We can also use another formula to estimate the contribution of the large set of parallel

lines. Again, if there are m lines, we keep those for which the index of appearance is

in the congruence class modulo 3 that has the largest contribution. At the end, we will

multiply the result by 3. We still have at least one point not on each equation for f1 and

for f2 between any two lines and thus (4.6.73) and (4.6.74) with their corresponding

consequences on (4.6.67) and (4.6.68) respectively. The idea is simply to apply Lemma

4.5 to the set xm, xm+s, xi, xj+1 described at the beginning of this section that containsa line ` with length l and denominator q and two other points. We thus have s1 := s = l

q

and s2 = 1 and then

(4.6.86) s ≤ max

(C15

3 C44

2∆2K

5l, 6α2K3, 24δ1δ2K

)from which we deduce that

(4.6.87)s

K≤ max

((16C15

3 C44

∆2L5

Q4

)1/5

,

(24α2L

2

Q2

)1/3

, 24δ1δ2

)

176

and thus

(4.6.88)J2(L,Q)

X≤ 9

2max

((16C15

3 C44

∆2L5

Q4

)1/5

,

(24α2L

2

Q2

)1/3

, 24δ1δ2

).

The third term gives total contribution of

(4.6.89) 54L(L+ 1)δ1δ2X

points. We now assume that one of the two rst terms holds. A rst possibility is to sum

(4.6.88) over one of the domains (4.6.75), by using Lemma 4.24, to obtain respectively

(4.6.90)9

2

28/5

24/5 − 1

(16C15

3 C44∆2

)1/5(

1088C31

5C64

δ1δ

λ2,1

)4/5X

L3/5,

(4.6.91)9

2

24/3

22/3 − 1(24α2)1/3

(1088C3

1

5C64

δ1δ

λ2,1

)2/3X

L2/3

and

(4.6.92)9

2

28/5

24/5 − 1

(16C15

3 C44∆2

)1/5(

1088

5C3i

δ2i

λ2,i

)4/5X

L3/5,

(4.6.93)9

2

24/3

22/3 − 1(24α2)1/3

(1088

5C3i

δ2i

λ2,i

)2/3X

L2/3.

We then optimize with (4.6.79) or the other possible estimate. We get to a contribution

of respectively

(4.6.94) ϕ4

(2176C2

4C31

λ2,0δ1δ

λ2,1

)3/8(

9

2

(16C15

3 C44∆2

)1/5(

1088C31

5C64

δ1δ

λ2,1

)4/5)5/8

ρ5/8M,

(4.6.95) ϕ5

(2176C2

4C31

λ2,0δ1δ

λ2,1

)2/5(

9

2(24α2)1/3

(1088C3

1

5C64

δ1δ

λ2,1

)2/3)3/5

ρ3/5M

for the rst possibility and

(4.6.96) ϕ4

(1088C5

i δ2i

)3/8

(9

2

(16C15

3 C44∆2

)1/5(

1088

5C3i

δ2i

λ2,i

)4/5)5/8

ρ5/8M,

177

(4.6.97) ϕ5

(1088C5

i δ2i

)2/5

(9

2(24α2)1/3

(1088

5C3i

δ2i

λ2,i

)2/3)3/5

ρ3/5M

for the second one. We have set

ϕ4 :=

(1

2

(3

5

)5/8

+ 23/5

(5

3

)3/8)

2

(24/5 − 1)5/8

and

ϕ5 :=

(1

2

(2

3

)3/5

+ 22/3

(3

2

)2/5)

24/5

(22/3 − 1)3/5.

As a second possibility, we optimize the summation in Q between (4.6.78) and a term in

(4.6.88) (for both possible estimate as above) and then sum in L. We get respectively

(4.6.98) ϕ6

(6528C2

4C31

λ2,0δ1δ

λ2,1

)2/7(9

2

(16C15

3 C44∆2

)1/5)5/7(

16C31δ1

λ2,1

)3/14

ρ5/7M,

(4.6.99) ϕ7

(6528C2

4C31

λ2,0δ1δ

λ2,1

)1/4(9

2(24α2)1/3

)3/4(16C3

1δ1

λ2,1

)1/8

ρ3/4M

for the rst possibility and

(4.6.100) ϕ6

(3264C5

i δ2i

)2/7(

9

2

(16C15

3 C44∆2

)1/5)5/7(

16C31δ1

λ2,1

)3/14

ρ5/7M,

(4.6.101) ϕ7

(3264C5

i δ2i

)1/4(

9

2(24α2)1/3

)3/4(16C3

1δ1

λ2,1

)1/8

ρ3/4M

for the second one. We have set

ϕ6 :=

(1

4

(2

5

)5/7

+ 24/5

(5

2

)2/7)

23/7

23/7 − 1

ϕ7 :=

(1

4

(1

3

)3/4

+ 22/331/4

)21/4

21/4 − 1.

We end this section with few comments. One can sum (4.6.77), or the other possibility,

over L ≥ 2. Also, one can sum (4.6.79) or the other possibility over L ≤ L0. Similarly,

one can sum the estimates from (4.6.90) to (4.6.93) over L ≥ 2.

The very best result is given by summing over the domain Q ≤ L/2, L ≤ L0 and one of

the three possibility in (4.6.75) the minimal result between the corresponding version

of (4.6.76) and (4.6.78) with (4.6.88). It is an easy computation if all the parameters

are given. This is what is stated in Theorem 4.1 as ΞiX where Ξi is dened in (4.2.4).

What precedes can be seen as inequalities that may save us from these computations.

178

4.6.4 The exceptional cases

The basis of many of the previous computations was to combine the density of integer

points inside a structured object, a line or a plane with some characteristics, together

with the density of such an object in a larger scale. The assumption that there are at

least two similar objects is thus necessary. In this section, we focus on the possibility

that an object occur only once. All the work has been already done in the previous

sections and we will basically just follow them in order of appearance and state the

results.

Starting with section 4.6.2.1 and looking at the estimate (4.6.21), we deduce from

Lemma 4.21 that the contribution is either

(4.6.102) ≤ (8∆1/3 + 16α)L1

or

(4.6.103) ≤(

8∆1/3

Y 1/3+ 16

α

Y

)L2

depending if the proper major arc is short or long respectively. There is nothing else

in this section.

In the section 4.6.2.2 there are two events that have to be handled. The rst one is the

situation described in (4.6.30) and like previously we either have a contribution of

(4.6.104) ≤(

5C64LQ∆ +

15

2C6

4

Qα

L

)L1

or

(4.6.105) ≤(

5C64LQ∆ +

15

2C6

4

Qα

L

)L2

Y.

The same comments also apply here and throughout the section 4.6.2.2. The second

situation is described in (4.6.34) and the result is a contribution of

(4.6.106) ≤

((8L2∆

Q

)1/2

+ 2α

)L1

or

(4.6.107) ≤(

8L2∆

Q

)1/2L2

Y 1/2+ 2α

L2

Y.

179

It remains to sum over Q and then L in the rst possibility and over the domain (4.6.36)

in the second one (for the rst term at least). The result from (4.6.104) to (4.6.107)

leads to

(4.6.108) ≤(

5C64

64C31δ1

3λ2,1

∆ +15

2C6

4αL)(

L1 +L2

Y

)and

(4.6.109)

≤(

34816C31

(21/2 − 1)25C64

)1/2(δ1δ0∆

λ2,1

)1/2(L1 +

L2

Y 1/2

)L+ αL(L+ 1)

(L1 +

L2

Y

).

The estimation of S2 is now completed.

For section 4.6.3, at most 24 lines have not been counted in the argument. Now, we

have seen in the proof of Lemma 4.15 the inequality 1 ≤ 4C21λ1,1Q (see the hypothesis

a 6= 0 and Remark 4.4). We deduce that the contribution of this line is at most

(4.6.110) 243

2

L0

Q≤ 144C2

1M(16C31δ1λ2,1)1/2.

Theorem 4.1 is then proved.


The basic idea is to consider the set S ′i ⊆ S as dene in Lemma 4.16. We concentrate

only on this estimate and at the end we multiply by three and add (4.4.23).

By proceeding as in section 4.6.1, we are lead to the formula

(4.7.1) |S ′i| ≤ 7S2 + 26S3 + 16

with S2 and S3 having respectively the same denition. The dierence is that this

time we want to remove major arcs with less than 14 points. So that we nd that the

worst case is two consecutive major arcs with exactly 14 points and we get to the 26 of

(4.7.1). The other terms are done the same way. There is no possibility of lines from

the construction in Lemma 4.16.

The term S3 is bounded by (4.6.6). The term S2 is done the exact same way as in

section 4.6.2.1 and the other method that we nd in section 4.6.2.3. A small dierence

is that instead of 5t+ 1, we now just have t+ 1 (see the two rst paragraphs of section

4.6.2.1) and the worst case is t = 3 again with a constant 4 that replace 16 in all the

computations. To this we have to add the possibility of only one major arc that we nd

in inequality (4.6.102) and (4.6.103) again with 4 instead of 16. Theorem 4.2 is proved.

180


The purpose of this theorem is to estimate the number of solutions in S when condition

(4.2.3) does not hold but (4.2.6) does. That is, the function f2 can be locally constant,

but f1 cannot. So the idea is to apply Lemma 4.17 for each intervals in which f2 stay

at δ2 of an integer value and to sum this over all the possible values.

As in the proof of Theorem 4.4, an interval in which |f2(x)| ≤ δ2 is of length at most

2C22δ2λ1,2

and the distance in x between the beginning of two consecutive intervals such

that |f2(x)| ≤ δ2 is at least 1C2

2λ1,2. We deduce that there are at most (C2

2λ1,2X + 2)

intervals of length at most 2C22δ2λ1,2

to consider. By a direct application of Lemma 4.17,

we have

S ≤ (C22λ1,2X + 2)

(6

(C3

1λ2,1

4

)1/3

2C22

δ2

λ1,2

+ 115C121 δ12C2

2

δ2

λ1,2

+ 6

)

+9C6

1

2

∑j≤C2

2λ1,2X+2

maxAj

s.

Here, only the last term needs to be worked out to end the proof, all the others go in

the statement. That is, it remains only to estimate the contribution of the proper linear

major arcs with the most points. If C22λ1,2X ≤ 2 then there are at most 4 intervals and

in this case we can write ∑j≤4

maxAj

s ≤ 16

q0

(C3δ1

λ2,1

)1/2

by using Remark 4.2 and q0 as dened in Lemma 4.16. If instead we have C22λ1,2X ≥ 2,

then we use Lemma 4.18. We have, in the notation of the lemma, that K ≤ 2C22λ1,2X,

d ≥ 1C2

2λ1,2and L3 ≤ 2C2

2δ2λ1,2

. We nd

K∑j=1

maxAj

s ≤ (512C51KMδ1)1/2 +

8C51Mδ1

dL3L

≤ (1024C51C

22Mλ1,2δ1)1/2 + 16C5

1C42Mδ1δ2L

which ends the proof of Theorem 4.3.

Remark 4.7. One could use another idea to estimate the contribution of the gradients

that correspond to only one linear major arc. Indeed, assume that we have two con-

secutive such proper linear major arcs on the lines `1 and `2, where the equation for

the function f2 are two dierent constants. Let a1x+b1q1

and a2x+b2q2

be respectively the

equation of `1 and `2 for the function f1. Let also respectively f(x1), f(x2) and f(x3),

181

f(x4) be the extremities of the proper major arc on `1 and on `2. A direct computation

reveals that D(x1, x2, x3, x4) 6= 0 if and only if a1

q16= a2

q2. So that ideas from the proof of

Theorem 4.1 can be used.

4.9 Examples and concluding remarks

Consider the function f(x) := (x, f1(x), f2(x)), where fi(x) = cixγi , with i = 1, 2,

ci ∈ R\0 and γi ∈ R. Then, conditions (4.1.2) and (4.1.4) hold unless γ1 = γ2

or at least one of γ1, γ2 is 0, 1 or 2. Running a Maple program that computes, given

λ0,1, λ0,2, δ1 and δ2, the limiting term in the inequality of Theorem 4.1. A quick search

has shown that about half the terms at least can be the limiting one (or is of equal

contribution), and also all the three values of i are important.

To illustrate the range of applicability and the strength of the result, we present here

two examples. Given δ1 and δ2, we are seeking for the number of integer solutions to

the system

‖xγ1 − t1‖ < δ1, ‖xγ2 − t2‖ < δ2 where γ1 :=

√8

3, γ2 :=

√3

2and (t1, t2) ∈ [0, 1)2

with x ∈ [M, 2M ]. In our notation, we have

f(x) = (x, f1(x), f2(x)) = (x, xγ1 − t1, xγ2 − t2).

One easily veries the conditions (4.1.2) and (4.1.4) hold for f . One also veries

that all the conditions of Theorem 4.1 hold. We nd that there are respectively M11/15−γ2/30 logM andM2/3−γ2/30 logM solutions for the choices δ1 = δ2 = 1

M1/5 and

δ1 = δ2 = 1M1/4 . These estimates are provided by the term

(δ2λ0,2

) 130

(δ1δΩ)12α

3100 M logM .

We choose one of the δi (i = 1, 2) to be 1x1/5 and the other to be 1

x1/4 , we then get

M7/10 log2M solutions, provided by the term α1/20 αM log2M .

We now consider the number of integer solutions x ∈ [ z1/3

4, z

1/3

2] to the system

x3 + y31 = z1, y2x

2 = z2 where z1, z2 ∈ [z, z + h] and y1, y2 ∈ N.

It is interesting to note that such an x xes both z1 and z2 if h < z2/3/16. In this

example, (4.1.9) does not have a signicant meaning as both terms are essentially

equal. So we only write f(x) = (x, f1(x), f2(x)) = (x, (z − x3)1/3, z/x2) and verify that

all the required conditions to use Theorem 4.1 are reunited if h x2/3, in which case we

have δ1 = δ2 hx2/3 . If we choose h = z5/9, then we nd that the number of solutions

182

is at most z5/27 and that the limiting term is given by ΞiM (for any i) and in fact

one can track it back to (4.6.81). If we choose h = z1/2 instead, then Theorem 4.1 can

still be used, but it gives the same as Theorem 4.2 which is z1/6.

One could remove the hypothesis (4.2.3) at the price of adding terms in the result. The

methods for that are all in the paper already. Here we have chosen to add Theorems

4.2, 4.3 and 4.4 instead. Various improvements are possible. Maybe one can treat

the special plane (see Remark 4.3) more eciently. We can let i vary somehow in Ξ

with some additional terms when we do the sum over L and Q. Also, despite of the

conditions we assume in Theorem 4.1, if only one function, f1 or f2, is a polynomial of

degree two then a similar special theorem can be shown with almost the same methods.

It is possible to distinguish two cases: |a′| ≥ 1 and a′ = 0 in Lemma 4.20. In the rst

case, we add the constribution ∆1/31 X + α1X to the contribution of the minor arcs

and in the second case we are left with simpler equations for the major arcs. These

equations can be treated by a lemma similar to Lemma 4.8 but with the rst and second

derivatives of g instead. The result is an improvement of many terms.

The nal statement would need a way to work with three proper linear major arcs at

the same time in a lemma similar to Lemma 4.5. We keep it or any other method for

future papers. A lot of methods, if not all, can be generalized to higher dimension n.

In particular, the minor arc is easy to estimate and the methods from [5] produce a

result that is nontrivial for δi (i = 1, . . . , n) really small.

183

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[2] M. Branton and P. Sargos, Points entiers au voisinage d'une courbe plane à

très faible courbure, Bull. Sci. Math. 118 (1994), no. 1, 15− 28.

[3] M. Filaseta and O. Trifonov, The distribution of fractional parts with appli-

cations to gap results in number theory, Proc. London Math. Soc. (3) 73 (1996),

no. 2, 241− 278.

[4] M. N. Huxley, The integer points close to a curve. III. Number theory in progress,

Vol. 2 (Zakopane-Ko±cielisko, 1997), 911− 940, de Gruyter, Berlin, 1999.

[5] M. N. Huxley, The integer points in a plane curve, Funct. Approx. Comment.

Math. 37 (2007), part 1, 213− 231.

[6] M. N. Huxley, Area, lattice points and exponential sums, London Mathematical

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184

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185

Conclusion

Among other things, we have established in Chapter 1 an asymptotic estimate for each

variance cited in the introduction that are related to a xed function f ∈ Ck.

Given k and f , we write

γf :=∏p

(1− 1

p

)(∑a≥0

1

pa

(|g(pa)|2 + 2<

∑i≥1

g(pa)g(pa+i)

pik

)).

Theorem. Let f be a function in Ck. Then for each β ∈ 0, . . . , h− 1 there exists a

positive constant θk ≤ 1k+2

such that

T (h,M ; β) = 2ζ(1/k − 1)

1/k − 1γfh

1/k(M + 1) +Oε

(Mhθk+ε +Mh1/2k+ε

)+Oε

(h2M1/k logN + h1+εM2/(k+1) logN

)for each ε > 0.

Theorem. Let f be a function in Ck. Assume that x = qM where M is a real number

with M ≥ 1. Then there exists a positive constant θk ≤ 1k+2

such that

W(x; q, f) = 2ζ(1/k − 1)

1/k − 1γf (q)qM

1/k +R(x; q, f),

where

γf (q) :=∏p-q

((1− 1

p

)∑j≥0

1

pj

(|g(pj)|2 + 2<g(pj)

∑i≥1

g(pj+i)

pik

))

×∏pα‖q

(r−1)k<α≤rk

(1− 1

p

)( r−1∑j=0

1

pjk

(|g(pj)|2 + 2<g(pj)

∑i≥1

g(pj+i)

pik

)

+1

pα−α/k

∑i≥r

1

pi

(|g(pi)|2 + 2<g(pi)

∑l≥1

g(pi+l)

plk

))

186

and

R(x; q, f)ε M12k

+εq1+ε +M θkq1+ε +x1+1/k log x

q1−ε

for each ε > 0.

A basic answer to the guiding question of Chapter 2 is given by the next two theorems.

Theorem. For every integer k ≥ 2, there exists a positive constant c1,k such that

gk(h)− h

ζ(k)< c1,k

h2/(k+1)

(log h)2k/(k+1).

Theorem. For every k ≥ 2 we have

gk(h)− h

ζ(k)> c2,k

h1/k


for some positive constant c2,k.

Our main motivation during the construction of this chapter has been to improve the

upper bound in the rst of these theorems, because we think that the truth should be

close to the following conjecture.

Conjecture. For every integer k ≥ 2, we have

gk(h) ≤ h

ζ(k)+ c(ε, k)h1/k+ε

for every ε > 0.

We have been able to get closer to this last conjecture. For example, we have established

the following theorem.

Theorem. Let k ≥ 2 be a xed integer. Assume that a very technical conjecture holds.

Then,

gk(h) ≤ h

ζ(k)+O(h

32k+1

+ε)

for each xed ε > 0.

We have addressed the similar problem with the interval replaced by an arithmetic

progression as well.

187

In Chapter 3 we have established two main theorems. They are essentially the best

possible answer to our problem.

Theorem. For every positive integer n with k = ω(n) ≥ 74,

(1) τ(n) <

(1 +

log n

k log k

)k.

Theorem. The largest integer n with k = ω(n) ≥ 44 for which (1) does not hold is the

integer n made up of the rst 44 primes and has the exponent vector

(354, 223, 152, 125, 102, 95, 86, 83, 77, 72, 71, 67, 65, 64, 63, 61, 59, 59, 57, 57, 56,

55, 55, 54, 53, 52, 52, 52, 51, 51, 50, 49, 49, 49, 48, 48, 48, 47, 47, 47, 46, 46, 46, 46),

that is

n = 2354 · 3223 · · · 19346.

In Chapter 4, we have found many complicated results. We present here a simplied

version of our main theorem.

Let's write

(2) λi1,j |f(i1)j (x)| and |f (i2)

j (x)| λi2,j (x ∈ X )

where i1 ∈ 0, . . . , 3, i2 ∈ 0, . . . , 4 and j ∈ 1, 2. Here, 0 < λi,j = Xλi+1,j for

i ∈ 0, 1, 2, 3 and j ∈ 1, 2. Also, we dene

∆i(x) := f(i)1 (x)f

(i+1)2 (x)− f (i+1)

1 (x)f(i)2 (x) (x ∈ X , i = 1, 2)

and assume that

(3) ∆i |∆i(x)| ∆i, (x ∈ X , i = 1, 2)

where ∆i = λ1,iλ2,i+1 = λ1,i+1λ2,i.

Theorem. Let X ∈ [X, 2X] be an interval and assume that 0 ≤ δ1, δ2 ≤ 14. Consider

two functions f1, f2 ∈ C4(X ) satisfying (2) and (3). Assume the previous notation and

α0 := δ1 + δ2, α := α0 + δ1λ1,2 + δ2λ1,1, δ := min(δ1, δ2), β0 :=δ1δ

λ2,1

, β1 :=δ2

1

λ2,1

,

188

β2 :=δ2

2

λ2,2

, Ω := λ1,2 + 1 +λ2,2

λ2,1

, Y := max

(λ1,1

λ1,2

,λ1,2

λ1,1

, λ1,1,1

λ1,1

, λ1,2,1

λ1,2

).

Assume also thatδ1

λ0,1

δ2

λ0,2

1.

Then for a xed i ∈ 0, 1, 2 we have

S ∆1/62 X + (α0∆1)1/3X + (α0λ2,1)1/3X + (α0λ2,2)1/3X + 1

+α1/20 αX log2X + min

((δ2

λ0,2

)2/15

α1/50 αX,α

1/30 α

X

Y 2/3

)log2X

min

((δ2

λ0,2

) 130

(δ1δΩ)12α

3100 X,α

130 (δ1δΩ)

12X

Y16

)logX

+α

Y

(δ2

λ0,2

)1/3

X log2X +

(δ1α0Ω

Y

)1/2(δ2

λ0,2

)1/3

X logX.

δ1/41 δ

1/22 λ

1/42,1X logX + λ

1/32,i

(δ1

λ2,1

)1/2

∆1/62 X + ΞiX.

where

Ξi :=∑′

1≤Q≤L(

δ1λ2,1

)1/2

QL2/βi

min

(βi max

(∆

1/42

Q1/2

L,α

1/22

Q

L3/2,δ1δ2

Q2

L3

),max

(∆

1/52

L

Q4/5,α

1/32

L2/3

Q2/3

),λ2,iβi

Q2

L

)

and where the apostrophe ′ indicates that the values taken by Q and L are only powers

of 2. The implied constants can be computed in term of f1(x) and f2(x).

189

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