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COUNTING LATTICE POINTS BY PYTHAGORAS’ THEOREM

MAURICE CRAIG

1. Introduction

Simple geometric arguments applied to point–lattices can be surprisingly effec-tive for obtaining results about modular forms. A striking instance is van der Pol’suse [37] of the lattice Z2 to deduce the relation

θ34 = θ0

4 + θ24

connecting the basic theta–series [39, p. 467]. Further examples include the twinidentities

θ2(1, q) = θ2(1, q5) + 4qθ(q2, q5)θ(q4, q5), (1)

θ(τ, q)θ(τ2, q) = θ2(1, q5)− qθ(q2, q5)θ(q4, q5), (2)

(τ = exp(2πi/5)) and the formula (essentially Watson’s quintuple–product [36])

θ(−z, q)θ(−z2q2, q2) = θ(−q2, q6)[θ(z−3q2, q3)− z2θ(z3q2, q3)]. (3)

Besides geometry, the proofs usually involve algebraic elements, such as use of thetriple–product formula [1]

θ(z, q) =∏

(1− q2n)(1 + q2n−1z)(1 + q2n−1z−1)

for the theta–series

θ(z, q) =∑

znq[n].

(I write n2 as [n] to avoid a second–order superscript. Further, unless indicated,summations are for all n ∈ Z and products over all n ≥ 1.)

The present paper investigates the theta–series of certain point–lattices associ-ated with the cyclotomic field Q(τ), τ = exp(2πi/p), where p denotes an odd ratio-nal prime number. This field can be embedded in Euclidean space of dimension psuch that its integer ring R = Z[τ ] maps to the intersection of the point–lattice Zp

with a union of p parallel hyper–planes. Elementary means then yield explicit ex-pressions, by values of the function θ(z, q), for the series corresponding to the idealsPk(k ≥ 0). Here, P is the principal ideal generated by the element ϖ = τ − 1.Thus P is the first–degree prime ideal divisor of p, and Pk = Rϖk.

By way of illustration, the small primes p = 3, 5, 7 receive extended treatment.In these cases (and also when p = 13), the general formulae are equivalent toexpressions by values of the eta–function

η(ω) = q1/12θ(−q, q3) = q1/12∏

(1− q2n), q = exp(πiω).

Copyright c⃝2006 Maurice Craig.

1

2 MAURICE CRAIG

These expressions are familiar from the work, of Ramanujan [31–33] and his suc-cessors, on congruence properties of the partition function. For completeness, thecorresponding formulae for p = 13 are derived also, though by a different approach.Their deduction from the general formulae would be arduous. A further short sec-tion treats the divisors of 3 in the field of ninth roots of unity.

Apart from these results themselves, various modular identities and Fourier–coefficient congruences arise that, despite their easy derivations, seem neither ob-vious nor obviously related to quadratic forms.

As implied above, the prerequisites are minimal. Besides those mentioned al-ready, the Fourier–series expansion∑

exp[−πω(x+ n)2] = ω−1/2∑

exp(−πn2/ω) exp(2πinx) (4)

and its extension to many variables are often relevant. For simplicity, this materialis deferred to the later sections where it becomes essential.

The section immediately below introduces the geometric representation. Thethird section derives the main theta–series formulae in the prime–cyclotomic sit-uation. Later sections specialise these results to the smallest values of p. TheConclusion considers briefly the related topic of series–product identities.

2. Geometric embedding

Let K be a number field of degree n whose algebraic closure has automorphismgroup G. For ξ, ψ ∈ K the expression

(ξ, ψ) =∑

ξgψgc

defines an inner product, summation being over all g ∈ G, and c ∈ G denotingthe restriction to K of complex conjugation. When ψ = ξ we obtain a positivequadratic form [8]

(ξ, ξ) =∑

|ξg|2. (5)

Gauss [18, p. 395] used the term “general measure”. From its mode of formationit might be called the “total norm” of ξ.

Apply these considerations when K is the field Kp = Q(τ), τ = exp(2πi/p) withp an odd prime. We can write ξ ∈ Kp non–uniquely in the form

ξ =∑

xjτj , 0 ≤ j < p. (6)

with xj ∈ Q. Parseval’s theorem for the discrete Fourier transform gives s2+(ξ, ξ) =pt, and hence

(ξ, ξ) = pt− s2 (7)

where

s =∑

xj , t =∑

xj2. (8)

The non–uniqueness in equation (6) is because the distinct powers of τ are linearlydependent, summing to zero. We can eliminate this defect and obtain an integral

COUNTING LATTICE POINTS BY PYTHAGORAS’ THEOREM 3

basis by requiring that e.g. x0 = 0 or xp−1 = 0. However, for ξ ∈ R it proves moreconvenient to impose instead the symmetrical constraint

|s| ≤ r = (p− 1)/2,

achieved by subtracting from ξ a suitable integer multiple of∑τ j = 0. When

this condition holds, I say that ξ is in standard form. The geometric representa-tion of ξ is then the point x, of the lattice Zp, lying on one of the p hyper–planes∑xj = s = 0,±1, . . . ,±r.

This description differs from the one used in [8]. Its usefulness derives from thesimple interpretation that it provides for our quadratic form. In fact, by equa-tion (7) we see that (ξ, ξ)/p is just the square of the displacement of the point xfrom F , the foot of the normal, from the origin, to the hyper–plane

∑xj = s.

This close connection, with the easily analysed lattice Zp, is the source of the basicformulae in the next section.

The final cyclotomic prerequisite is a test for ξ ∈ Pk. When p = 3 we can write(for x, y, z ∈ Z)

x+ yτ + zτ2 = (x+ y + z) + (y + 2z)ϖ + zϖ2

so that x + y + z ≡ 0(mod3) is the condition for ξ ∈ P. To require ξ ∈ P2 = 3Radds the further constraint y + 2z ≡ 0(mod3). In the general case, the conditionsfor ξ ∈ Pk are (interpret 00 as p0 = 1)∑

jm−1xj ≡ 0(modp), 1 ≤ m ≤ k < p. (9)

This theorem (slightly modified) appears already as a footnote in Smith’s Report[35, p. 135]. If ξ is in standard form, the first congruence is just the equation s = 0.This observation is crucial, because it allows us to represent P by a prime sectionof Zp. For instance, when p = 3 equations (7)–(8) give (ξ, ξ)/3 = x2+y2+ z2, withx+ y + z = 0.

3. Pythagoras’ theorem in Zp

(a) The theta–series for the quadratic form (ξ, ξ) is∑q(ξ,ξ), summed over all inte-

gers of the relevant ideal. For Pk with k > 0, it is a series in qp. This fact makesit expedient to define

Θk(q) =∑

q(ξ,ξ)/p, ξ ∈ Pk

so that Θk(qp) is the true series for Pk.

Lemma 1 below shows that the essentially distinct series Θk(q) have 1 ≤ k ≤ r.Write

δ =∑

(j|p)τ j

for the quadratic Gauss sum, where (j|p) is the Legendre symbol.

Lemma 1. If N = mr + k ≥ 0 for 1 ≤ k ≤ r then ΘN (q) = Θk(qP ) with P = pm.

4 MAURICE CRAIG

Proof. For g ∈ G we have δg = ±δ. From equation (7) with s = 0, t = p − 1 wederive |δ|2 = p, so δR = Pr. Also (ξδ, ξδ) = p(ξ, ξ), whence the quadratic form forPr+k is just Θk(q

p).

Proposition 1. We have

pθ(1, qp)Θ1(q) =∑A

θp(τA, q), (10)

p2θ(1, qp)Θ2(q) =∑A

θp(τA, q) + p(p− 1)∏j

θ(τ j , q), (11)

and generally

pkθ(1, qp)Θk(q) =∑A

∑B

. . .∑C

∏j

θ(τf(j), q). (12)

Here

f(j) = A+Bj + . . .+ Cjk−1 (13)

and A,B, . . . , C traverse independently the distinct residues modulo p.

Proof. Let k = 1. From θp(z, q) =∑zsqt we get

∑θp(τA, q) = p

∑qt, with

right–side summation over all points x ∈ Zp satisfying the first congruence in (9).By Pythagoras’ theorem we can replace this sum by θ(1, qp)

∑qt, the summation

restricted now to p–tuples with s = 0. For, the points contributing to the formersum are just the translates, of the points that produce the latter sum, by someinteger multiple of the vector

u = (1, 1, . . . , 1)T

normal to the hyper–plane s = 0.

When k = 2 we must restrict to points x satisfying also the second congru-ence (9), say S =

∑jxj ≡ 0(modp). To this end, in the formula∏j

θ(z1zj2, q) =

∑zs1z

S2 q

t (0 ≤ j < p)

replace z2 by τB and sum to produce∑B

∏j

θ(z1τBj , q) = p

∑zs1q

t

where only points x with S ≡ 0( mod p) contribute. Putting τA for z1 and summingagain gives ∑

A

∑B

∏j

θ(τA+Bj , q) = p2∑

qt (s ≡ S ≡ 0).

Now, the terms on the left with B = 0 supply the first sum on the right of equa-tion (11). The residues A+Bj with B = 0 are just p−1 permutations of those withB = 1, which themselves are p 1permutations of those with A = 0. Thus we haveaccounted for the second sum. Moreover, by Pythagoras’ theorem the quantity onthe right agrees with the left side of equation (11).

The derivation of equation (12) follows the same pattern.


Remark 1. Stronger results hold. For example, equation (10) can be strengthenedto read

pθ(zp, qj)Θ1(q) =∑A

θp(τAz, q).

Before summing the formula θp(z, q) =∑zsqt, replace z by τAz, rather than merely

τA. This remark extends to Propositions 2, 3 and 7 and to Lemma 4.

Corollary 1. We have

pΘ2(q)−Θ1(q) = (p− 1)ηp(ω)/η(pω).

Proof. (Cf. [19, p. 325]) The equation∏j

θ(τ jz, q) = θ(zp, qp)ηp(ω)/η(pω)

follows by applying the triple–product identity to each factor on the left–side.

Similar reasoning produces, as follows, an expression for Θ0 different to the oneimplied by Proposition 1 combined with Θ0(q

p) = Θr(q).

Proposition 2. [7, p. 110] With outer summation over all integers k for which|k| ≤ r, we have

Θ0(q) =∑[∑

τ−jkθp(τ j , q)]/[∑

τ−jkθ(τ j , q1/p)]. (14)

Proof. . The method, in geometrical language, is to retract each of the p planess = k to the origin, then replicate it by all its translates s = l with l ≡ k(modp).Thus, beginning from

θ(z, q) = zk∑

zn−kq[n], θp(z, q) = zk∑

zs−kqt

we obtain both ∑τ−jkθ(τ j , q) = p

∑n≡k( mod p)

q[n], (15)

∑τ−jkθ(τ j , q) = p

∑s≡k( mod p)

qt. (16)

Now, for a fixed integer k, the expression q−[l]/p∑qt is the same for all integers

l ≡ k(modp), the sum being understood to comprise points x for which s = l. For,the points on both hyper–planes s = k, s = l determine the same integers ξ andhence the same values (ξ, ξ) (equations (6)–(8)). Write the result as∑

qt = q[l]/p∑

qt−[k]/p

where s = l in the left–hand sum, s = k on the right, then sum both sides overall l ≡ k(modp). The resulting double–sum on the left can be written simply asone over all s ≡ k(modp), for which equation (15) gives an expression by θ–values.Equation (16) gives similar help with the remaining member, to produce∑

q(ξ,ξ)/p =∑

τ−jkθp(τ j , q)/∑

τ−jkθ(τ j , q1/p),

where s = k on the left.

(b) The complexity of equation (12) increases with k. Next I develop for Θk analternative formula of progressively decreasing complexity.

6 MAURICE CRAIG

Lemma 2. A basis for the solution space of the congruences (9) is given by the

vectors x(l) ∈ (Z/pZ)p where, for 0 ≤ l ≤ p− k − 1, we have x(l)j = jl.

Proof. Recall that every x ∈ Z satisfies the congruence xp − x ≡ 0(modp). ByNewton’s power–sum formulae we conclude that (with summation over 0 ≤ j < p)∑

jm−1 ≡ 0(modp),

for 1 ≤ m < p − 1. Consequently, by the congruence test (9), x(0) ∈ Pp−1 =pR,x(1) ∈ Pp−2, and so on. In this way we construct solutions of (9), regarded asa system of k linear equations over the field Z/pZ. For each k < p the system hasfull rank, because a linear relation connecting the x(l) would imply a polynomialequation with degree less than the number of roots.

Write

θj,p(q) =∑

n≡j( mod p)

q[n].

Thus θj,p(q) =∑q[np+j] and we have (0 ≤ k < p, summation for 0 ≤ j < p)

θ(τk, q) =∑

τ jkθj,p(q). (17)

The summation ranges in equations (15) and (17) can be contracted to 1 ≤ j ≤ r,giving

pθk,p(q) = θ(1, q) +∑j

(τ jk + τ−jk)θ(τ j , q), (18)

θ(τk, q) = θ0,p(q) +∑j

(τ jk + τ−jk)θj,p(q). (19)


θ(1, qp)Θp−1(q) = [θ0,p(q)]p + 2[θ1,p(q)]

p + . . .+ 2[θr,p(q)]p, (20)

θ(1, qp)Θp−2(q) =∑

[θj,p(q)]p + p(p− 1)θ0,p(q)[θ1,p(q) . . . θr,p(q)]

2, (21)

and generally

θ(1, qp)Θp−k(q) =∑A

∑B

. . .∑C

∏θf(j),p(q), (22)

with f(j) as in equation (13).

Proof. The usual Pythagorean argument yields θ(1, qp)Θp−k(q) =∑qt, summa-

tion being over all x ∈ Zp satisfying the relevant congruences. For k = 1,x must becongruent to a multiple of u. Thus, we can write xj = A+ pyj for suitable integersyj . Each residue A contributes to the right side a term [θA,p(q)]

p.

For k = 2 the congruences (9) have the further basic solution x(1). Thus, xj =A+Bj+ pyj and we get instead

∑A

∑B

∏j θA+Bj,p(q). By an argument like that

used for equation (11), we see that the additional contribution is p(p−1)∏

j θj,p(q).The general case is similar.

Simplifications of equations (12) and (22) analogous to (10)–(11) and (20)–(21)depend specifically on p, so will be considered only as need arises.



Θp−2(q)−Θp−1(q) = p(p− 1)ηp(p2ω)/η(pω). (23)

Further, if p > 3 then

Θr−1(q)−Θr(q) = p(p− 1)ηp(pω)/η(ω). (24)

Proof. Noting that θj,p(q) = q[j]θ(q2pj , q[p]), from the triple–product identity wederive

θj,p(q1/p) = q[j]/p

∏(1− q2pn)(1 + q(2n−1)p−j)(1 + q(2n−1)p+j).

Fix n and consider the values of the last two factors for all j in the range 1 ≤ j ≤ r.They include all expressions 1+q2m−1 with (n−1)p < m ≤ np except the one withm = pn− r. The product over n for this range of j is thus∏

θj,p(q1/p) = q([p]−1)/24

∏[(1− q2pn)r(1 + q2n−1)/(1 + q(2n−1)p)].

Square, multiply by θj,p(q1/p) = θ(1, qp), then apply Proposition 3.

Propositions 1 and 3 are dual, related by modular transformation, for which seeLemma 17 below. The dual of Proposition 2 is not required below, so for brevityis omitted.

Rangachari [34] proved certain results in this section after finding them implicitin late work of Ramanujan [33].

4. First Interlude

I proceed next to interpret the formulae when p = 3, 5, 7, while taking note ofsuch modular–form relations as the investigation allows. Partitioning of the termsin a q–series F (q) =

∑anq

n, according to the residue class of the exponent n, willplay a role. If Fj(q) denotes the j–part in a partition modulo p, then

F (τkq) =∑j

τ jkFj(q)

whence, by Fourier inversion,

pFj(q) =∑j

τ−jkF (τkq).

The product∏

k F (τkq) equals the cyclic determinant, ∆(F0(q), . . . , Fp−1(q))

say, with first row as indicated. Write Ω for the operator that separates from F (q)its p–part: ΩF (q) = F0(q).

With

ϕ(q) =∏

(1− q2n).

an easy application of the triple–product formula yields:

Lemma 3. [27, p. 79; 28, p. 5] We have∏ϕ(τkq) = ϕp+1(qp)/ϕ(q[p]).

Similar result are valid for some other infinite products, as in the following result.

8 MAURICE CRAIG

Lemma 4. We have∏k

θ(τ j , τkq) = θ(1, qp)θp(τ j , qp)/θ(1, q[p]). (25)

For certain series F (q) the p–partition is lacunary, i.e. some Fj(q) vanish iden-tically. Thus, by a series–product identity due to Jacobi [23, 24], the 3–partition ofϕ3(q) is lacunary, a fact that leads [28, p. 15] to the parametrisation of a certainplane cubic curve by the parts in the 3–partition of ϕ(q).

Lemma 5. The p–partition for Θr(q) is always lacunary, with ΩΘr(q) = Θ1(qp).

Proof. Note that Θr(q) and Θ1(qp) are the respective theta–series for R and P.

By equation (7), (ξ, ξ) can represent only integers congruent to the negatives ofsquares modulo p. Moreover, it is divisible by p if and only if ξ ∈ P.

As regards truncated circulants, for later use I note the particular formulae

∆(A,B, 0) =A3 +B3,

∆(A, 0, B, C, 0) =A5 +B5 + C5 − 5A3BC + 5A(BC)2, (26)

∆(A,B,D, 0, C, 0, 0) =A7 + 14A4BCD − 7A3∑

B3C]

+ 14A2∑

B2C3 − 7A[∑

BC5 + (BCD)2]

+[∑

B7 + 7BCD∑

B3C]. (27)

Permutation of the linear factors shows that ∆(A,B, 0, 0, C) also has the value (26).Summations in equation (27) are over a cyclic permutation of the letters B,C,D.

Finally, to simplify the eta–formulae I use the abbreviations

h = η(ω), H = η(pω), J = η(p2ω).

The meanings of H and J , like Ω, are context–dependent, varying with p.

5. The prime p = 3


Θ1(q) = h3/H + 9J3/H, (28)

Θ2(q) = h3/H + 3J3/H. (29)

Proof. By corollaries 1 and 2,

3Θ2(q)−Θ1(q) = 2h3/H,

Θ1(q)−Θ2(q) = 6J3/H.

Corollary 3. [12]. We have Θ2(q) = Θ1(q3). Further, the equation

Θ1(q) = Θ1(q3) + 6q2ϕ3(q9)/ϕ(q3)

gives a 3–partition of Θ1(q).

Proof. Lemma 1 gives Θ1(q) = Θ0(q3) and Θ2(q) = Θ0(q

9). Now invoke Proposi-tion 4. Corollary 4. [28, p. 15; 38, p. 255]. We have

(H/J)12 = [(h/J)2 + 3]3 − 27.


Proof. Write equation (29) as h3 = HΘ2(q)− 3J3 or, by Corollary 3,

ϕ3(q) = ϕ(q3)Θ1(q3)− 3q2ϕ3(q9).

Lemma 3 gives

ϕ12(q3)/ϕ3(q9) = [ϕ(q3)Θ1(q3)]3 − 27q6ϕ9(q9).

Replace q3 by q, divide by ϕ3(q) and invoke equation (29) to produce

(h3/H + 9J3/H)3 = h9/H3 + 27H9/h3, (30)

equivalent to the stated result, but in the form needed for Proposition 12. This completes the work for p = 3. See [4, 26] for an extensive coverage from a

different perspective. See also Section 8, treating p2 = 9.

6. The prime p = 5


Θ1(q) = h5/H + 25H5/h, (31)

Θ2(q) = h5/H + 5H3/h. (32)

Proof. By Corollaries 1 and 2,

5Θ2(q)−Θ1(q) = 4h5/H,

Θ1(q)−Θ2(q) = 20H5/h. The next results prepare for determining the 5–partitions of Θ1(q) and Θ2(q).

(The exponent (n|5) in Lemma 6 below is a Legendre symbol.)

Lemma 6. Let

R(q) = q2/5∏

(1− q2n)(n|5), (33)

and ν = R(q5), X = h/J, Y = H/J . We have then

X = 1/ν − 1− ν, (34)

Y 6 = 1/ν5 − 11− ν5, (35)

Proof. Equations (34)-(35) specialise [2, Lemma 6] to the case p = 5. The leftside of equation (35) results from applying Lemma 3 to X. The right side is thecirculant ∆(−1, 0,−ν, ν−1, 0).

Elimination of ν between equations (34) and (35) yields:

Corollary 5. [38, p. 256] We have

Y 6 = X5 + 5X4 + 15X3 + 25X2 + 25X. Now write Pj = πjJ

5/H , where

π1 = 2/ν2 − ν3, π2 = 1/ν4 − 3ν,

π4 = 2ν2 + 1/ν3, π3 = ν4 + 3ν.

Lemma 7. We have the 5–partitions

h5/H = H5/J + 5(P1 − P2 − P3 + P4), (36)

H5/h = 5J5/H + P1 + P2 + P3 + P4. (37)

10 MAURICE CRAIG

Proof. Raise equation (34) to the fifth power. By equation (35) we get

X5 = Y 6 + 5(π1 − π2 − π3 + π4)

from which follows (36) by writing h5/H = (J5/H)X5. Likewise, from H5/h =(H5/J)/X and

Y 6/X = 5 + π1 + π2 + π3 + π4

we obtain equation (37).

Proposition 6. We have the 5–partitions

Θ1(q) = (H5/J + 125J5/H) + 10(3P1 + 2P2 + 2P3 + 3P4), (38)

Θ2(q) = (H5/J + 25J5/H) + 10(P1 + P4), (39)

the 2–part and 3–part in equation (39) being zero and the 5–part Θ1(q5).

Proof. Combine Lemma 7 with Proposition 5.

Propositions 5 and 6 imply the following easy results, mentioned here for latercomparison with Lemma 14.


ΩΘ2(q) = Θ3(q),

Ω[Θ1(q)−Θ2(q) = 5[Θ3(q)−Θ4(q)].

The next developments use the abbreviations

a =∑

q[5n], b =∑

q[5n+1], c =∑

q[5n+2].

Thus, in the notation of equations (17)–(18),

a = θ0,5(q), b = θ1,5(q), c = θ2,5(q).

Lemma 8. We have

θ2(1, q5) = a2 + 4bc, (40)

θ(τ, q5)θ(τ2, q5) = a2 − bc. (41)

Proof. The function θ2(1, q) is the theta–series∑∑

q[x]+[y] for the lattice of Gauss-ian integers x+iy. Points (k−2y, y) on the line x+2y = k lie at distances d from theorigin, where 5d2 = (5y−2k)2+k2. Thence θj,5(q

1/5)θ2j,5(q1/5) is the contribution

from points with k ≡ j( mod 5). Add the results for each residue class j then replaceq by q5. A similar argument applies to θ(τ, q)θ(τ2, q) =

∑∑τx+2yq[x]+[y].

Remark 2. Formulae (40)–(41) are effectively (1)–(2) of the Introduction. In placeof this Pythagorean proof, one may note that the kernel of the ring homomorphismx + iy → 2x − y(mod5) of Z[i] onto Z/5Z is the principal ideal (1 + 2i)Z[i]. Soevery integer has the form

x+ iy = (m+ in)(1 + 2i) + j

with j = 0,±1or±2. Now 5(x2+y2) = (5m+j)2+(5n−2j)2 and Lemma 8 followseasily. A similar proof of equation (3) can be based on the decomposition (cf. [25])

x+ y√−2 = (m+ n

√−2)(1−

√−2) + j, j = 0,±1.


Lemma 9. We have the 5–partitions

θ(1, q5)Θ1(q) = a5 + 2b5 + 2c5 + 120a(bc)2

+ 10(3Q1 + 2Q2 + 2Q3 + 3Q4), (42)

θ(1, q5)Θ2(q) = a5 + 2b5 + 2c5 + 20a(bc)2 + 10(Q1 +Q4), (43)

where

Q1 = ac4 + 2b3c2 + 2a2b2c, Q2 = a3b2 + 2ab3c+ 2bc4,

Q4 = ab4 + 2b2c3 + 2a2bc2, Q3 = a3c2 + 2abc3 + 2b4c.

Proof. Equations (42)–(43) come by taking k = 3, 4 in equation (22). Alternatively(as may be easier for hand calculation), substitute in equations (10)–(11) the valuesof the θ(τ j , q) given by equation (19). By inspection the q–series for Qj includesonly terms with exponents ≡ j(mod5). Lemma 10. For 1 ≤ j ≤ 4 we have Qj = θ(1, q5)Pj.

Proof. Compare equations (42)–(43) with equations (38)–(39). For the next result, three different arguments (i)–(iii) are offered. The solution

of (44) as a diophantine equation is given in [10].

Proposition 7. [31, vol. 2, p. 234] We have the identity

a(b5 + c5) + (abc)2 = a4bc+ 2(bc)3. (44)

Proof. (i) Equation (19) for j = 0 gives the 5–partition θ(1, q) = a+2b+2c. Fromequation (25) we deduce the formula

θ6(1, q5)/θ(1, q25) = ∆(a, 2b, 0, 0, 2c).

The denominator on the left is simply a. By equations (26) and (40) we obtain

(a2 + 4bc)3 = a(a5 + 32b5 + 32c5 − 20a3bc+ 80ab2c2),

a form of equation (44).

(ii) Subtract equation (10) from equation (11) then replace q by q5 to produce

θ(1, q25)[5Θ2(q5)−Θ1(q

5)] = 4∏j

θ(τ j , q5).

Now, equations (20)–(21) take the respective forms

θ(1, q5)Θ4(q) = a5 + 2b5 + 2c5,

θ(1, q5)Θ3(q) = a5 + 2b5 + 2c5 + 20a(bc)2,

so

θ(1, q5)[5Θ2(q5)−Θ1(q

5)] = 4(a5 + 2b5 + 2c5 − 5ab2c2),

Eliminating the expression [5Θ2(q5)−Θ1(q

5)] we obtain

a(a5 + 2b5 + 2c5 − 5ab2c2) = [θ(1, q5)θ(τ, q5)θ(τ2, q5)]2

and, by equations (40)–(41), the right member is (a2 + 4bc)(a− bc)2.

(iii) According to Proposition 2, the 4–part of Θ2(q) is 2G where, with

λ = τ + 1/τ, µ = τ2 + 1/τ2,

12 MAURICE CRAIG

we have

5bG = θ5(1, q5) + λθ5(τ, q5) + µθ5(τ2, q5).

By Lemma 4 we deduce that

5(bG/a)θ(1, q5) =∏k

θ(1, τkq) + λ∏k

θ(τ, τkq) + µ∏k

θ(τ2, τkq).

The first term on the right is ∆(a, 2b, 0, 0, 2c). The second term is

λ∆(a, λb, 0, 0, µc) = λa5 − (8λ− 5)b5 − (3λ+ 5)c5 + 5λa3bc+ 5a(bc)2,

the third term being its conjugate. Consequently

(bG/5a)θ(1, q5) = 2b5 + c5 − a3bc+ 3a(bc)2,

whereas, by equation (43), we have θ(1, q5)G = 5Q4. Lemma 11. We have

Q2Q3 −Q1Q4 = (a2 + 4bc)(a2 − bc)2, (45)

Q21 −Q3Q4 = bc2(a2 + 4bc)(a2 − bc)(b3 − ac2), (46)

Q24 −Q1Q2 = b2c(a2 + 4bc)(a2 − bc)(c3 − ab2). (47)

Proof. Regarding a, b, c temporarily as indeterminates, write

F = a(b5 + c5) + (abc)2 − a4bc− 2(bc)3.

We find by direct computation that

Q2Q3 −Q1Q4 = ab2c2(a5 + 2b5 + 2c5 − 5ab2c2)

so

Q2Q3 −Q1Q4 − 2(bc)2F = (a2 + 4bc)(a2 − bc)2.

Again,

Q21 − ac3F = abc4(a4 + 3a2bc+ 6b2c2) + b4c2(4a4 + 7a2bc+ 4b2c2),

Q3Q4 − 2b3cF = abc4(2a4 + 6a2bc+ 2b2c2) + b4c2(3a4 + 4a2bc+ 8b2c2),

whence

Q21 −Q3Q4 + (2b3 − ac2)cF = bc2(a2 + 4bc)(a2 − bc)(b3 − ac2).

A further equation results from symmetry, on interchange of b with c, and Qj withQ5−j . Then formulae (45)–(47) follow by virtue of equation (44). Proposition 8. We have

ν = bc2(b3 − ac2)/(a2 − bc), − 1/ν = b2c(c3 − ab2)/(a2 − bc).

Proof. From the definition of the πj we easily compute the values

π2π3 − π1π4 = 1/ν5 − 11− ν5,

π21 − π3π4 = (1/ν5 − 11− ν5)ν,

π24 − π1π2 = −(1/ν5 − 11− ν5)/ν.

But Qj = θ((1, q5)Pj and Pj = (J5/H)πj so

ν = (Q21 −Q3Q4)/(Q2Q3 −Q1Q4),−1/ν = (Q2

4 −Q1Q2)/(Q2Q3 −Q1Q4),

whence the results follow by Lemma 11.


Remark 3. Proposition 8 is tantamount to a pair of modular identities found byGarvan. They appeared as equations (1.32)–(1.33) in the draft [14] (along with anappeal, immediately preceding the Acknowledgements, for an elementary proof). Inthe printed version [15] they appear as the identities (4.9)–(4.10), now less eye–catching in an umbral disguise.

This completes the work for p = 5, the simplest typical case. As the difficultiesincrease rapidly with p, the treatment when p = 7 will be sufficiently elaborate forillustrative purposes without developments analogous to those above. Here I merelynote the possibility of such an investigation based, once again, on [2, Lemma 6].

7. The prime p = 7

Write the functions θj,7(q) =∑q[7n+j] for 0 ≤ j ≤ 3 as a, b, c, d respectively. Like

a, b, c when p = 5, every three of these four functions are algebraically dependent.Thus, the analogue of equation (44) is an infinite set of such relations. The aim ofthe next few results is to establish two basic relations (52)–(53). As abbreviations(cf. the right side in equation (27)), write

P =∑

b3c, Q =∑

b2c3, R =∑

bc5, S = bcd, T =∑

b7,

where∑

denotes a sum of three terms obtained by cyclically permuting b, c, d.These five functions of b, c, d are mutually related. For example, even withoutinterpreting b, c, d as functions of q, we have the algebraic identities (cf. [11, p.117])

P (R+ S2) = Q2 + ST, Q(T + 5PS) = P 3 +R2 + 3RS2 + 9S4.

The following result is an analogue of Lemma 8. Like the matrix (1 4 | 1 -1)in equations (40)–(41), the matrix of coefficients for a4, P, aS in equations (48)–(50) is essentially self–inverse. As presented below, the proof conceals a quaternionproduct formula (cf. Remark 2).

Lemma 12. We have

θ4(1, q7) = a4 + 8P + 24aS, (48)

θ3(τ, q7)θ(τ2, q7) + θ3(τ2, q7)θ(τ3, q7) + θ3(τ3, q7)θ(τ, q7)

= 3a4 − 4P + 9aS, (49)

θ(1, q7)θ(τ, q7)θ(τ2, q7)θ(τ3, q7) = a4 + P − 4aS, (50)

Proof. Every vector in Z4 can be expressed in the formXYZW

=

2 −1 −1 −11 2 −1 11 1 2 −11 −1 1 2

xyzw

+

00uv

for unique x, y, z, w, u, v ∈ Z with |u|, |v| ≤ 3. In fact, noting that the coefficientmatrix is orthogonal, we see that u, v are determined by the congruences

u+ v ≡ 2X + Y + Z +W (mod7),

u− v ≡ −X + 2Y + Z −W (mod7).

14 MAURICE CRAIG

The matrix equation then produces unique integer values for x, y, z, w.

Next, by series rearrangement we find that

θ(A, q7)θ(B, q7)θ(C, q7)θ(D, q7) =∑u

∑v

CuDv[∑

x

(A2BCD)xq[7x+u+v]∑y

(A−1B2CD−1)yq[7y+u−v]

×∑z

(A−1B−1C2)zq[7z+2u+v]∑w

(A−1BC−1D2)wq[7w−u+2v]].

Specialising A,B,C,D to make A2BCD etc. all unity gives the simpler result

θ(τ3j+2k, q7)θ(τ2j−3k, q7)θ(τ−j , q7)θ(τ−k, q7) =∑u

∑v

τ−ju−kv

×∑

x

q[7x+u+v]∑y

q[7y+u−v]∑z

q[7z+2u+v]∑w

q[7w−u+2v]. (51)

The quantities in square brackets have residues u(1, 1, 2,−1) + v(1,−1, 1, 2). Theyare given for u = 0 by the columns of the first of the matrices

0 1 2 3 −3 −2 −10 −1 −2 −3 3 2 10 1 2 3 −3 −2 −10 2 −3 −1 1 3 −2

,

1 2 3 −3 −2 −1 01 0 −1 −2 −3 3 22 3 −3 −2 −1 0 1−1 1 3 −2 0 2 −3

and, for u = 0, by those of the second and its non–zero multiples modulo 7.

If j = k = 0, the former matrix implies a right–side contribution of a4+2P . Thelatter matrix contributes P+4aS, so the complete right side is a4+2P+6(P+4aS),which gives equation (48). If j = 0, k = 1 the contribution from the latter matrixchanges to

(τ + . . .+ τ6)(P + 4aS) = −(P + 4aS),

which yields equation (50).

For equation (49) take j = k = 1. Writing u in place of u+ v, for the right sidein equation (51) we get∑

u

∑v

τ−u∑

x

q[7x+u]∑y

q[7y+u2v]∑z

q[7z+2u−v]∑w

q[7w−u+3v].

The terms with u = 0 supply a4+6aS. Those with u = 1 give τ−1(3b3c+d3b+3aS),which becomes τ−2(3c3d + b3c + 3aS) for u = 2, and so on. But the left side ofequation (51) is now θ3(τ, q7)θ(τ2, q7), so addition of conjugate equations producesequation (49).

The following result comes by imitating the first two proofs of Proposition 7.

Proposition 9. Connecting the functions a, b, c, d are the formulae

Sa3 −Qa+ 2S2 = 0, (52)

Qa2 − (R+ 3S2)a+ PS = 0. (53)


Proof. Equation (19) gives the 7–partition formula

θ(σ, q) = a+ λb+ µc+ νd,

where λ = σ + 1/σ, µ = σ3 + 1/σ3, ν = σ2 + 1/σ2 with σ denoting an arbitraryseventh root of unity. From equation (25) we deduce the formula

θ(1, q7)θ7(σ, q7)/θ(1, q49) = ∆(a, λb, νd, 0, µc, 0, 0),

the right side of which can be expanded by means of equation (27). We make twoapplications of this result as follows.

(i) With σ = 1 the left side is (a2 + 8P + 24aS)2/a by equation (48). Thence

Sa5 − 2Pa4 + 7Qa3 − (7R+ 16S2)a2 + 2(T + 4PS)a− P 2 = 0. (54)

(ii) Instead, compute the mean for all seven values σ = τk. The right side reducesto

a7 + 28Sa4 + 84Qa2 − 70(R+ S2)a+ 2T + 112PS.

By equation (10) the left side yields θ(1, q7)Θ1(q7). However, from equation (22)

with k = 3 we have

θ(1, q7)Θ1(q7) = a7 + 42Qa2 + 84S2a+ 2T + 42PS,

whence

2Sa4 + 3Qa2 − (5R+ 11S2)a+ 5PS = 0. (55)

Next, in analogy with the second proof for Proposition 7, from equations (10)–(11) we obtain by subtraction

a[7Θ2(q7)−Θ1(q

7)] = 6∏j

θ(τ j , q).

By equation (50) the right side is 6(a4 + P − 4aS)2/θ(1, q7). However, equa-tions (20)–(21) imply that

θ(1, q7)[7Θ2(q7)−Θ1(q

7)] = 6(a7 − 7Qa2 + 35S2a+ 2T − 7PS),

so by elimination we derive

8Sa5 − 2Pa4 − 7Qa3 + 19S2a2 + (2T + PS)a− P 2 = 0.

Subtracting equation (54) eliminates T . Simplification yields

Sa4 − 9Qa3 + (R+ 5S2)a− PS = 0.

This equation forms with (55) a pair easily seen equivalent to (52)–(53).

Remark 4. A further identity of modest degree linking a, b, c, d is

Ra2 − 2PSa+ 4QS − P 2 = 0.

I have not investigated its relation to equations (52)–(53).


Θ1(q) = h7/H + 72(hH)3 + 73H7/h, (56)

Θ2(q) = h7/H + 7(hH)3 + 72H7/h, (57)

Θ3(q) = h7/H + 7(hH)3 + 7H7/h. (58)

16 MAURICE CRAIG

Proof. By the triple product identity

b(q1/7)c(q1/7)d(q1/7) = q2θ(q2, q7)θ(q4, q7)θ(q6, q7)

= q2∏

[(1− q14n)3(1 + q2n−1)/(1 + q14n−7)].

Again,

θ(τ, q)θ(τ2, q)θ(τ3, q) = =∏[(1− q2n)3(1 + q14n−7)/(1 + q2n−1)]

so

b(q1/7)c(q1/7)d(q1/7)θ(τ, q)θ(τ2, q)θ(τ3, q) = (hH)3

Replacing q by q7 and applying equation (50) gives

θ(1, q7)(HJ)3 = S(a4 + P − 4aS)

= a(R− 3S2)

by Proposition 9.

However, equation (22) for k = 2, 3 gives

θ(1, q7)[Θ1(q7)−Θ2(q

7)] = 42(Qa2 + S2a+ PS)

= 42a(R+ 4S2) (59)

by equation (53) , so that

Θ1(q7)−Θ2(q

7) = 42[(HJ)3 + 7aS2/θ(1, q7)].

From equations (21))–(22) we have

Θ2(q7)−Θ3(q

7) = 42aS2/θ(1, q7), (60)

which is 42J7/H by equation (23). Replacing q7 by q we get

Θ1(q)−Θ2(q) = 42[(hH)3 + 7H7/h].

Proposition 10 follows from this result, read in conjunction with Corollaries 1and 2.

The next objective is the analogue of Corollary 6. By equation (22)

θ(1, q7)Θ7−k(q) =∑∏

j

∑k

q[kp+f(j)], (61)

with the (multiple) outer summation over the coefficients in equation (13).

Lemma 13. The residues contributing to Ω[Θ2(q) − Θ3(q)] are generated by thevalues of

f(j) = A+Bj + j4, (62)

each repeated 42 times. For Ω[Θ1(q)−Θ2(q)] they are given by

f(j) = A+ 3D2j + Cj2 +Dj3 + j5, (63)

also repeated 42 times. Here, the symbols A,B,C,D run independently over allresidue classes modulo 7.


Proof. Take k = 4 in equation (61). By Lemma 2, f(j) is quartic in j, with the con-tribution to θ(1, q7)Θ3(q) from elements in P4 coming from the cubic polynomials.Thus, Θ2 −Θ3 is built from non–zero multiples of monic polynomials

f(j) = A+Bj + Cj2 +Dj3 + j4.

Contributors to Ω[Θ2 − Θ3] satisfy the further condition∑

[f(j)]2 ≡ 0(mod7).However, we have

∑jk ≡ 0(mod7) for 0 ≤ k ≤ 5 and

∑jk+6 ≡

∑jk(mod7), so

this condition reduces to C ≡ 3D2(mod7). We have then

f(j) ≡ (A− 2BD −D4) + (B + 3D3)(j + 2D) + (j + 2D)4(mod7),

whose distinct values are the same as those of the polynomial (62). A factor 7comes from additive shifts applied to j. Also, every non–zero multiplier is a fourthpower, so can be absorbed within j4, which implies a further factor 6.

Similar reasoning supplies the Ω[Θ1 −Θ2] result.

Lemma 14. (Cf. Corollary 6) We have

ΩΘ3(q) = Θ4(q), (64)

Ω[Θ2(q)−Θ3(q)] = 7[Θ4(q)−Θ5(q)], (65)

Ω[Θ1(q)−Θ2(q)] = 42[Θ4(q)−Θ5(q)] + 49[Θ5(q)−Θ6(q)]. (66)

Proof. Equation (64) follows from Lemmas 1 and 5. Equation (62) reduces, to amanual calculation, the task of listing the 49 vectors [f(0), . . . , f(6)] needed to showthat

θ(1, q7)Ω[Θ2(q)−Θ3(q)]/42 = 2Sa4 + 3Qa2 + (2R+ 17S2)a+ 5PS.

Equation (63) leads by a longer calculation to the formula

θ(1, q7)Ω[Θ1(q)−Θ2(q)]/42 = 8Sa4 + 26Qa2 + (8R+ 131S2)a+ 34PS.

Lemma 10 reduces the right sides, in the last two equations, to the respectiveforms

7a(R+ 4S2), 42a(R+ 4S2) + 49aS2.

Now apply equations (59)–(60).

By combining Lemma 14 with Proposition 10 we obtain:


Ω[h7/H] = H7/J + 72(HJ)3,

Ω[(hH)3] = −7(HJ)3,

Ω[H7/h] = 7(HJ)3 + 72J7/H.

Remark 5. These results are the analogues of

Ω[h5/H] = H5/J, Ω[H5/h] = 5J5/H

evident from equations (36)–(37). The last equation in each case is responsible forthe well known partition congruences discovered by Ramanujan [32].

18 MAURICE CRAIG

8. The prime power p2 = 9

This section builds on the results for p = 3 to establish eta–function formulae forthe ideal divisors of 3 in the field K9 = Q(τ), τ = exp(2πi/9). Thus τ6+ τ3+1 = 0and every element has the form, for ξj ∈ K3,

ξ = ξ0 + ξ1τ + ξ2τ2 (67)

The integer ring of K9 is R = Z[τ ]. So ξ is integral when all ξj are integers of K3,and conversely.

If ξj = a+ bτ3 + cτ6 write sj = a+ b+ c and tj = a2 + b2 + c2. We suppose ξjnormalized so |sj | ≤ 1. In K9 the first–degree prime divisor of 3 is P = R(τ − 1).The theta–series for Pk is Φk(q

3), where I define

Φk(q) =∑

q(ξ,ξ)/3

summed over all ξ ∈ Pk. We have P3 = (τ3 − 1)R = δR with δ =√−3, so

Φk+3(q) = Φk(q3) and we need consider only k = 0, 1, 2.

Our first result comes immediately from the congruences

τ ≡ 1(modP), τ2 ≡ −τ − 1(modP2).

Lemma 15. The condition for ξ ∈ P is s0 + s1 + s2 ≡ 0(mod3). The conditionsfor ξ ∈ P2 are s0 ≡ s1 ≡ s2(mod3).

To express the next result, write the theta–series for the integer ring of K3 as∑q3t−[s] = φ0 + 2φ1,

where φj gathers the terms with |s| = j. The left side is known to be Θ1(q).

Lemma 16. We have

Φ0(q) = φ03 + 6φ0

2φ1 +12φ0φ12 + 8φ1

3,

Φ1(q) = φ03 +6φ0φ1

2 + 2φ13,

Φ2(q) = φ03 + 2φ1

3.

Proof. With c denoting complex conjugation, equation (67) implies

ξ1+c =∑∑

ξjξkτj−k.

Now, Aut(K9/Q) is generated by c together with the 3–cycle τ → τ4, whose fixedfield is K3. Noting that the three conjugates of τ j sum to 3 when 3 divides j andto zero otherwise, we see that

(ξ, ξ)/3 =∑

(3tj − s2j ).

The theta–series for R is therefore (φ0 + 2φ1)3. For P we must restrict to

±(s0, s1, s2) = (0, 0, 0), (1, 1, 1), (1,−1, 0) etc. with∑sj ≡ 0(mod3). The three

vectors listed on the right side contribute respectively the terms φ30, φ

31 and φ0φ

21,

the complete expression being as stated above. Finally, the only solutions of thecongruences s0 ≡ s1 ≡ s2(mod3) have all sj = 0, 1 or −1.


Proposition 12. With h,H, J retaining their meanings for p = 3, we have

Φ0(q) = h9/H3 + 33H9/h3, (68)

Φ1(q) = h9/H3 + 32H9/h3, (69)

Φ2(q1/3) = h9/H3 + 34H9/h3. (70)

Proof. Equation (68) follows from Φ0(q) = Θ31(q) together with equations (28) and

(30). Next, by Corollary 3 we have φ0 = Θ31(q), φ1 = 3J3/H, so

φ30 + 2φ3

1 = (H9/J3 + 27J9/H3) + 54J9/H3

by equations (28) and (30), which yields equations (70).

Finally, equation (29) and Corollary 3 give

φ0 = h3/H + 3J3/H, φ1 = 3J3/H.

Thence by Lemma 16

H3Φ1(q) = (h3 + 3J3)3 + 6(h3 + 3J3)(3J3)2 + 2(3J3)3

= h9 + 9J3(h6 + 9h3J3 + 27J6).

However, the modular equation (30) can be written

h12 + 27H12 = (h4 + 9hJ3)3

= h12 + 27h3J3(h6 + 9h3J3 + 27J6).

Thus (hH)3Φ1(q) = h12 + 9H12, equivalent to equation (69).

9. Second Interlude

Results like those of Propositions 5 and 10 are valid also for p = 13, when de-duction of eta–formulae for the Θj(q) from their expressions by values of θ(z, q)becomes too laborious. The results are therefore derived another way, namely, bythe technique of Corollary 3 in [11, p. 118]. The same approach works also forp = 5 or 7 (and for p2 = 9), but is not necessarily preferable. Relative to Proposi-tions 1–3, this method has the weakness that, besides being non–elementary, it iseither inapplicable (when 12/r ∈ Z, i.e. p− 1 does not divide 24) or else attains itsobjective without establishing any base–camps useful for further excursions.

To apply deeper results about modular forms we require further informationabout the theta–series. Write Θkω = Θk(q), where q = exp(πiω), and let Ψω =hp/H for an odd prime p > 3. The following result comes from [9].

Lemma 17. For 1 ≤ k ≤ r the series Θkω is a modular form of weight r andof Nebentypus for the congruence subgroup Γ0(p) of the modular group Γ. Further,with l = r − k + 1 we have Θk−1/pω = pl−1/2(ω/i)rΘlω.

Corollary 7. [21, p. 285–286], [22, p. 938–939] The modular form Ψω is likewiseof weight r and of Nebentypus for Γ0(p).

Proof. Combine Lemma 17 with Corollary 1.

20 MAURICE CRAIG

Proposition 13. Assume 12/r ∈ Z. For 1 ≤ k ≤ r we have

Θk(q) =∑

aj(hp/H)(H/h)12j/r.

Here, summation is over integers j such that 0 ≤ 12j/r ≤ p + 1, the aj beingsuitable constants with a0 = 1.

Proof. The ratio Θkω/Ψω is a modular form of weight zero, i.e. a modularfunction. It is therefore a rational function of the Hauptmodul (H/h)12/r for Γ0(p)and can be expanded as a Laurent series. But Θk(q) is a power series in q withconstant term unity. Thus

Θkω/Ψω =∑

aj(H/h)12j/r,

summed for j ≥ 0 and with a0 = 1.

To this equation apply the modular inversion ω → −1/pω. The standard formula

η(−1/ω) = η(ω)√ω/i combines with Lemma 17 to produce

Θlω = pk(Hp/h)∑

aj(√pH/h)−12j/r.

Here Hp/h and (H/h)12/r are power series in q2 with no constant term, thesmallest powers of q having respective exponents (p2 − 1)/12 and 2. Thus 2j ≤(p2 − 1)/12.

Remark 6. Lemma 17 is stronger than needed for Proposition 13 but has a furtherapplication to checking Table 1 below. Namely, modular inversion must interchangethe eta–function expressions for Θk and Θl.

10. The prime p = 13

By Proposition 13, each Θk(q) for 1 ≤ k ≤ 6 is h13/H times a seventh–degreepolynomial in (H/h)2 with constant term unity. Thus, the work reduces to com-puting and comparing initial segments of q–series. Identities of Euler and Jacobi[29, p. 123] supply convenient q–series expansions, for η(ω) and η3(ω), from whichto develop other powers. To compute the series coefficients for Θk(q) I ran Fortranprograms with nested loops, first to generate elements ξ as in (6), then to countthose passing exclusion tests based on the congruences (9). Results are summarisedin Table 1.

p = 13 Θ1/Ψ Θ2/Ψ Θ3/Ψ Θ4/Ψ Θ5/Ψ Θ6/Ψ1 1 1 1 1 1 1χ p2 p p p p pχ2 2232p2 2232p 2 · 3p 2 · 3p 2 · 3p 2 · 3pχ3 2 · 19p3 2 · 19p2 2211p 225p 225p 225pχ4 225p4 225p3 225p2 2211p 2 · 19p 2 · 19pχ5 2 · 3p5 2 · 3p4 2 · 3p3 2 · 3p2 2232p 2232pχ6 p6 p5 p4 p3 p2 p2

χ7 p6 p5 p4 p3 p2 p

Table 1. Coefficients for the expansion in powers of χ = (H/h)2 for Θj/Ψ, whereΨ = hp/H and p = 13.


It follows from [20, p. 115–116] that, unlike the cases p = 5, 7, the series Θk(q)for p = 13 do not span the space of modular forms of weight r and Nebentypus.

For the next result, write f(q) ≡ 0(mod6) to mean that every coefficient bN inthe q–series f(q) =

∑bNq

N is an integer divisible by 6. Further, if Q(x) is a pos-itive quadratic form, let a(N) denote the numbers of solutions of the Diophantineequation Q(x) = N . The a(N) are the so–called representation numbers for Q.

Proposition 14. [17, p. 470] For

f(q) =∏

[(1− qn)/(1− q13n)] + q∏

[(1− q13n)/(1− qn)]

we have df/dq ≡ 0(mod6).

Proof. Suppose f(q) =∑cNq

N . From the last column in Table 1 we find

Θ6(q) ≡ f(q)∏

(1− qn)6 + q3∏

(1− q13n)62

(mod6).

The binomial theorem for prime modulus p = 2 or 3 shows the cofactor of f(q) tobe a power series in qp(modp). The constant term is unity, whence the reciprocalis likewise of that form. Now, except a(0), all representation numbers a(N) of aquadratic form are even, so Θ6 ≡ 1(mod2). It follows that NcN ≡ 0(mod2).

For modulus 3, I show that either 3 divides (ξ, ξ), or else ξ is one of three elementswith the same measure. Observe that τ → τ3 defines an automorphism of order 3fixing the quartic subfield L = Q(λ) of K13, where λ = τ + τ3 + τ9. Thus everyinteger ξ ∈ L belongs to a triplet of elements with the same measure. On the otherhand, λ and its conjugates comprise an integral basis for L. So for ξ ∈ L, both sand t are divisible by 3, whence (ξ, ξ) = 13t− s2 ≡ 0(mod3).

11. Conclusion

The present investigation helps round out the work in [8] by determining gen-erating functions for the representation numbers of the quadratic forms discussedthere. By 7 April, 1980 I knew empirically the results of Proposition 10, havingfound those of Proposition 5 at the end of February. The Proposition 12 formulaefollowed soon after.

There was little further progress till, while browsing in Macquarie UniversityLibrary, I happened upon Reference [7]. Struck instantly by the geometric sig-nificance of its equations (56)–(57), I borrowed the library copy, with a returndate 17 May, 1991. The crucial novelty is the use of ratios of theta–series. Theseformulae atop page 110 are the progenitors of all the ratio formulae in Section 3above, and hence of the “elementary” proofs for the theta–formulae. Equation (44)(Proposition 7) emerged 7 October, 1991, but Proposition 8, not until 17 May, 1998.

To return to the opening theme of the Introduction, quadratic forms need not beenvisaged as mere passive beneficiaries of results, about modular forms, obtained byanalytic means. Rather, suitably chosen quadratic forms can yield modular–formproperties whose enunciations may show no trace of the underlying connection.Examples of such results include Corollary 7 as well as Propositions 7 and 14.

22 MAURICE CRAIG

A further particular type is as follows. Expressions given above for theta–seriesare in terms of the eta–function, whereas Hecke’s examples [20] are generally interms of Dirichlet series, or the q-series to which they readily convert. Series–product identities for modular forms follow by comparing the alternative expres-sions for the same theta–series. The method can be especially effective when severalseries, forming a basis for the relevant space of modular forms, are taken together.Thus, by comparing Proposition 10 with Hecke’s Beispiel 9 [20, p. 122], we obtainan alternative proof of Proposition 3 in [16, p. 10].

Further results of this type were proved by Bailey [3], Carlitz [5, 6], Garvan[13] and Raghavan [30]. It may be remarked that the last formula in [30] relatesto the theta–series for a prime divisor of 5 in K20. Application of the modulartransformation z → −1/5z to this formula, followed by removal of an extraneousfactor, can be used to derive Corollary 5 above. The penultimate formula [30, p.229] concerns the senary form (cf. Proposition 10) constructed as the sum of threecopies of the binary form 2(x2 + xy + 2y2).

Acknowledgments

Drs F. Garvan and M. Hirschhorn kindly supplied preprints and other advanceinformation on unpublished related developments. Prof. Hirschhorn recountedProposition 14 to me by phone on 18 June, 1991. Knowing Table 1 already, I couldfax him the proof above next day.

References

[1] G. E. Andrews, A simple proof of Jacobi’s triple product identity. Proc.Amer. Math. Soc. 16 (1965), 333–334.

[2] A. O. L. Atkin, and P. Swinnerton–Dyer, Some properties of partitions.Proc. Lond. Math. Soc. (3) 4 (1954), 84–106.

[3] W. N. Bailey, A further note on two of Ramanujan’s formulae. Quart. J.Math. Oxford (2) 3 (1952), 158–160.

[4] J. M. Borwein, P. B., Borwein and F. G. Garvan, Some cubic modularidentities of Ramanujan. Trans. Amer. Math. Soc. 343 (1994), 35–47.

[5] L. Carlitz, Note on some partition formulae. Quart. J. Math. (2) 4 (1953),168–172.

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[26] M. Hirschhorn, F. Garvan and J. Borwein, Cubic analogues of the Jacobiantheta function θ(z, q). Can. J. Math. 45 (1993), 673–694.

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[28] O. Kolberg, An elementary discussion of certain modular forms. Univ.Bergen Naturv. Rekke Nr. 16 (1959).

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24 MAURICE CRAIG

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