Analytic Number Theory Solutions

Sean LiCornell Universitysxl6@cornell.edu

Jan. 2013

Introduction

This document is a work-in-progress solution manual for Tom Apostol’s Intro-duction to Analytic Number Theory. The solutions were worked out primarilyfor my learning of the subject, as Cornell University currently does not offer ananalytic number theory course at either the undergraduate or graduate level.However, this document is public and available for use by anyone. If you are astudent using this document for a course, I recommend that you first try workout the problems by yourself or in a group. My math documents are stored ona math blog at www.epicmath.org.

2 Arithmetical Functions and Dirichlet Multi-plication

Notationϕ(n) Euler totient functionµ(n) Mobius functionΛ(n) von Mangoldt functionλ(n) Louiville’s function∗ Dirichlet multiplicationI(n) identity function, I(1) = 1, I(n) = 0 if n 6= 1u(n) unit function, u(n) = 1 for all nN(n) the function for which N(n) = n for all nfp(x) Bell series of f modulo p

1

2.1. Find all integers such that

(a) ϕ(n) = n/2.

We use Theorem 2.4, which states

ϕ(n)

n=∏p|n

(1− 1

p

).

If 2 is the only prime factor, this product is equal to 1/2. If there were anyother prime factor, then the denominator would be greater than 2. Henceϕ(n) = n/2 if and only if n = 2k for some k ≥ 1.

(b) ϕ(n) = ϕ(2n).

Applying Theorem 2.4, we have∏p|n

(1− 1

p

)= 2

∏p|2n

(1− 1

p

).

All the prime factors other than 2 are the same in both products, so the onlyprime that matters is 2. If n is even, then both products are equal, which isa contradiction as there is an additional factor of 2 on the right-hand-side(RHS). If n is odd, then the product on the RHS is half as large, whichcancels out the factor of 2. Hence ϕ(n) = ϕ(2n) if and only if n is odd.

(c) ϕ(n) = 12.

Factor n = pa11 · · · pakk . Then Theorem 2.4 gives the product formula

ϕ(n) =

k∏i=1

pai−1i (pi − 1).

Since 12 = 22 · 3, we enumerate through all posibilities of pi and ai thathave 2, 3, 4, 6, or 12 as factors in the product. Of these, m = 2, 4, 6, and 12satisfy m+ 1 is prime, so can come from the pi − 1 term.

Let pi−1 = 12. Then pi = 13 and thus n = 13 or n = 2 ·13. Let pi−1 = 6.Then pi = 7 and we need a factor of 2. It can come from pj − 1 = 2, thusn = 3 · 7 = 21 or n = 2 · 3 · 7 = 42. But the factor of 2 could also comefrom 22−1 = 2, thus n = 22 · 7 = 28. Let pi − 1 = 4, then pi = 5 and weneed a factor of 3. Since 3 + 1 = 4 is not prime, it must come from p2−1j ,but this is impossible as it adds in another factor of 3− 1 = 2. Finally, letpi − 1 = 2. Then pi = 3, and we need a factor of 6. The only way to dothis is to have n = 22 · 32 = 36. Hence the integers with ϕ(n) = 12 aren = 13, 21, 26, 28, 36, and 42.

2.2. For each of the following statements either give a proof or exhibit a coun-terexample.

2

(a) If (m,n) = 1 then (ϕ(m), ϕ(n)) = 1.

False. Clearly (5, 9) = 1, but (ϕ(5), ϕ(9)) = (2, 6) = 2.

(b) If n is composite, then (n, ϕ(n)) = 1.

False. 6 is composite, but (6, ϕ(6)) = (6, 2) = 2.

(c) If the same primes divide m and n, then nϕ(m) = mϕ(n).

True. Proof follows by rearrangement of Theorem 2.4.

2.3. Prove thatn

ϕ(n)=∑d|n

µ2(d)

ϕ(n).

Let n = pa11 · · · pakk . Note that the nonzero terms of the sum come from d = 1

and divisors which are products of distinct prime divisors of n. Moreover, eachnumerator for a nonzero term is equal to µ(n)2 = 1. Thus the RHS is equal to

∑d|n

µ2(d)

ϕ(n)=1 + 1/ϕ(p1) + · · ·+ 1/ϕ(pk) + 1/ϕ(p1p2) + · · ·

+ 1/ϕ(pk−1pk) + · · ·+ 1/ϕ(p1 · · · pk).

By Theorem 2.4, we have

1

ϕ(p1 · · · pj)=

1

p1 · · · pj∏ji=1

pi−1pi

=1

(p1 − 1) · · · (pj − 1).

Hence the RHS equals 1(p1−1)···(pj−1) times

1 +

k∑i=1

(pi − 1) +

k−1∑i=1

k∑j=i+1

(pi − 1)(pj − 1) + · · ·+ (p1 − 1) · · · (pj − 1).

This is the result of the expansion

((p1 − 1) + 1) · · · ((pk − 1) + 1) = p1 · · · pk,

so the RHS equalsp1 · · · pk

(p1 − 1) · · · (pj − 1).

But from Theorem 2.4, the LHS equals

n

ϕ(n)=

n

n∏p|n

p−1p

=∏p|n

p

p− 1,

3

and thus the equation holds.

2.4. Prove that ϕ(n) > n/6 for all n with at most 8 distinct prime factors.

From Theorem 2.4, this is equivalent to showing

1

6<ϕ(n)

n=∏p|n

p− 1

p

when there are at most 8 p’s. It is easy to see that the product is minimizedwhen the smallest 8 primes are selected. Hence∏

p|n

p− 1

p≥ 1

2· 2

3· 4

5· 6

7· 10

11· 12

13· 16

17· 18

19≈ 0.171 >

1

6.

2.5. Define v(1) = 0, and for n > 1 let v(n) be the number of distinct primefactors of n. Let f = µ ∗ v and prove that f(n) is either 0 or 1.

Proof. Factor n = pa11 · · · pakk . We induct on m = a1 + · · · + ak for m ≥ 2.

Note that for m = 0 we have f(1) = µ(1)v(1) = 0, and for m = 1 we havef(p1) = µ(1)v(p1) + µ(p1)v(1) = 1 + 0 = 1. Now we claim that f(n) = 0 whenm ≥ 2, and the proof is by induction.

The base case of the induction, m = 2, has two possibilities: either n = p21 orn = p1p2. In the first possibility we have

f(p21) = µ(1)v(p2) + µ(p)v(p) = 1− 1 = 0,

and the second possibility we have

f(p1p2) = µ(1)v(p1p2) + µ(p1)v(p2) + µ(p2)v(p1) = 2− 1− 1 = 0.

Now suppose it is true for m, we shall show it is true for m + 1. Let n′ be anumber with the prime factorization as such. There are two cases: either n′

contains an additional prime factor, i.e. n′ = npk+1 = pa11 · · · pakk pk+1, or n′ has

a prime factor raised to a higher power, i.e. n′ = pa11 · · · pai+1i · · · pakk for some

i, 1 ≤ i ≤ k. In the first case, a divisor of n′ is either of divisor of n, or is pk+1

times a divisor of n. Hence

f(n′) =∑d|n

(µ(d)v

(npk+1

d

)+ µ(dpk+1)v

(nd

)).

Since v(npk+1

d ) = v(nd ) + 1, the first sum is just f(n) +∑d|n µ(d) = f(n) as m

was assumed to be ≥ 2. And since µ(dpk+1) = −µ(d), we have∑d|n

µ(dpk+1)v(nd

)=∑d|n

−µ(d)v(nd

)= 0

4

by the induction hypothesis. Hence the induction step holds when the additionalpower is from a new prime.

If the power of an existing prime is raised, say ai → ai + 1, we abbreviates = n′/pai+1

i . Consider each term in the sum f(n′), grouping terms by j in the

form µ(pji t)v(n/(pji t)), with 0 ≤ j ≤ ai + 1 and t|s. Summing the terms withj = ai + 1 results in 0 as ai + 1 ≥ 2. For terms with j < ai, the terms have thesame value as that for f(n), i.e. µ(pji t)v(n/(pji t)) = µ(pji t)v(n/(pj+1

i t)). Finally,the terms with j = ai satisfy

µ(pji t)v(n/(pji t)) = µ(pji t)v(n/(pj+1i t)) + µ(pji t)

as the value of v is one higher. Hence the total sum for terms with j ≤ ai is∑d|n

µ(pji t)v(n/(pj+1i t)) +

∑d|n,paii |d

µ(paii d)v(d/(paii )).

The first sum is 0 by the induction hypothesis, and the second sum is 0 as well:if ai > 1 then µ = 0, and if ai = 1 then it is the sum

∑d|s µ(d) = 0. Thus the

induction step holds and f(n′) = 0.

2.6. Prove that ∑d2|n

µ(d) = µ2(n)

and, more generally,∑dk|n

µ(d) =

{0 if mk|n for some m > 1,

1 otherwise.

The last sum is extended over all positive divisors d of n whose kth power alsodivide n.

Proof. Let n = pa11 · · · pajj . If all a1, . . . , aj < k, then sum is empty except for

the d = 1 term, which gives the value of 1.

Suppose ai ≥ k for a set of i of size m, say ai1 . . . , aim . Then∑dk|n

µ(d) = 1 +

(m

1

)(−1) +

(m

2

)(−1)2 + · · ·+

(m

m

)(−1)m = (1− 1)m = 0.

2.7. Let µ(p, d) denote the value of the Mobius function at the gcd of p and d.Prove that for every prime p we have

∑d|n

µ(d)µ(p, d) =

1 if n = 1,

2 if n = pa, a ≥ 1,

0 otherwise.

5

Proof. The n = 1 case is trivial. If n = pa with a ≥ 1, then the sum is inthe form

∑ai=0 µ(pi)µ(pi, i). The first two terms each contribute +1, and the

remaining terms are all 0. Hence the sum is 2.

Now suppose neither of these cases are true, then n = paqb11 · · · qbkk . Note that

each factor is in the form pαqβ1

1 · · · qβkk , where 0 ≤ α ≤ a and 0 ≤ βi ≤ bi. We

split the sum into groups based on α. For α = 0, the sum is∑d|qb11 ···q

bkk

µ(d)µ(p, d) =∑

d|qb11 ···qbkk

µ(d) = 0

as µ(p, d) = µ(1) = 1. For α = 1, the sum is equal to∑d|qb11 ···q

bkk

µ(pd)µ(p, dp) =∑

d|qb11 ···qbkk

−µ(d) = 0

as µ(p, dp) = µ(p) = −1. Finally, for α ≥ 2, µ = 0 as p2 is a square divisor.Hence the sum is 0.

2.8. Prove that ∑d|n

µ(d) logm d = 0

if m ≥ 1 and n has more than m distinct prime factors.

Proof. We prove the claim by induction. Let m = 1 and there is more than 1distinct prime factor. Then n = pa11 · · · p

akk , k ≥ 2. Note that µ(d) 6= 0 only if d

is the product of distinct primes, i.e. d = pi1 . . . pij where i1, . . . , ij are a subsetof 1, . . . , k. Finally, let x = p1 · · · pk. Then∑

d|n

µ(d) log d =∑d|x

µ(d) log(pi1 · · · pij )

=∑d|x

µ(d)(log pi1 + · · ·+ log pij )

=∑d|x

∑pi|d

µ(d) log pi

=∑pi|x

log pi∑d|xpi|d

µ(d).

But the condition that pi|x, pi|d is equivalent to the condition that d|(x/pi).Since n was assumed to have more than 1 distinct prime factor, d|(x/pi) is notequal to 1, hence the double sum is equal to∑

pi|x

log pi∑

d|(x/pi)

µ(d) = 0

6

as each∑d|(x/pi) µ(d) is 0.

Now assume that∑d|n µ(d) logm d = 0 for m ≥ 1, with more than m distinct

prime factors. Now suppose n has more than m+ 1 distinct prime factors, i.e.n = pa11 · · · p

akk , k ≥ m+ 2. Let x = p1 · · · pk. The only nonzero µ(d) come from

d which are the product of distinct primes, i.e. d = pi1 . . . pij where i1, . . . , ijare a subset of 1, . . . , k. Then∑

d|n

µ(d) logm+1 d =∑d|x

µ(d) logm+1(pi1 · · · pij )

=∑d|x

µ(d) logm d(log pi1 + · · ·+ log pij )

=∑d|x

∑pi|d

µ(d) logm d log pi

=∑pi|x

log pi∑d|xpi|d

µ(d) logm d

=∑pi|x

log pi∑

d|(x/pi)

µ(d) logm d.

Since k ≥ m+2, the number of distinct prime factors of x/(pi) is at least m+1.Thus by the induction hypothesis,∑

d|(x/pi)

µ(d) logm d = 0

and hence the double sum is also 0.

2.9. If x is real, x ≥ 1, let ϕ(x, n) denote the number of positive integers ≤ xthat are relatively prime to n. Prove that

ϕ(x, n) =∑d|n

µ(d)[xd

]and

∑d|n

ϕ(xd,n

d

)= [x].

Proof. For the first part, note that

ϕ(x, n) =

x∑k=1

[1

(n, k)

]=

x∑k=1

∑d|(n,k)

µ(d) =

x∑k=1

∑d|nd|k

µ(d).

Let k = qd, then since 1 ≤ k ≤ x, we have 1 ≤ q ≤ [x/d], so the sum equals

∑d|n

[x/d]∑q=1

µ(d) =∑d|n

µ(d)

[x/d]∑q=1

1 =∑d|n

µ(d)[xd

].

7

For the second part, use the same technique as the proof on p. 26: partitionthe numbers 1, . . . , [x] into disjoint sets based on their gcd with n, the resultfollows.

2.10.−2.12. In Exercises 10, 11, and 12, d(n) denotes the number of positivedivisors of n.

2.10. Prove that∏t|n t = nd(n)/2.

Proof. If t is a divisor of n, then so is n/t. Thus∏t|n t =

∏t|n n/t, and∏

t|n

t

2

=

∏t|n

t

∏t|n

n

t

=∏t|n

n = nd(n).

2.11. Prove that d(n) is odd if and only if n is a square.

Proof. If n is not a square, then every divisor t of n has a corresponding divisorn/t, thus d(n) is even. If n is a square, then every divisor except n1/2 has acorresponding divisor. Along with n1/2, this means d(n) is odd.

2.12. Prove that∑t|n d(t)3 =

(∑t|n d(t)

)2.

Proof. We may enumerate through all factors of n as follows. Let n = pa11 · · · pakk .

Then every divisor is of the form pb11 · · · pbkk where 0 ≤ bi ≤ ai. Hence the LHS

sum isa1∑b1=0

· · ·ak∑bk=0

(b1 · · · bk)3

while the RHS sum is (a1∑b1=0

· · ·ak∑bk=0

b1 · · · bk

)2

.

This is a known arithmetic equality and can be proven by induction on m =a1 + · · ·+ ak.

2.13. Product form of the Mobius inversion formula. If f(n) > 0 for all n andif a(n) is real, a(1) 6= 0, prove that

g(n) =∏d|n

f(d)a(n/d) if and only if f(n) =∏d|n

g(d)b(n/d),

where b = a−1, the Dirichlet inverse of a.

8

Proof. Rewrite the relation g(n) =∏d|n f(d)a(n/d) as

elog g(n) = e∑d|n a(n/d) log f(d).

Taking the log of both sides (as f > 0) gives

log g(n) =∑d|n

a(n/d) log f(d),

so that log g = a ∗ log f . Now Dirichlet multiply both sides by b, resulting in

b ∗ log g = b ∗ (a ∗ log f) = (b ∗ a) ∗ log f = log f

as b ∗ a = I by the hypothesis. Exponentiating gives the desired result for theonly if direction. The other direction follows by symmetry.

2.14. Let f(x) be defined for all rational x in 0 ≤ x ≤ 1 and let

F (n) =

n∑k=1

f

(k

n

), F ∗(n) =

n∑k=1

(k,n)=1

f

(k

n

)

(a) Prove that F ∗ = µ ∗ F , the Dirichlet product of µ and F .

Proof.

u ∗ F ∗ =∑d|n

F ∗(d) =∑d|n

d∑k=1

(k,d)=1

f

(k

d

)=∑d|n

d∑k=1

(k,d)=1

f

(k(nd

)n

).

We see that each k nd is unique, and moreover they run through every integerfrom 1 to n. Thus the sum equals precisely F (n), so that u ∗ F ∗ = F .Dirichlet mulplication by µ gives

µ ∗ F = µ ∗ (u ∗ F ∗) = (µ ∗ u) ∗ F ∗ = F ∗.

(b) Use (a) or some other means to prove that µ(n) is the of the primitive nthroots of unity:

µ(n) =

n∑k=1

(k,n)=1

e2πik/n.

Let f(x) = e2πix. Then F (n) =∑nk=1 e

2πix is equal to 1 when n is 1, andis 0 when n > 1, as all the terms cancel. Then

µ = µ ∗ F =

n∑k=1

(k,n)=1

e2πik/n

as F = I.

9

2.15. Let ϕk(n) denote the sum of the kth powers of the numbers ≤ n andrelatively prime to n. Note that ϕ0(n) = ϕ(n). Use Exercise 14 or some othermeans to prove that ∑

d|n

ϕk(d)

dk=

1k + · · ·+ nk

nk.

Proof. ∑d|n

ϕk(d)

dk=∑d|n

ϕk(d)(n/d)k

nk=

1

nk

∑d|n

ϕk(d)(n/d)k.

It suffices to show that the sum includes each mk, 1 ≤ m ≤ n, once. This is asin Exercise 2.14(a).

2.16. Invert the formula in Exercise 15 to obtain, for n > 1,

ϕ1(n) =1

2nϕ(n), and ϕ2(n) =

1

3n2ϕ(n) +

n

6

∏d|n

(1− p).

Derive a corresponding formula for ϕ3(n).

Proof. The following summation formulas are significant:∑ni=1 i = n(n+ 1)/2,∑n

i=1 i2 = n(n+ 1)(2n+ 1)/6, and

∑ni=1 i

3 = n2(n+ 1)2/4. By Exercise 2.15,

ϕ1(n)

n=∑d|n

1 + · · ·+ d

dµ(nd

)=∑d|n

d(d+ 1)

2dµ(nd

)=

1

2

∑d|n

(d+ 1)µ(nd

)=

1

2

∑d|n

dµ(nd

)=

1

2ϕ(n).

Hence ϕ1(n) = 12nϕ(n).

10

Next, we have

ϕ2(n)

n2=∑d|n

12 + · · ·+ d2

d2µ(nd

)=∑d|n

d(d+ 1)(2d+ 1)

6d2µ(nd

)=

1

6

∑d|n

(2d+ 3 +

1

d

)µ(nd

)

=1

6

∑d|n

2dµ(nd

)+∑d|n

1

dµ(nd

)=

1

3ϕ(n) +

1

6

∑d|n

d

nµ(d)

=1

3ϕ(n) +

1

6n

∑d|n

dµ(d)

=1

3ϕ(n) +

1

6n

∏p|n

(1− p).

Hence ϕ2(n) = 13n

2ϕ(n) + n6

∏pn

(1− p).

Finally, we have

ϕ3(n)

n3=∑d|n

13 + · · ·+ d3

d3µ(nd

)=∑d|n

d2(d+ 1)2

4d3µ(nd

)=

1

4

∑d|n

(d+ 2 +

1

d

)µ(nd

)

=1

4

∑d|n

dµ(nd

)+∑d|n

1

dµ(nd

)=

1

4ϕ(n) +

1

4n

∏p|n

(1− p).

Hence ϕ3(n) = 14n

3ϕ(n) + 14n

2∏pn

(1− p).

2.17. Jordan’s totient Jk is a generalization of Euler’s totient defined by

Jk(n) = nk∏p|n

(1− p−k).

11

(a) Prove that

Jk(n) =∑d|n

µ(d)(nd

)kand nk =

∑d|n

Jk(d).

Proof. Write n = pa11 · · · pamm . Since µ(d) is nonzero only when d is theproduct of distinct primes, we have

Jk(n)

nk=∑d|n

µ(d)

(1

d

)k=1−

(1

pk1+ · · ·+ 1

pkm

)+(

1

pk1pk2

+ · · ·+ 1

pkm−1pkm

)+ · · ·+ (−1)m

1

pk1 · · · pkm=∏p|n

(1− p−k

),

which shows the first result. The second result follows by Mobius inversionof the first: Jk = Nk ∗ µ implies Nk = Jk ∗ u.

(b) Determine the Bell series for Jk.

The Bell series for µ is µp(x) = 1− x for all p, and Bell series for Nkp (x) =

1+pkx+p2kx2+· · · = 11−pkx . By Theorem 2.25, the Bell series of a Dirichlet

product is the product of the Bell series, so

(Jk)p(x) = µp(x) ·Nkp (x) =

1− x1− pkx

.

2.18. Prove that every number of the form 2a−1(2a − 1) is perfect if 2a − 1 isprime.

Proof. A number n is perfect if σ(n) = 2n. Every divisor of n = 2a−1(2a− 1) isin the form 2i or 2i(2a − 1) for 0 ≤ i ≤ a− 1. Hence

σ(n) =

a−1∑i=0

(2i(2a − 1) + 2i) =

a−1∑i=0

2i2a = 2aa−1∑i=0

2i = 2a(2a − 1) = 2n.

2.19. Prove that if n is even and perfect then n = 2a−1(2a− 1) for some a ≥ 2.

Proof. Write n = 2a−1c where c is odd, a − 1 ≥ 1. Suppose c is prime. Thenthe factors of n are in the form 2i and 2ic, so that

σ(n) =

a−1∑i=0

2i(c+ 1) = (c+ 1) (2a − 1) = 2ac.

12

Since (c+ 1, c) = 1 and we have c|2a − 1 and c+ 1|2a. Moreoever, 2a - 2a−1, sothat 2a|c+ 1, thus c+ 1 = 2a and c = 2a − 1.

Now suppose c is the product of distinct odd primes, so that n = 2a−1p1 · · · pk,so

σ(n) =

a−1∑i=0

2i(1 + p1 + · · · pk + p1p2 + · · · pk−1pk + · · ·+ p1 · · · pk

= (2a − 1)∏p|n

(1 + p)

Since (2a, 2a− 1) = 1 for a ≥ 2, and (p1 · · · pk, 2) = 1, we have p1 · · · pk = 2a− 1and (1+p1) · · · (1+pk) = 2a, which is possible only if k = 1, i.e. that c consists ofa single prime. If there was more than one prime, then

∏p|n(1+p)−

∏p|n p > 1,

contradicting that 2a − (2a − 1) = 1. Thus c = pi = 2a − 1.

Finally, suppose c has some prime factor with a power higher than 1, say pb11 ,b1 > 1. Then the sum of divisors is even greater than in the previous case, asσ(n) =

∑a−1i=0

∑b1j=0 2ipj1 · · · > 2ac = 2n. Thus c must be a prime of the form

2a−1(2a − 1).

2.20. Let P (n) be the product of the positive integers which are ≤ n andrelatively prime to n. Prove that

P (n) = nϕ(n)∏d|n

(d!

dd

)µ(n/d).

Proof. From the product form of the Mobius inversion formula (Exercise 2.13),

P (n)

nϕ(n)=∏d|n

(d!

dd

)µ(n/d)if and only if

n!

nn=∏d|n

P (d)

dϕ(d).

Note thatn!

nn=

1

n· 2

n· · · n

n=

n∏k=1

k

n.

Now put all the fractions 1/n, · · · , n/n in reduced form. Grouping same de-nominators together, we have

n∏k=1

k

n=∏d|n

d∏a=1

(a,d)=1

a

d=∏d|n

P (d)

dϕ(d),

which is the desired result.

13

2.21. Let f(n) = [√n] − [

√n− 1]. Prove that f is multiplicative but not

completely multiplicative.

Proof. It is clear that f(n) equals 1 if n is a square and 0 otherwise. Let n = ab,where (a, b) = 1. Thus the prime factorizations of a and b consist of disjoint setsof primes. Then clearly if both a and b are squares (i.e. each prime in the primefactorization is raised to an even power), the product ab is also a square. Andif one of a or b is not a square then the product ab is not a square, hence thefunction f(n) is multiplicative. It is not completely multiplicative, as f(4) = 1,but f(2)f(2) = 0.

2.22. Prove thatσ1(n) =

∑d|n

ϕ(d)σ0

(nd

),

and derive a generalization involving σα(n). (More than one generalization ispossible.)

We have σ1 = N ∗ u, σ0 = u ∗ u, and ϕ = µ ∗N . Then

σ0 ∗ ϕ = (u ∗ u) ∗ (µ ∗N) = u ∗ (u ∗ µ) ∗N = u ∗N = σ1,

which is the desired result. In general, σα = u ∗ Nα. Since Nα is completelymultiplicative, it follows that

σα = u ∗Nα = (u ∗Nα−1) ∗ ((Nα−1)−1 ∗Nα) = σα−1 ∗ (µNα−1 ∗Nα).

But

µNα−1 ∗Nα =∑d|n

µ(d)dα−1(nd

)α= nα−1

∑d|n

µ(d)n

d

= nα−1ϕ(n).

Hence one generalization is

σα = σα−1 ∗Nα−1ϕ.

2.23. Prove the following statement or exhibit a counter example. If f ismultiplicative, then f(n) =

∏d|n f(d) is multiplicative.

False. Let f = N , i.e. f(n) = n. This is clearly multiplicative, but F (2)F (3) =2 ∗ 3 = 6, but F (6) = 2 ∗ 3 ∗ 6 = 36, so F is not multiplicative.

2.24. Let A(x) and B(x) be fomal power series. If the product A(x)B(x) isthe zero series, prove that at least one factor is zero. In other words, the ringof formal power series has no zero divisors.

14

Proof. Assume both A(x) and B(x) are not zero, then let m be the smallestinteger such that a(m) 6= 0, and let n be the smallest integer such that b(n) 6= 0.Each coefficient in A(x)B(x) is zero, so the xm+n must be zero. But thiscoefficient equal to the sum

a0bm+n + · · ·+ ambn + · · ·+ am+nb0 = 0.

Since we assumed that every ai = 0 with i < m and every bj = 0 with j < n,every term above except the ambn term is equal to zero. Removing all thoseterms results in ambn = 0, contradicting that both am and bn are not equal tozero.

2.25. Assume f is multiplicative. Prove that:

(a) f−1(n) = µ(n)f(n) for every squarefree n.

Proof. Let g(n) = µ(n)f(n). Since n is squarefree, n = p1 · · · pk, the productof distinct primes, and if d is a divisor of n, then (d, n/d) = 1. Then

(g ∗ f)(n) =∑d|n

µ(d)f(d)f(nd

)= f(n)

∑d|n

µ(d) = f(n)I(n) = I(n)

since f(1) = 1 and I(n) = 0 for n > 1. Hence g(n) = f−1(n) for squarefreen.

(b) f−1(p2) = f(p)2 − f(p2) for every prime p.

Proof. Write f2(p) = f(p2). Let g(p) = f(p)2 − f2(p). Then

(g ∗ f2)(p) =∑d|p

(f(d)2 − f2(d))f2(p/d)

= (f(1)2 − f2(1))f2(p) + (f(p)2 − f2(p))f2(1)

= f(p)2 − f2(p) = g(p).

However, since f was an arbitrary multiplicative function, it must be thatg(p) = 0. This implies g ∗ f2 is the identity on prime p, so that g = f−12 .

2.26. Assume f is multiplicative. Prove that f is completely multiplicative if,and only if, f−1(pa) = 0 for all primes p and all integers a ≥ 2.

Proof. Assume f−1(pa) = 0 for all primes p and all integers a ≥ 2. Sincef ∗ f−1 = I we have ∑

d|paf(d)f−1(pa/d) = 0

for prime p and a ≥ 2. Since f−1(pb) = 0 for all 2 ≤ b ≤ a, the terms are 0except for d = pa−1 and d = pa, so the sum is equal to

f(pa)f−1(1) + f(pa−1)f−1(p) = 0.

15

From Exercise 2.25(a), we have that f−1(p) = µ(p)f(p) = −f(p), hence

f(pa) = f(pa−1)f(p).

By induction, this implies f(pa) = f(p)a, thus f is completely multiplicative.

Conversely, suppose f is completely multiplicative. Then by Theorem 2.17,

f−1(pa) = µ(pa)f(pa).

Since a ≥ 2 implies µ(pa) = 0, we have f−1(pa) = 0.

2.27.

(a) If f is completely multiplicative, prove that

f · (g ∗ h) = (f · g) ∗ (f · h)

for all arithmetical functions g and h, where f · g denotes the prdinaryproduct, (f · g)(n) = f(n)g(n).

Proof. If f is completely multiplicative and d is a divisor of n, thenf(d)f(n/d) = f(n). Hence

((f · g) ∗ (f · h))(n) =∑d|n

f(d)g(d)f(nd

)h(nd

)= f(n)

∑d|n

g(d)h(nd

)= (f · (g ∗ h))(n)

(b) If f is multiplicative and the relation in (a) holds for g = µ and h = µ−1,prove that f is completely multiplicative.

Proof. We have f · (µ ∗ u) = (f · µ) ∗ (f · u).

I(n) =∑d|n

µ(d)f(d)f(n/d)

for all n. Letting n = pa be a prime power, if a ≥ 2 then all the terms arezero except for d = 1 and d = p, else µ(d) = 0. Hence we have

f(pa)− f(a)f(pa−1) = 0,

which by induction on a implies that f(pa) = f(p)a, hence f is completelymultiplicative.

2.28.

16

(a) If f is completely multiplicative, prove that

(f · g)−1 = f · g−1

for every arithmetical function g with g(1) 6= 0.

Proof. Let h = f · g−1. Then h ∗ (f · g) = (f · g−1) ∗ (f · g). By Exercise2.27(a), if f is completely multiplicative then

(f · g−1) ∗ (f · g) = f · (g ∗ g−1) = f · I = I.

Hence h = (f · g)−1. The condition that g(1) 6= 0 is used to guarantee thatg−1 exists.

(b) If f is multiplicative and the relation in (a) holds for g = µ−1, prove that fis completely multiplicative.

Proof. Suppose (f · u)−1 = f · µ. Then (f · u) ∗ (f · µ) = I. For each n = pa

a prime power with a ≥ 2,

((f · u) ∗ (f · µ))(n) =∑d|n

µ(d)f(d)f(nd

)= 0,

and since µ(d) 6= 0 only for d = 1 and d = p, we have

f(pa)− f(a)f(pa−1) = 0,

which by induction on a implies that f(pa) = f(p)a, hence f is completelymultiplicative.

2.29. Prove that there is a multiplicative arithmetical function g such that

n∑k=1

f((k, n)) =∑d|n

f(d)g(nd

)for every arithmetical function f . Here (k, n) is the gcd of n and k. Use thisidentity to prove that

n∑k=1

(k, n)µ((k, n)) = µ(n).

Proof. The technique on p. 26 shows that g = ϕ. For the second part, thefunction in question is f = N · µ, so it remains to show that f ∗ ϕ = µ. Now Nis completely multiplicative, so its inverse is µ ·N . Since ϕ = N ∗ µ, we have

f ∗ ϕ = (N · µ) ∗ (N ∗ µ) = (N−1 ∗N) ∗ µ = I ∗ µ = µ.

2.30. Let f be multiplicative and let g be any arithemtical function. Assumethat

f(pn+1) = f(p)f(pn)− g(p)f(pn−1)

17

for all primes p and all n ≥ 1. Prove that for each prime p the Bell series for fhas the form

fp(x) =1

1− f(p)x+ g(p)x2.

Also prove the converse.

Proof. With the recursion formula, we have

fp(x) = 1 + f(p)x+ (f(p)f(p)− g(p))x2 + (f(p)f(p2)− g(p)f(p))x3 + · · ·= 1 + f(p)fp(x)x− g(p)fp(x)x2.

Rearranging givesfp(x)(1− f(p)x+ g(p)x2) = 1,

so that

fp(x) =1

1− f(p)x+ g(p)x2.

These steps are all reversible, so the converse holds.

2.31. (Continuation of Exercise 30.) If g is completely multiplicative prove thatthe first statement of Exercise 30 implies

f(m)f(n) =∑

d|(m,n)

g(d)f(mnd2

),

where the sum is extended over the positive divisors of the gcd(m,n).

Proof. First we consider the case m = pa and n = pb. Assume without lossof generality that a ≥ b. We prove by induction on b. For b = 1 the firststatement of Exercise 30 gives f(pa)f(p) = f(p)f(pa+1) + g(p)f(pa−1). Nowsince g(1) = 1, we have

∑d|p

g(d)f

(pa+1

d2

)= g(1)f(pa+1) + g(p)f(pa−1)

18

as desired. Now suppose the equation is true for 1, . . . , b− 1. Then

f(pa)f(pb) =f(pa)(f(p)f(pb−1)− g(p)f(pb−2))

=(f(pa+1) + g(p)f(pa−1))f(pb−1)− g(p)f(pa)f(pb−2)

=∑d|pb−1

g(d)f

(pa+b

d2

)+ g(p)

∑d|pb−1

g(d)f

(pa+b−2

d2

)

− g(p)∑d|pb−2

g(d)f

(pa+b−2

d2

)

=

b−1∑i=0

g(pi)f(pa+b−2i) + g(p)

b−1∑i=0

g(pi)f(pa+b−2(i+1)

)− g(p)

b−2∑i=0

g(pi)f(pa+b−2(i+1)

)=

b−1∑i=0

g(p)if(pa+b−2i) +

b∑i=1

g(p)if(pa+b−2i)−b−1∑i=1

g(p)if(pa+b−2i)

=

b∑i=0

g(p)if(pa+b−2i)

=∑d|pb

g(d)f

(pa+b

d2

).

Hence the equation is true when m and n are prime powers. Since f is multi-plicative, it follows that the equation is true for all m and n.

2.32. Prove thatσα(m)σα(n) =

∑d|(m,n)

dασα

(mnd2

).

Proof. First we show that for prime p and b ≥ 1,

σα(pb+1) = σα(p)σα(pb)− pασα(pb−1).

Recall that for prime p we have,

σα(pb) =

{pα(b+1)−1pα−1 if α 6= 0,

b+ 1 if α = 0,

19

The case α = 0 is trivial to show, so assume that α 6= 0. Then

σα(p)σα(pb)− pασα(pn−1) =(p2α − 1)(p(b+1)α − 1)

(pα − 1)2− pα(pbα − 1)

pα − 1

=(p2α − 1)(p(b+1)α − 1)

(pα − 1)2− pα(pbα − 1)(pα − 1)

(pα − 1)2

=p(b+2)α+1 − p(b+2)α − pα + 1

(pα − 1)2

=p(b+2)α − 1

pα − 1

= σα(pb+1).

Hence the first statement in Exercise 30 holds for f = σα and g = Nα, andsince Nα is completely multiplicative, the result of Exercise 31 gives

σα(m)σα(n) =∑

d|(m,n)

dασα

(mnd2

).

2.33. Prove that Liouville’s function is given by the formula

λ(n) =∑d2|n

µ( nd2

).

Proof. The sum consists of just one nonzero term, as µ(n/d2) = 0 except whenn/d2 is the product of distinct primes. Let n = pa11 · · · p

akk . Now we list the

odd ai as ai1 , . . . , aij . Then the nonzero term is n/d2 = pi1 · · · pij . If j iseven, then µ(pi1 · · · pij ) = 1, and if j is odd then µ(pi1 · · · pij ) = −1. Similarly,λ(n) = (−1)a1+···+ak = (−1)j , so the equation holds.

2.34. This exercise describes an alternate proof of Theorem 2.16 which statesthat the Dirichlet inverse of a multiplicative function is multiplicative. Assumeg is multiplicative and let f = g−1.

(a) Prove that if p is prime then for k ≥ 1 we have

f(pk) = −k∑t=1

g(pt)f(pk−t).

Proof. Since f = g−1 we have g ∗ f = I. In particular, if p is a prime andk ≥ 1, then

(f ∗ g)(pk) =∑d|pk

g(d)f(pk/d) = 0.

The divisors of pk are in the form pt, 0 ≤ t ≤ k, so pulling the t = 0 termout of the sum and rearranging gives the desired result.

20

(b) Let h be the uniquely determined multiplicative function which agrees withf at the prime powers. Show that h ∗ g agrees with the identity function Iat the prime powers and deduce that h ∗ g = I. This shows that f = h sof is multiplicative.

Proof. Let p be prime and k ≥ 1. Then

(h ∗ g)(pk) =∑d|pk

h(d)g(pk/d)

=

k∑t=0

h(pt)g(pk−t)

=k∑t=0

f(pt)g(pk−t)

= (f ∗ g)(pk) = 0.

This and the fact that (h ∗ g)(1) = 1 imply h ∗ g = I, and since Dirichletinverses are unique, this implies f = h and f is multiplicative.

2.35. If f and g are multiplicative and if a and b are positive integers witha ≥ b, prove that the function h given by

h(n) =∑da|n

f( nda

)g( ndb

)is also multiplicative. The sum is extended over those divisors d of n for whichda also divides n.

Proof. Let (m,n) = 1. If da|m and (d′)a|n, then (dd′)a|mn and

f(m/da)f(n/(d′)a) = f(mn/(dd′)a).

Hence

h(m)h(n) =

∑d|m

f(mda

)g(mdb

)∑d|n

f( nda

)g( ndb

)=∑d|mn

f(mnda

)g(mndb

)= h(mn),

which demonstrates that h is multiplicative.

21

2.36.−2.40. In Exercises 36−40, we study µk, the Mobius function of order k,defined as follows:

µk(1) = 1,

µk(n) = 0 if pk+1|n for some prime p,

µk(n) = (−1)r if n = pk1 · · · pkr∏i>r

paii , 0 ≤ ai < k,

µk(n) = 0 otherwise.

Prove the properties of the functions µk described in the following exercises.

2.36. If k ≥ 1 then µk(nk) = µ(n).

Proof. Clearly if n = 1 then µk(1) = µ(1). If n contains any prime square factor,then both sides are 0 as pk+1|p2k and p2k|nk, so that pk+1|nk. Finally, supposen = p1 · · · pj is the product of distinct primes. Then µk(nk) = µ(pk1 · · · pkj ) =

(−1)j = µ(n).

2.37. Each function µk is multiplicative.

Proof. This is equivalent to showing that

µk(m)µk(n) = µk(mn)

where (m,n) = 1. Clearly if m or n equals 1 then the equation holds. If m or nhas a prime factor to the (k + 1)th power as a factor, the the product mn alsohas a prime factor to the (k + 1)th power, so that both sides of the equationare 0. Finally, suppose neither of these cases is true, then let j be the numberof prime factors p of m such that pk is also a factor of m, and let k be the samefor n. Since (−1)j(−1)k = (−1)j+k, the equation holds.

2.38. If k ≥ 2 we have

µk(n) =∑dk|n

µk−1

( ndk

)µk−1

(nd

).

Proof. Write n = pk1 · · · pkr∏i>r p

aii , ai 6= k. If n equals one then the equation is

evident, and if any ai is greater than k, then the sum is 0 as one of µk−1(n/dk)or µk−1(n/d) will be 0 for every d, so the sum is 0. Finally, suppose all ai < kfor all i > r. Then µk(n) = (−1)r. The sum runs through all products ofp1, . . . , pr. Thus the sum is equal to

µk−1(n)2 +∑i

µk−1

(n

pki

)µk−1

(n

pi

)+∑i,j

µk−1

(n

pki pkj

)µk−1

(n

pipj

)

+ · · ·+ µk−1

(n

pk1 · · · pkr

)µk−1

(n

p1 · · · pr

).

22

Note that if ai = k − 1 for any i, then the contribution is canceled out when itis multiplied twice (i.e. (−1)(−1) = 1). Thus we may assume that none of theai are equal to k − 1. Now every term except the last term has some factor inthe form pki , and thus are all equal to 0. Hence the sum equals

µk−1

(n

pk1 · · · pkr

)µk−1

(n

p1 · · · pr

)= µk−1

(n

p1 · · · pr

)= (−1)r,

which is the desired result.

2.39. If k ≥ 1 we have|µk(n)| =

∑dk+1|n

µ(d).

Proof. Note that |µk(n)| = 1 if there is no prime p such that pk+1|n, and|µk(n)| = 0 otherwise. It is easy to see that if there is no prime p such thatpk+1|n, then the sum is just one term that is equal to µ(1) = 1. Otherwise,

let n = pa11 · · · pamm , and let n′ = p[a1/(k+1)]1 · · · p[am/(k+1)]

1 , where [x] denotes thegreatest integer less than or equal to x. Then the sum is equal to∑

d|n′µ(d) = 0

as n′ was assumed to be not equal to 1.

2.40. For each prime p the Bell series for µk is given by

(µk)p(x) =1− 2xk + xk+1

1− x.

Proof. Since

µk(pa) =

1 if a < k,

−1 if a = k,

0 if a > k,

we have(µk)p(x) = 1 + x+ · · ·+ xk−1 − xk.

Now (1− x)(1 + x+ · · ·+ xk−1 − xk) = 1− 2xk + xk+1, so that

(µk)p(x) =1− 2xk + xk+1

1− x.

23