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On the number of representations of certain integers as sums of 11 or 13 squares
ARTICLE IN PRESS
Journal of Number Theory 103 (2003) 135–162
http://www.elsevier.com/locate/jnt
On the number of representations of certain integers as sums of 11 or 13 squares Shaun Cooper Institute of Information and Mathematical Sciences, Massey University Albany, Private Bag 102-904, North Shore Mail Centre, Auckland, New Zealand Received 22 August 2001; revised 22 May 2003 Communicated by D. Goss Dedicated to Sri Chinmoy on the occasion of his 70th birthday
Abstract Let rk ðnÞ denote the number of representations of an integer n as a sum of k squares. We prove that for odd primes p; ð p1Þ=2 4 9 p þ 352 r11 ð p2 Þ ¼ 330 31 ð p þ 1Þ 22ð1Þ 31 Hð pÞ;
where Hð pÞ is the coefficient of qp in the expansion of q
N N Y Y ð1 q2j Þ28 ð1 ðqÞ j Þ16 ð1 q2j Þ4 þ 32q2 : j 8 j¼1 j¼1 ð1 ðqÞ Þ
This result, together with the theory of modular forms of half integer weight is used to prove that 0 9Ilp =2m 29Il=2mþ9 1 Y p9Ilp =2mþ9 1 1 4 n p p ; r11 ðnÞ ¼ r11 ðn0 Þ p 29 1 p9 1 p9 1 p Q where n ¼ 2l p plp is the prime factorisation of n; n0 is the square-free part of n; and n0 is of the form 8k þ 7: The products here are taken over the odd primes p; and ðnpÞ is the Legendre symbol. We also prove that for odd primes p; 11 5 13936 r13 ð p2 Þ ¼ 4030 691 ð p þ 1Þ 26p þ 691 tð pÞ;
E-mail address: [email protected]. 0022-314X/$ - see front matter r 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.jnt.2003.06.001
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Q PN j 24 n where tðnÞ is Ramanujan’s t function, defined by q N j¼1 ð1 q Þ ¼ n¼1 tðnÞq : A conjectured formula for r2kþ1 ð p2 Þ is given, for general k and general odd primes p: r 2003 Elsevier Inc. All rights reserved. MSC: Primary 11E25; Secondary 05A15; 33D15; 33E05 Keywords: Hecke operator; Modular forms of half integer weight; Sums of squares
1. Introduction Let rk ðnÞ denote the number of representations of n as a sum of k squares. The generating function for rk ðnÞ is N X j¼N
!k q
j2
¼
N X
rk ðnÞqn :
n¼0
The study of rk ðnÞ has a long and interesting history. Generating functions which yield the value of rk ðnÞ for k ¼ 2; 4; 6 and 8 were found by Jacobi [18]. Glaisher [12] proved formulas for k ¼ 10; 12; 14; 16 and 18; and Ramanujan [25, Eqs. (145)–(147)] stated a general formula for rk ðnÞ for arbitrary even values of k: Ramanujan’s formula was proved by Mordell [23] and an elementary proof was given by Cooper [4]. The problem of determining rk ðnÞ for odd values of k is more difficult, and not so well known. The value of r3 ðnÞ was found by Gauss [11, Section 291]. Formulas for r5 ðnÞ and r7 ðnÞ for square-free values of n were stated without proof by Eisenstein [8,10]. These formulas were extended to all non-negative integers n (again without proof) by Smith [30]. The Paris Academy of Sciences, apparently unaware of Smith’s work [30], proposed as its Grand Prix des Sciences Mathe´matiques competition for 1882 the problem of completely determining the value of r5 ðnÞ: The prize was awarded jointly to Smith [32] and Minkowski [22], who both gave formulas as well as proofs. An interesting account of this competition and the controversy surrounding it has been given by Serre [29]. Others to have worked on this problem are (chronologically) Stieltjes [33], Hurwitz [16], Hardy [13,14], Lomadze [20,21] and Hirschhorn and Sellers [15]. More information can be found in [7, Chapters VI–IX]. Eisenstein [9,31] stated that the sequence of formulas for rk ðnÞ ceases for kX9: As a result of computer investigations [5], I found that there are in fact simple, closedform formulas for r9 ðnÞ; but only for certain values of n: The purpose of this article is to state and prove the corresponding results for r11 ðnÞ: These results were also initially discovered as a result of computer investigations. There appear to be no similar formulas for rk ðnÞ for k ¼ 13; 15; 17; y: Formulae can be given, but they all involve more complicated number theoretic functions.
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2. Summary of results Q Throughout this article p will always denote an odd prime, and p will always denote a product over all odd primes p: Accordingly, let us denote the prime factorisation of n by Y n ¼ 2l pl p ; p
where l and lp are all non-negative integers, only finitely many of which are nonzero. Let n0 denote the square-free part of n: Hirschhorn and Sellers [15] with the help of their referee proved that r3 ðnÞ ¼ r3 ðn0 Þ
YpIlp =2mþ1 1 p
p1
0 Ilp =2m n p 1 ; p1 p
ð2:1Þ
where n0 is the square-free part of n: The analogous results for r5 ðnÞ; r7 ðnÞ and r9 ðnÞ were proved by Cooper [5]. For sums of five squares we have 3Il=2mþ3 2 1 23Il=2m 1 þ e 23 1 23 1 Y p3Ilp =2mþ3 1 n0 p3Ilp =2m 1 p ; p3 1 p3 1 p p
r5 ðnÞ ¼ r5 ðn0 Þ
ð2:2Þ
where 8 > <0 e ¼ 4 > : 16=7
if n0 1 ðmod 8Þ; if n0 2 or 3 ðmod 4Þ;
ð2:3Þ
if n0 5 ðmod 8Þ:
The result for sums of seven squares is 5Il=2mþ5 2 1 25Il=2m 1 þe r7 ðnÞ ¼ r7 ðn Þ 25 1 25 1 5Ilp =2m 0 Y p5Ilp =2mþ5 1 1 2 n p p ; p5 1 p5 1 p p 0
ð2:4Þ
where 8 > <8 e¼ 0 > : 64=37
if n0 1 or 2 ðmod 4Þ; if n0 3 ðmod 8Þ; 0
if n 7 ðmod 8Þ:
ð2:5Þ
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For squares, these formulas reduce to lp Yplp þ1 1 ð p1Þ=2 p 1 2 ð1Þ r3 ðn Þ ¼ 6 ; p1 p1 p
r5 ðn2 Þ ¼ 10
3lþ3 2 1 Y p3lp þ3 1 p3lp 1 p ; 23 1 p3 1 p3 1 p
25lþ5 1 25l 1 þ 8 25 1 25 1 5lp Y p5lp þ5 1 1 ð p1Þ=2 2 p ð1Þ p : p5 1 p5 1 p
r7 ðn2 Þ ¼ 14
ð2:6Þ
ð2:7Þ
ð2:8Þ
Eqs. (2.6) and (2.7) seem to have been first explicitly stated by Hurwitz [17] and [16], respectively, and (2.8) is due to Sandham [27]. For sums of nine squares we have 0 7Ilp =2m 7Il=2mþ7 1 Y p7Ilp =2mþ7 1 1 0 2 3 n p p r9 ¼ r9 ðn Þ ð2:9Þ 27 1 p7 1 p7 1 p p in the case that n0 5 ðmod 8Þ: Results for when n0 c5 ðmod 8Þ were also given in [5], but these involve additional and more complicated number theoretic functions. The first purpose of this article is to prove that for odd primes p; ð p1Þ=2 4 9 r11 ð p2 Þ ¼ 330 p þ 352 31 ð p þ 1Þ 22ð1Þ 31 Hð pÞ;
where Hð pÞ is the coefficient of qp in the expansion of q
N N Y Y ð1 q2j Þ28 ð1 ðqÞ j Þ16 ð1 q2j Þ4 þ 32q2 : j 8 j¼1 j¼1 ð1 ðqÞ Þ
This result, together with the theory of modular forms of half integer weight, is then used to prove 0 9Ilp =2m 29Il=2mþ9 1 Y p9Ilp =2mþ9 1 1 4 n p p r11 ðnÞ ¼ r11 ðn0 Þ ð2:10Þ 29 1 p9 1 p9 1 p p in the case that n0 7 ðmod 8Þ: Some results for when n0 c7 ðmod 8Þ will also be given, but just as for sums of nine squares, these involve additional and more complicated number theoretic functions. A conjecture for the value of r11 ðn0 Þ for square-free values of n0 satisfying n0 7 ðmod 8Þ was given in [5, p. 471]. The second purpose of this article is to prove that for odd primes p; 11 5 13936 r13 ð p2 Þ ¼ 4030 691 ð p þ 1Þ 26p þ 691 tð pÞ;
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where tðnÞ is Ramanujan’s t function, defined by q
N N Y X ð1 q j Þ24 ¼ tðnÞqn : j¼1
ð2:11Þ
n¼1
We also compute the eigenvalues and eigenfunctions of the Hecke operators Tp2 : There does not appear to be a formula like (2.1), (2.2), (2.4), (2.9) or (2.10) that expresses r13 ðnÞ in terms of just r13 ðn0 Þ and the prime factorisation of n; for any special cases of n0 : Finally, we offer a conjectured formula for the value of r2kþ1 ð p2 Þ; for general k and a general odd prime p:
3. Definitions Let jqjo1 and set ða; qÞN ¼
N Y ð1 aq j1 Þ: j¼1
Let N X
fðqÞ ¼
2
qj ;
j¼N
cðqÞ ¼
N X
q jð jþ1Þ=2 :
j¼0
We will use the notation ½qn f ðqÞ :¼ the coefficient of qn in f ðqÞ: Let s and t be non-negative integers. We will always assume t is a multiple of 4: Put fs;t ðqÞ ¼ fðqÞs ð2q1=4 cðq2 ÞÞt ; and let us write fs;t ðqÞ ¼
N X
fs;t ðnÞqn ;
n¼0
so that fs;t ðnÞ is the coefficient of qn in the series expansion of fs;t ðqÞ: Geometrically, fs;t ðnÞ counts the number of lattice points, that is, points with integer coordinates, on the sphere s X j¼1
x2j þ
sþt X
ðxj 1=2Þ2 ¼ n:
j¼sþ1
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Let rs;t ðnÞ denotes the number of representations of n as a sum of s þ t squares, of which s are even and t are odd. Then rs;t ðnÞ has the generating function N X sþt sþt rs;t ðnÞqn ¼ ð2qÞt fðq4 Þs cðq8 Þt ¼ fs;t ðq4 Þ: t t n¼0 Let z ¼ fðqÞ2 ; x ¼ 16q
cðq2 Þ4 fðqÞ4
:
Then [3] or [24, Chapter 16, Entry 25(vii)] zðsþtÞ=2 xt=4 ¼ fs;t ðqÞ; 1x¼
fðqÞ4 fðqÞ4
:
By Jacobi’s triple product identity [3, p. 35], z¼
ðq; qÞ4N ðq2 ; q2 Þ2N
;
xð1 xÞ ¼ 16q
ðq2 ; q2 Þ24 N ðq; qÞ24 N
ð3:1Þ
:
Therefore z5 xð1 xÞ ¼ 16q
ðq2 ; q2 Þ14 N ðq; qÞ4N
ð3:2Þ
;
z6 xð1 xÞ ¼ 16qðq2 ; q2 Þ12 N; z10 x2 ð1 xÞ2 ¼ 256q2
ðq2 ; q2 Þ28 N ðq; qÞ8N
z12 xð1 xÞ ¼ 16qðq; qÞ24 N ¼ 16
ð3:3Þ ð3:4Þ
;
N X ð1Þnþ1 tðnÞqn ;
ð3:5Þ
n¼1
where tðnÞ is Ramanujan’s t-function defined in Eq. (2.11). Define HðqÞ and HðnÞ by 2 2 4 2 HðqÞ ¼ qðq; qÞ16 N ðq ; q ÞN þ 32q
ðq2 ; q2 Þ28 N ðq; qÞ8N
¼
N X n¼1
HðnÞqn :
ð3:6Þ
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Then by (3.1), (3.4) and (3.5), 1 10 z xð1 xÞð1 þ 2x 2x2 Þ: HðqÞ ¼ 16
ð3:7Þ
We will require some facts about Hecke operators on modular forms of half integer weight. All of the facts below can be found in [19]. * 0 ð4ÞÞ is the vector space consisting of all linear combinations of Fact 1. Mð2kþ1Þ=2 ðG f2kþ14j;4j ðqÞ; j ¼ 0; 1; y; Ik=2m [19, Proposition 4, p. 184]. Fact 2. The Hecke operators Tp2 ; where p is any prime, map Mð2kþ1Þ=2 ðG* 0 ð4ÞÞ into itself [19, p. 206]. That is, * 0 ð4ÞÞ ) Tp2 f AMð2kþ1Þ=2 ðG * 0 ð4ÞÞ: f AMð2kþ1Þ=2 ðG Fact 3. Suppose f ðqÞ ¼
N X
* 0 ð4ÞÞ: aðnÞqn AMð2kþ1Þ=2 ðG
n¼0
If p is an odd prime, then [19, Proposition 13, p. 207] implies ! N N N X X X ð1Þk n 2 n k1 Tp2 f ¼ að p nÞq þ p aðn=p2 Þqn ; aðnÞqn þ p2k1 p n¼0 n¼0 n¼0 while if p ¼ 2; the note in [19, p. 210] implies T22 f ¼
N X
að4nÞqn :
n¼0
4. Some preliminary lemmas Lemma 4.1. Let N X 1 5 z ð5 xÞð1 xÞ ¼ aðnÞqn ; 4 n¼0 N X 1 5 z ð4x þ x2 Þ ¼ bðnÞqn ; 64 n¼0 N X 1 5 z xð1 xÞ ¼ cðnÞqn ; 16 n¼0
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N X 1 z6 ð1 xÞð1 x þ x2 Þ ¼ dðnÞqn ; 8 n¼0 N X 1 6 z ð1 xÞð1 x2 Þ ¼ eðnÞqn : 8 n¼0
Q Let n ¼ 2l p plp be the prime factorisation of n: For any prime p 1 ðmod 4Þ; let xp and yp be the unique pair of positive integers satisfying x2p þ y2p ¼ p;
2jxp ;
and define yp by tan yp ¼
yp ; xp
p 0oyo : 2
Then að0Þ ¼ 5=4;
bð0Þ ¼ 0;
cð0Þ ¼ 0;
dð0Þ ¼ 1=8;
eð0Þ ¼ 1=8
and for nX1; " # 4ðlp þ1Þ Y ð1Þðlp þ1Þð p1Þ=2 lp ð p1Þ=2 p aðnÞ ¼ ð1Þ ; p4 ð1Þð p1Þ=2 p
bðnÞ ¼ 24l
Y p4ðlp þ1Þ ð1Þðlp þ1Þð p1Þ=2 p4 ð1Þð p1Þ=2
p
cðnÞ ¼
8 > < 2 > : n ð1Þ
; if lp is odd for any p 3 ðmod 4Þ;
0 l
Q p1 ðmod 4Þ
sin 4ð1 þ lp Þyp sin 4yp
8 > > > <
Q p5lp þ5 1 p5 1 p dðnÞ ¼ > 30 25l þ 63 Q p5lp þ5 1 > > : 25 1 p5 1 p 8 > > > <
Q p5lp þ5 1 p5 1 p eðnÞ ¼ 5lþ5 > 2 63 Q p5lp þ5 1 > > : 25 1 p5 1 p
otherwise;
if n is odd; if n is even;
if n is odd; if n is even:
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Proof. From [3] or [24, Chapter 17, Entry 17(viii)] we have N 1 5 5 X ð1Þk ð2k þ 1Þ4 q2kþ1 z ð5 xÞð1 xÞ ¼ þ : 4 4 k¼0 1 q2kþ1
Therefore að0Þ ¼ 5=4 and X ð1Þðd1Þ=2 d 4 aðnÞ ¼ djn d odd
" # 4ðlp þ1Þ Y ð1Þðlp þ1Þð p1Þ=2 lp ð p1Þ=2 p ¼ ð1Þ : p4 ð1Þð p1Þ=2 p Next, from [3] or [24, Chapter 17, Entry 17(iii)] we have N X 1 5 k 4 qk z ð4x þ x2 Þ ¼ : 64 1 þ q2k k¼1
Therefore bð0Þ ¼ 0 and bðnÞ ¼
X
ð1Þðd1Þ=2
djn d odd
¼ 24l
n 4 d
Y p4ðlp þ1Þ ð1Þðlp þ1Þð p1Þ=2 p
p4 ð1Þð p1Þ=2
:
For the coefficients cðnÞ; by (3.2) and one of the Macdonald identities for BC2 (for example, see [6]) we have 1 5 ðq2 ; q2 Þ14 1 N z xð1 xÞ ¼ q ¼ 4 16 ðq; qÞN 6
X
2
abða2 b2 Þqða þb
2
Þ=5
:
a2 ðmod 5Þ b1 ðmod 5Þ
Therefore cð0Þ ¼ 0 and cðnÞ ¼
¼
1 6
X
abða2 b2 Þ
a2 þb2 ¼5n a;b1 ðmod 5Þ
8 > <0
2 > : n ð1Þ
if lp is odd for any p 3 ðmod 4Þ; l
Q p1 ðmod 4Þ
sin4ð1 þ lp Þyp sin4yp
otherwise:
The second part of this formula expressing cðnÞ in terms of yp and lp was stated (without proof) by Ramanujan [25, Eq. (156)].
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Lastly, from [3] or [24, Chapter 17, Entries 14(iii) and (vi)] we have N 1 1 X ð1Þk k5 qk z6 ð1 xÞð1 x þ x2 Þ ¼ ; 8 8 k¼1 1 þ qk N 1 6 1 X ð1Þk k5 qk z ð1 xÞð1 x2 Þ ¼ : 8 8 k¼1 1 qk
Therefore, dðnÞ ¼
X
ð1Þðdþn=dÞ d 5
djn
8 Q p5lp þ5 1 > > > < p p5 1 ¼ > 30 25l þ 63 Q p5lp þ5 1 > > : 25 1 p5 1 p
eðnÞ ¼
X
if n is odd; if n is even;
ð1Þd d 5
djn
8 Q p5lp þ5 1 > > if n is odd; > < p p5 1 ¼ 5lþ5 > 2 63 Q p5lp þ5 1 > > if n is even: : 25 1 p5 1 p
&
Lemma 4.2. If p is an odd prime then að pÞ ¼ ð1Þð p1Þ=2 p4 þ 1; dð pÞ ¼ p5 þ 1; eð pÞ ¼ p5 þ 1:
Proof. These follow right away from Lemma 4.1.
&
Lemma 4.3. Each of aðnÞ; bðnÞ; cðnÞ; dðnÞ and eðnÞ are multiplicative. That is, aðmnÞ ¼ aðmÞaðnÞ; if m and n are relatively prime; and similarly for b; c; d and e: Proof. These follow right away from Lemma 4.1.
&
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Lemma 4.4. Under the change of variables z-zð1 xÞ1=2 ;
x-
x ; x1
the functions z10 ð5x5 þ 411x4 2118x3 þ 4262x2 4275x þ 1710Þ and z12 ð43870208x6 131830116x5 þ 162785487x4 105780950x3 þ 162565995x2 131610624x þ 43870208Þ are invariant and z6 xð1157 1157x þ 1008x2 Þ changes sign. Proof. This follows by straightforward calculation.
&
Lemma 4.5. The q-expansions of z10 ð5x5 þ 411x4 2118x3 þ 4262x2 4275x þ 1710Þ and z12 ð43870208x6 131830116x5 þ 162785487x4 105780950x3 þ 162565995x2 131610624x þ 43870208Þ contain only even powers of q; and the q-expansion of z6 xð1157 1157x þ 1008x2 Þ contains only odd powers. Proof. This follows from the Change of Sign Principle [3, p. 126] or [17, Chapter 24, Entry 14(xii)]. & Lemma 4.6. ½qp z10 ð1 þ xÞð1 þ 14x þ x2 Þð1 34x þ x2 Þ ¼ 264ð p9 þ 1Þ; ½qp z12 ½441ð1 þ 14x þ x2 Þ3 þ 250ð1 þ xÞ2 ð1 34x þ x2 Þ2 ¼ 65520ð p11 þ 1Þ:
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Proof. From [3] or [24, Chapter 17, Entries 13(iii) and (iv)], followed by [2] or [24, Chapter 15, Entry 12(iii)] we have z10 ð1 þ xÞð1 þ 14x þ x2 Þð1 34x þ x2 Þ ¼ MðqÞNðqÞ ¼ 1 264
N X k 9 qk : 1 qk k¼1
The coefficient of qp in this is readily seen to be 264ð p9 þ 1Þ: Also, from [3] or [24, Chapter 17, Entries 13(iii) and (iv)] followed by [2] or [24, Chapter 15, Entry 13(i)] we have 441z12 ð1 þ 14x þ x3 Þ3 þ 250z12 ð1 þ xÞ2 ð1 34x þ x2 Þ2 ¼ 441MðqÞ3 þ 250NðqÞ2 ¼ 691 þ 65520
N X j 11 q j : 1qj j¼1
The coefficient of qp in this is readily seen to be 65520ð p11 þ 1Þ: & 5. The value of r11 ð p2 Þ for odd primes p Theorem 5.1. If p is an odd prime and Hð pÞ is defined by Eq. (3.6), then ð p1Þ=2 4 9 r11 ð p2 Þ ¼ 330 p þ 352 31 ð p þ 1Þ 22ð1Þ 31 Hð pÞ:
Proof. Let us write p2 ¼
11 X
x2i :
i¼1
There are three possibilities: either one, five or nine of the xi are odd, and the others are even. Accordingly, we have 22 X 22 X r6;4 ð p2 j 2 Þ þ r2;8 ð p2 j 2 Þ 5 9 j odd j odd j odd X 2 2 10 22 16q4 fðq4 Þ6 cðq8 Þ4 ¼ ½q p j 22fðq4 Þ10 þ 5 4 j odd 10 22 2 8 4 2 8 8 16 q fðq Þ cðq Þ þ 9 8 X 11 11 2 2 ¼2 ½qð p j Þ=4 fðqÞ10 þ 16qfðqÞ6 cðq2 Þ4 1 5 j odd
r11 ð p2 Þ ¼ 22
X
r10;0 ð p2 j 2 Þ þ
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þ
11
9
162 q2 fðqÞ2 cðq2 Þ8
147
p X 11 11 11 ½q jð pjÞ z5 þ xþ x2 1 5 9 j¼0
¼
p X ½q jð pjÞ z5 ð11 þ 462x þ 55x2 Þ:
¼
j¼0
In general, we can also prove that p X r2kþ1 ð p Þ ¼ ½q jð pjÞ zk 2
j¼0
Ik=2m X i¼0
2k þ 1 4i þ 1
! xi :
ð5:2Þ
Now 5 2 528 5 264 5 z5 ð11 þ 462x þ 55x2 Þ ¼ 11 5 z ð5 xÞð1 xÞ þ 5 z ð4x þ x Þ þ 5 z xð1 xÞ:
Using this together with Lemmas 4.1–4.3 gives r11 ð p2 Þ ¼
p X
44 5 að jð p
4224 jÞÞ þ 33792 5 bð jð p jÞÞ þ 5 cð jð p jÞÞ
j¼0 88 44 ¼ 88 5 að0Þ 5 að0Það pÞ þ 5
p X
að jÞað p jÞ
j¼0
þ 33792 5
p X j¼0
bð jÞbð p jÞ þ 4224 5
p X
cð jÞcð p jÞ
j¼0
2 p 1 5 ¼ 22ð1Þð p1Þ=2 p4 þ 44 5 ½q 4z ð5 xÞð1 xÞ 2 p 1 5 2 2 4224 p 1 5 þ 33792 5 ½q 64 z ð4x þ x Þ þ 5 ½q 16z xð1 xÞ p 10 4 2 ¼ 22ð1Þð p1Þ=2 p4 þ 11 4 ½q z ð2x þ 20x 12x þ 5Þ:
Now 4 11 4 ð2x
5 þ 20x2 12x þ 5Þ ¼ 124 p1 ðxÞ þ 22 31 p2 ðxÞ þ p3 ðxÞ
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where p1 ðxÞ ¼ ð1 þ xÞð1 þ 14x þ x2 Þð1 34x þ x2 Þ; p2 ðxÞ ¼ xð1 xÞð1 þ 2x 2x2 Þ; 1 ð5x5 þ 411x4 2118x3 þ 4262x2 4275x þ 1710Þ: p3 ðxÞ ¼ 124
This, together with Lemmas 4.5, 4.6 and Eq. (3.7) gives 5 ð264Þð p9 þ 1Þ þ 22 r11 ð p2 Þ ¼ 22ð1Þð p1Þ=2 p4 124 31 16Hð pÞ þ 0 ð p1Þ=2 4 9 ¼ 330 p þ 352 31 ð p þ 1Þ 22ð1Þ 31 Hð pÞ:
&
Remark 5.3. The coefficients HðnÞ have some interesting properties. 1. H is multiplicative. That is, HðmnÞ ¼ HðmÞHðnÞ for any pair of relatively prime positive integers m and n: 2. jHðnÞjpn9=2 dðnÞ; where dðnÞ is the number of divisors of n: In particular, jHð pÞjp2p9=2
ð5:4Þ
for primes p: Ramanujan [25, Section 28] made some conjectures concerning the orders of some similar functions. See his Eqs. (157), (160) and (163). See also Berndt’s commentary [26, p. 367] for references and more information.
6. The Hecke operator Tp2 when p ¼ 2 for eleven squares Theorem 6.1. T4 f11;0 ¼ f11;0 þ 330f7;4 þ 165f3;8 ; T4 f7;4 ¼ 336f7;4 þ 176f3;8 ; T4 f3;8 ¼ 320f7;4 þ 192f3;8 : That is, f11;0 ð4nÞ ¼ f11;0 ðnÞ þ 330f7;4 ðnÞ þ 165f3;8 ðnÞ; f7;4 ð4nÞ ¼ 336f7;4 ðnÞ þ 176f3;8 ðnÞ; f3;8 ð4nÞ ¼ 320f7;4 ðnÞ þ 192f3;8 ðnÞ:
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Proof. Let us consider f11;0 first. By Facts 1 and 2, we have T4 f11;0 ¼ a1 f11;0 þ a2 f7;4 þ a3 f3;8 ; for some constants a1 ; a2 ; a3 : By Fact 3, we have
T4 f11;0 ¼
N X
f11;0 ð4nÞqn :
n¼0
Therefore N X n¼0
f11;0 ð4nÞqn ¼ a1
N X
f11;0 ðnÞqn þ a2
n¼0
N X
f7;4 ðnÞqn þ a3
n¼0
N X
f3;8 ðnÞqn :
n¼0
Equating coefficients of 1; q and q2 gives a1 ¼ 1; 22a1 þ 16a2 ¼ 5302; 220a1 þ 224a2 þ 256a3 ¼ 116380 and so a1 ¼ 1;
a2 ¼ 330;
a3 ¼ 165:
This proves the first part of the theorem. The results for T4 f7;4 and T4 f3;8 follow similarly. & Corollary 6.2. The eigenfunctions and eigenvalues of T4 are w11;1 :¼ 511f11;0 682f7;4 þ 187f3;8 ; w11;2 :¼ 20f7;4 þ 11f3;8 ; w11;3 :¼ f7;4 f3;8 ;
l1 ¼ 1;
l2 ¼ 512;
l3 ¼ 16:
These coefficients satisfy many interesting properties. Some of these are summarised in
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Theorem 6.3. 32w11;1 ðnÞ 33w11;2 ðnÞ ¼ 0
if n 1 or 2 ðmod 4Þ;
512w11;1 ðnÞ 17w11;2 ðnÞ ¼ 0 15872w11;1 ðnÞ þ 495w11;2 ðnÞ ¼ 0
if n 3 ðmod 8Þ; if n 7 ðmod 8Þ;
w11;3 ðnÞ ¼ 0
if n 7 ðmod 8Þ:
Proof. A proof of Theorem 6.3 has been given by Barrucand and Hirschhorn [1]. & As a consequence of the previous results we have Theorem 6.4. r11 ð4l ð8k þ 7ÞÞ ¼
29lþ9 1 r11 ð8k þ 7Þ: 29 1
Proof. r11 ð4l ð8k þ 7ÞÞ ¼ f11;0 ð4l ð8k þ 7ÞÞ 1 ð31w11;1 ð4l ð8k þ 7ÞÞ þ 495w11;2 ð4l ð8k þ 7ÞÞ þ 11242w11;3 ð4l ð8k þ 7ÞÞÞ ¼ 31511 1 ð31w11;1 ð8k þ 7Þ þ 495 512l w11;2 ð8k þ 7Þ þ 11242 ð16Þl w11;3 ð8k þ 7ÞÞ; ¼ 31511
where we have applied Corollary 6.2 in the last two equalities. Applying Theorem 6.3 gives 1 r11 ð4l ð8k þ 7ÞÞ ¼ 31511 ð31w11;1 ð8k þ 7Þ 15872 512l w11;1 ð8k þ 7Þ þ 0Þ 1 ð512lþ1 1Þw11;1 ð8k þ 7Þ: ¼ 511
Taking l ¼ 0 in this gives r11 ð8k þ 7Þ ¼ w11;1 ð8k þ 7Þ and so 1 r11 ð4l ð8k þ 7ÞÞ ¼ 511 ð512lþ1 1Þr11 ð8k þ 7Þ ¼
as required.
&
29lþ9 1 r11 ð8k þ 7Þ; 29 1
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7. The Hecke operator Tp2 when p is an odd prime for sums of eleven squares By Facts 2 and 3 we have Tp2 f11;0 ¼
N X
f11;0 ð p2 nÞqn þ p4
n¼0
N N X X n f11;0 ðn=p2 Þqn f11;0 ðnÞqn þ p9 p n¼0 n¼0
¼ c1 f11;0 ðqÞ þ c2 f7;4 ðqÞ þ c3 f3;8 ðqÞ;
Tp2 f7;4 ¼
N X
f7;4 ð p2 nÞqn þ p4
n¼0
N N X X n f7;4 ðn=p2 Þqn f7;4 ðnÞqn þ p9 p n¼0 n¼0
¼ d1 f11;0 ðqÞ þ d2 f7;4 ðqÞ þ d3 f3;8 ðqÞ;
Tp2 f3;8 ¼
N X
f3;8 ð p2 nÞqn þ p4
n¼0
ð7:1Þ
ð7:2Þ
N N X X n f3;8 ðn=p2 Þqn f3;8 ðnÞqn þ p9 p n¼0 n¼0
¼ e1 f11;0 ðqÞ þ e2 f7;4 ðqÞ þ e3 f3;8 ðqÞ;
ð7:3Þ
for some constants c1 ; c2 ; c3 ; d1 ; d2 ; d3 ; e1 ; e2 ; e3 : Equating the constant terms in each of the three equations above gives c1 ¼ p9 þ 1; d1 ¼ 0; e1 ¼ 0: Equating the coefficients of q in (7.1)–(7.3) gives 22c1 þ 16c2 ¼ f11;0 ð p2 Þ þ 22ð1Þð p1Þ=2 p4 ; 22d1 þ 16d2 ¼ f7;4 ð p2 Þ þ 16ð1Þð p1Þ=2 p4 ; 22e1 þ 16e2 ¼ f3;8 ð p2 Þ: By Theorem 5.1 we have ð p1Þ=2 4 9 f11;0 ð p2 Þ ¼ 330 p þ 352 31 ð p þ 1Þ 22ð1Þ 31 Hð pÞ:
ð7:4Þ
By the same methods it can be shown that ð p1Þ=2 4 9 f7;4 ð p2 Þ ¼ 320 p þ 176 31 ð p þ 1Þ 16ð1Þ 31 Hð pÞ;
ð7:5Þ
9 320 f3;8 ð p2 Þ ¼ 320 31 ð p þ 1Þ 31 Hð pÞ:
ð7:6Þ
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It follows that 9 22 c2 ¼ 22 31 ð p þ 1Þ þ 31 Hð pÞ; 9 11 d2 ¼ 20 31 ð p þ 1Þ þ 31 Hð pÞ; 9 20 e2 ¼ 20 31 ð p þ 1Þ 31 Hð pÞ:
Next, equating coefficients of q4 in (7.1)–(7.3) gives f11;0 ð4p2 Þ þ 5302ð1Þð p1Þ=2 p4 ¼ 5302c1 þ 5376c2 þ 5120c3 ; f7;4 ð4p2 Þ þ 5376ð1Þð p1Þ=2 p4 ¼ 5302d1 þ 5376d2 þ 5120d3 ; f3;8 ð4p2 Þ þ 5120ð1Þð p1Þ=2 p4 ¼ 5302e1 þ 5376e2 þ 5120e3 : By Theorem 6.1 and Eqs. (7.4)–(7.6) we have f11;0 ð4p2 Þ ¼ f11;0 ð p2 Þ þ 330f7;4 ð p2 Þ þ 165f3;8 ð p2 Þ ð p1Þ=2 4 9 ¼ 158730 p þ 5632 31 ð p þ 1Þ 5302ð1Þ 31 Hð pÞ:
Similarly, ð p1Þ=2 4 9 f7;4 ð4p2 Þ ¼ 163840 p þ 2816 31 ð p þ 1Þ 5376ð1Þ 17 Hð pÞ ð p1Þ=2 4 9 f3;8 ð4p2 Þ ¼ 163840 p 5120 31 ð p þ 1Þ 5120ð1Þ 17 Hð pÞ:
It follows that 9 22 c3 ¼ 22 31 ð p þ 1Þ 31Hð pÞ; 9 11 d3 ¼ 11 31 ð p þ 1Þ 31Hð pÞ; 9 20 e3 ¼ 11 31 ð p þ 1Þ þ 31Hð pÞ:
The above results may be summarised as Theorem 7.7. Let a ¼ p9 þ 1;
b ¼ Hð pÞ:
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Then 0 1 Tp2 f11;0 31a 1B B T 2f C @ p 7;4 A ¼ @ 0 31 Tp2 f3;8 0 0
22a þ 22b 20a þ 11b 20a 20b
10 1 f11;0 22a 22b CB C 11a 11b A@ f7;4 A: f3;8 11a þ 20b
Corollary 7.8. The eigenfunctions and eigenvalues of Tp2 are z11;1 :¼ f11;0 2f7;4 ;
l1 ¼ a;
z11;2 :¼ 20f7;4 þ 11f3;8 ¼ w11;2 ; z11;3 :¼ f7;4 f3;8 ¼ w11;3 ;
l2 ¼ a;
l3 ¼ b:
That is, Tp2 z11;1 ¼ ð p9 þ 1Þz11;1 ; Tp2 z11;2 ¼ ð p9 þ 1Þz11;2 ; Tp2 z11;3 ¼ Hð pÞz11;3 :
Using Fact 3, together with Corollary 7.8, we have 2 4 n z11;1 ð p nÞ þ p z11;1 ðnÞ þ p9 z11;1 ðn=p2 Þ ¼ ð p9 þ 1Þz11;1 ðnÞ; p
2
z11;2 ð p nÞ þ p
4
z11;3 ð p2 nÞ þ p4
ð7:9Þ
n z11;2 ðnÞ þ p9 z11;2 ðn=p2 Þ ¼ ð p9 þ 1Þz11;2 ðnÞ; p
ð7:10Þ
n z11;3 ðnÞ þ p9 z11;3 ðn=p2 Þ ¼ Hð pÞz11;3 ðnÞ: p
ð7:11Þ
Iterating (7.9)–(7.11) we obtain Theorem 7.12. If p2 [n; then 9m 9mþ9 p 1 4 n p 1 z11;1 ð p nÞ ¼ p z ðnÞ; p9 1 p p9 1 11;1 9mþ9 9m p 1 4 n p 1 z11;2 ð p2m nÞ ¼ p z ðnÞ; p9 1 p p9 1 11;2 2m
z11;3 ð p2m nÞ ¼ ½c1 ðnÞrm1 þ c2 ðnÞrm2 z11;3 ðnÞ;
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where pffiffiffiffiffiffiffiffiffiffiffi Hð pÞ þ Dð pÞ ; r1 ¼ 2 pffiffiffiffiffiffiffiffiffiffiffi Hð pÞ Dð pÞ ; r2 ¼ 2 c1 ðnÞ ¼
4 n 1 Hð pÞ 2p ð p Þ pffiffiffiffiffiffiffiffiffiffiffi þ ; 2 2 Dð pÞ
c2 ðnÞ ¼
4 n 1 Hð pÞ 2p ð p Þ pffiffiffiffiffiffiffiffiffiffiffi ; 2 2 Dð pÞ
Dð pÞ ¼ Hð pÞ2 4p9 : Remark 7.13. Eq. (5.4) implies Dð pÞo0: It is possible to express r11 ðnÞ in terms of r11 ðno Þ and z11;3 ðno Þ; where no is the oddsquare free part of n; that is, no is the greatest divisor of n which is not divisible by an odd square. Let gp;m;m ¼
9m p9mþ9 1 4 m p 1 p ; p9 1 p p9 1
hp;m;m ¼ c1 ðmÞrm1 þ c2 ðmÞrm2 ; where c1 ; c2 ; r1 ; r2 are as for Theorem 7.12. If p2 [m then 1 r11 ð p2m mÞ ¼ 31 ½31z11;1 ð p2m mÞ þ 2z11;2 ð p2m mÞ þ 22z11;3 ð p2m mÞ 1 ½31gp;m;m z11;1 ðmÞ þ 2gp;m;m z11;2 ðmÞ þ 22hp;m;m z11;3 ðmÞ ¼ 31
¼
1 ½31gp;m;m z11;1 ðmÞ þ 2gp;m;m z11;2 ðmÞ þ 22gp;m;m z11;3 ðmÞ 31 þ 22 31 ½hp;m;m gp;m;m z11;3 ðmÞ
¼ gp;m;m r11 ðmÞ þ 22 31 ½hp;m;m gp;m;m z11;3 ðmÞ: By induction on j; it follows that if p1 ; p2 ; y; pj are distinct odd primes, and p2i [m for 1pipj; then 2m
2m
2m
r11 ð p1 1 p2 2 ?pj j mÞ ! " ! j j Y 22 Y ¼ gpi ;mi ;m r11 ðmÞ þ hp m ;m 31 i¼1 i i i¼1
j Y i¼1
!# gpi ;mi ;m
z11;3 ðmÞ:
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Consequently, if no is the odd-square free part of n and either p2mp jjn or p2mp þ1 jjn; then o 9m n p p 1 r11 ðnÞ ¼ r11 ðn Þ p p9 1 p9 1 p p " o 9m # Y Yp9mp þ9 1 22 p p 1 o 4 n p hp;mp ;no : þ z11;3 ðn Þ 31 p9 1 p9 1 p p p o
Yp9mp þ9 1
4
If no ¼ 4m ð8k þ 7Þ for some nonnegative integers m and k; then Theorem 6.3 implies z11;3 ðno Þ ¼ 0: In this case we have r11 ðnÞ ¼ r11 ðno Þ
Yp9mp þ9 1 p
p9 1
p4
o 9m n p p 1 : p9 1 p
This, together with Theorem 6.4 implies Q Theorem 7.14. Let n ¼ 2l p plp be the prime factorisation of n and let n0 be the square-free part of n: If n0 is of the form 8k þ 7 then r11 ðnÞ ¼ r11 ðn0 Þ
0 9Ilp =2m 29Il=2mþ9 1 Y p9Ilp =2mþ9 1 1 4 n p p : 29 1 p9 1 p9 1 p p
8. The value of r13 ð p2 Þ for odd primes p Theorem 8.1. If p is an odd prime and tð pÞ is defined by Eq. (2.11), then 11 5 13936 r13 ð p2 Þ ¼ 4030 691 ð p þ 1Þ 26p þ 691 tð pÞ;
Proof. We proceed in a similar way to Section 5. Taking k ¼ 6 in (5.2) gives r13 ð p2 Þ ¼
p X
½q jð pjÞ z6 f13 þ 1287x þ 715x2 þ x3 g:
ð8:2Þ
j¼0
Now z6 ð13 þ 1287x þ 715x2 þ x3 Þ ¼ 1014z6 ð1 xÞð1 x þ x2 Þ 1001z6 ð1 xÞð1 x2 Þ þ 2z6 xð1157 1157x þ 1008x2 Þ:
ð8:3Þ
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Therefore Lemmas 4.1–4.3 give r13 ð p2 Þ ¼ 8112
p X
dð jð p jÞÞ 8008
j¼0
p X
eð jð p jÞÞ þ 0
j¼0
¼ 16224dð0Þ þ 16224dð0Þdð pÞ 8112
p X
dð jÞdð p jÞ
j¼0
16016eð0Þ þ 16016eð0Þeð pÞ 8008
p X
eð jÞeð p jÞ
j¼0
1 8
¼ 26p5 8112½qp z6 ð1 xÞð1 x þ x2 Þ
2
8008½qp
2 1 6 z ð1 xÞð1 x2 Þ 8
1 ¼ 26p5 ½q p z12 ð1014ð1 xÞ2 ð1 x þ x2 Þ2 þ 1001ð1 xÞ2 ð1 x2 Þ2 Þ: 8 Now 18 ð1014ð1 xÞ2 ð1 x þ x2 Þ2 þ 1001ð1 xÞ2 ð1 x2 Þ2 Þ 31 p1 ðxÞ þ 871 ¼ 504691 691 p2 ðxÞ þ p3 ðxÞ;
ð8:4Þ
where p1 ðxÞ ¼ 441ð1 þ 14x þ x2 Þ3 þ 250ð1 þ xÞ2 ð1 34x þ x2 Þ2 ; p2 ðxÞ ¼ xð1 xÞ; 1 ð43870208x6 131830116x5 þ 162785487x4 105780950x3 p3 ðxÞ ¼ 174132
þ 162565995x2 131610624x þ 43870208Þ: Therefore Lemmas 4.5, 4.6 and Eq. (3.5) give 31 65520ð p11 þ 1Þ þ 871 r13 ð p2 Þ ¼ 26p5 þ 504691 691 16tð pÞ þ 0 11 5 13936 ¼ 4030 691 ð p þ 1Þ 26p þ 691 tð pÞ;
as required.
&
Remark 8.5. The coefficients tðnÞ satisfy some well-known interesting properties. 1. t is multiplicative. That is, tðmnÞ ¼ tðmÞtðnÞ for any pair of relatively prime positive integers m and n:
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2. jtðnÞjpn11=2 dðnÞ; where dðnÞ is the number of divisors of n: In particular, jtð pÞjp2p11=2 for primes p: Both properties were conjectured by Ramanujan [25, Eqs. (103), (104) and (105)]. The first was proved by Mordell and the second by Deligne. See Berndt’s commentary [26, p. 367] for references and more information.
9. The eigenvalues and eigenfunctions for the Hecke operators Tp2 for 13 squares In this section we give the analogues of the results in Sections 6.1 and 7 for the Hecke operator Tp2 for thirteen squares. The methods of proof are identical with those in Sections 6.1 and 7, and therefore we omit them. Theorem 9.1. T4 f13;0 ¼ f13;0 þ 715f9;4 þ 1287f5;8 þ 13f1;12 ; T4 f9;4 ¼ 744f9;4 þ 1296f5;8 þ 8f1;12 ; T4 f5;8 ¼ 768f9;4 þ 1280f5;8 ; T4 f1;8 ¼ 512f9;4 þ 1536f5;8 :
Corollary 9.2. The eigenfunctions of T4 are w13;1 :¼ 1414477f13;0 2298777f9;4 þ 908427f5;8 2015f1;12 ; w13;2 :¼ 256f9;4 þ 434f5;8 þ f1;12 ; w13;3 :¼ of9;4 ðo þ 2Þf5;8 þ 2f1;12 ; w13;4 :¼ of % 9;4 ðo % þ 2Þf5;8 þ 2f1;12 and the corresponding eigenvalues are l1 ¼ 1; l2 ¼ 2048; l3 ¼ 4o; l4 ¼ 4o; % where pffiffiffiffiffiffiffiffiffiffiffi o ¼ 3 þ 119: The coefficients in the expansions of f13;0 ; f9;4 ; f5;8 and f1;12 satisfy some interesting properties: Some of these are summarised in
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Theorem 9.3. If n 2 or 3 ðmod 4Þ then 8f9;4 ðnÞ 9f5;8 ðnÞ þ f1;12 ðnÞ ¼ 0; 64f13;0 ðnÞ 104f9;4 ðnÞ þ 39f5;8 ðnÞ ¼ 0: If n 1 ðmod 8Þ then f5;8 ðnÞ f1;12 ðnÞ ¼ 0; 2048f13;0 ðnÞ 3328f9;4 ðnÞ þ 1313f5;8 ðnÞ ¼ 0: If n 5 ðmod 8Þ then 64f9;4 ðnÞ 57f5;8 ðnÞ 7f1;12 ðnÞ ¼ 0; 14336f13;0 ðnÞ 23488f9;4 ðnÞ þ 9369f5;8 ðnÞ ¼ 0:
Proof. A proof of Theorem 9.3 has been given by Barrucand and Hirschhorn [1]. & For odd primes p we have Theorem 9.4. Let a ¼ p11 þ 1; Then 0 0 1 Tp2 f13;0 691a B B T 2f C 1 B 0 B p 9;4 C B B C¼ @ Tp2 f5;8 A 691@ 0 Tp2 f1;12
0
b ¼ tð pÞ:
871a þ 871b 871a 871b 256a þ 435b 256a 256b 256a 256b
0
z13;1 :¼ 691f13;0 871f9;4 þ 871f5;8 ;
l1 ¼ a;
z13;2 :¼ 256f9;4 þ 434f5;8 þ f1;12 ¼ w13;2 ;
z13;4 :¼ f5;8 f1;12 ¼ w13;4 ;
f13;0
1
B C ab C CB f9;4 C CB C: 434a þ 257b a b A@ f5;8 A f1;12 434a 434b a þ 690b
434a 434b
Corollary 9.5. The eigenfunctions and eigenvalues of Tp2 are
z13;3 :¼ f9;4 f5;8 ¼ w13;3 ;
10
l3 ¼ b; l4 ¼ b:
l2 ¼ a;
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That is, Tp2 z13;1 ¼ ð p11 þ 1Þz13;1 ; Tp2 z13;2 ¼ ð p11 þ 1Þz13;2 ; Tp2 z13;3 ¼ tð pÞz13;3 ; Tp2 z13;4 ¼ tð pÞz13;4 : 10. Sums of an odd number of squares It is interesting to compare the values of rk ð p2 Þ for various k: We have: r1 ð p2 Þ ¼ 2;
ð10:1Þ
r2 ð p2 Þ ¼ 4ð2 þ ð1Þð p1Þ=2 Þ;
ð10:2Þ
r3 ð p2 Þ ¼ 6ð p þ 1 ð1Þð p1Þ=2 Þ;
ð10:3Þ
r4 ð p2 Þ ¼ 8ð p2 þ p þ 1Þ;
ð10:4Þ
r5 ð p2 Þ ¼ 10ð p3 p þ 1Þ;
ð10:5Þ
r6 ð p2 Þ ¼ 12ð p4 þ ð1Þð p1Þ=2 p2 þ 1Þ;
ð10:6Þ
r7 ð p2 Þ ¼ 14ð p5 ð1Þð p1Þ=2 p2 þ 1Þ;
ð10:7Þ
r8 ð p2 Þ ¼ 16ð p6 þ p3 þ 1Þ;
ð10:8Þ
7 3 32 r9 ð p2 Þ ¼ 274 17 ð p þ 1Þ 18p þ 17 yð pÞ;
ð10:9Þ
ð p1Þ=2 4 8 2 p þ 1Þ þ 32 r10 ð p2 Þ ¼ 68 5 ð p þ ð1Þ 5 cð p Þ;
ð10:10Þ
ð p1Þ=2 4 9 p þ 352 r11 ð p2 Þ ¼ 330 31 ð p þ 1Þ 22ð1Þ 31 Hð pÞ;
ð10:11Þ
r12 ð p2 Þ ¼ 8ð p10 þ p5 þ 1Þ þ 16Oð p2 Þ;
ð10:12Þ
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11 5 13936 r13 ð p2 Þ ¼ 4030 691 ð p þ 1Þ 26p þ 691 tð pÞ:
ð10:13Þ
where N X
1 8 yðnÞqn ¼ qðq; qÞ8N ðq2 ; q2 Þ8N ¼ 16 z xð1 xÞ;
n¼1 N X
cðnÞqn 11 ¼ q
n¼1 N X
ðq2 ; q2 Þ14 N ðq; qÞ4N
1 5 ¼ 16 z xð1 xÞ;
2 2 4 2 HðnÞqn ¼ qðq; qÞ16 N ðq ; q ÞN þ 32q
n¼1
ðq2 ; q2 Þ28 N ðq; qÞ8N
1 10 ¼ 16 z xð1 xÞð1 þ 2x 2x2 Þ; N X
1 6 OðnÞqn ¼ qðq2 ; q2 Þ12 N ¼ 16z xð1 xÞ;
n¼1 N X
1 12 ð1Þnþ1 tðnÞqn ¼ qðq; qÞ24 N ¼ 16z xð1 xÞ:
n¼1
Observe that formulas (10.9)–(10.13) are significantly more complicated than (10.1)–(10.8), as anticipated by Eisenstein [8] and Smith [31]. Eq. (10.1) is trivial. Eqs. (10.2), (10.4), (10.6), (10.8), (10.10) and (10.12) follow readily from the well known formulas for r2 ðnÞ; r4 ðnÞ; r6 ðnÞ; r8 ðnÞ; r10 ðnÞ and r12 ðnÞ; see, for example, [4]. Eqs. (10.3), (10.5) and (10.7) follow right away from (2.6)–(2.8). Direct proofs of (10.5) and (10.7) are given in [5], and it is possible to prove (10.3) in the same way. Eqs. (10.11) and (10.13) are Theorems 5.1 and 8.1, respectively. Formulas for r9 ðn2 Þ; r11 ðn2 Þ and r13 ðn2 Þ for any integer n were given by Sandham [28]. Results for r2k ð p2 Þ can be written down immediately from the general formulas for r2k ðnÞ in, for example, [4]. For sums of an odd number of squares, we have Conjecture 10.14. r2kþ1 ð p2 Þ ¼ Ak ð p2k1 þ 1Þ ð4k þ 2Þð1Þkð p1Þ=2 pk1 þ ½qp
Ik=2m X
cj;k q j
j¼1
ðq; qÞ8k24j N ðq2 ; q2 Þ4k24j N
where Ak and cj;k are rational numbers and Ak þ c1;k ¼ 4k þ 2:
;
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References [1] P. Barrucand, M.D. Hirschhorn, Formulae associated with 5, 7, 9 and 11 squares, Bull. Austral. Math. Soc. 65 (2002) 503–510. [2] B.C. Berndt, Ramanujan’s Notebooks, Part II, Springer, New York, 1989. [3] B.C. Berndt, Ramanujan’s Notebooks, Part III, Springer, New York, 1991. [4] S. Cooper, On sums of an even number of squares, and an even number of triangular numbers: an elementary approach based on Ramanujan’s 1 c1 summation formula. in: B.C. Berndt, K. Ono (Eds.), q-Series with Applications to Combinatorics, Number Theory and Physics Contemporary Mathematics, Vol. 291, American Mathematical Society, Providence, RI, 2001, pp. 115–137. [5] S. Cooper, Sums of five, seven and nine squares, Ramanujan J. 6 (2002) 469–490. [6] S. Cooper, M. Hirschhorn, R. Lewis, Powers of Euler’s product and related identities, Ramanujan J. 4 (2000) 137–155. [7] L.E. Dickson, History of the Theory of Numbers, Vol. 2, Chelsea, New York, 1952. [8] G. Eisenstein, Lehrsa¨tze, Crelle’s J. 39 (1850) 180–182 (reprinted in Mathematische Werke, Band II, Chelsea, New York, 1975, pp. 620–622). [9] G. Eisenstein, Neue Theoreme der ho¨heren Arithmetik, Crelle’s J. 35 (1847) 117–136 (reprinted in Mathematische Werke, Band I, Chelsea, New York, 1975, pp. 483–502; p. 501). [10] G. Eisenstein, Note sur la repre´sentation d’un nombre par la somme de cinq carre´s, Crelle’s J. 35 (1847) 368 (reprinted in Mathematische Werke, Band II, Chelsea, New York, 1975, pp. 505). [11] C.F. Gauss, Disquisitiones Arithmeticae (Translated into English by A. A. Clarke, Yale Univ. Press, New Haven, 1966). 1801. W.C. Waterhouse (Ed.) Springer, New York, 1986. [12] J.W.L. Glaisher, On the numbers of representations of a number as a sum of 2r squares, where 2r does not exceed eighteen, Proc. London Math. Soc. 5 (2) (1907) 479–490. [13] G.H. Hardy, On the representations of a number as the sum of any number of squares, and in particular of five or seven, Proc. Nat. Acad. Sci. USA 4 (1918) 189–193. [14] G.H. Hardy, On the representation of a number as the sum of any number of squares, and in particular of five, Trans. Amer. Math. Soc. 21 (1920) 255–284. [15] M.D. Hirschhorn, J. Sellers, On representations of a number as a sum of three squares, Discrete Math. 199 (1999) 85–101. [16] A. Hurwitz, Sur la de´composition des nombres en cinq carre´s, Paris, C.R. Acad. Sci. 98 (1884) 504–507 (reprinted in Mathematische Werke, Band 2, Birkhauser, Basel, 1933, pp. 5–7). [17] A. Hurwitz, L’Interme´diaire des Math 14 (1907) 107 (reprinted in Mathematische Werke, Band 2, 751, Birkhauser, Basel, 1933, p. 751). [18] C.G.J. Jacobi, Fundamenta Nova Theoriae Functionum Ellipticarum, 1829 (reprinted in Gesammelte Werke, Vol. 1, Berlin, 1881, pp. 49–239; see pp. 159, 160, 164, 170; reprinted by Chelsea, New York, 1969). [19] N. Koblitz, Introduction to Elliptic Curves and Modular forms, Springer, New York, 1984. [20] G.A. Lomadze, On the representation of numbers by sums of squares, Akad. Nauk Gruzin. SSR. Trudy Tbiliss. Mat. Inst. Razmadze 16 (1948) 231–275 (Russian). [21] G.A. Lomadze, On the number of representations of natural numbers by sums of nine squares, Acta Arith. 68 (3) (1994) 245–253 (Russian). [22] H. Minkowski, Me´moire sur la the´orie des formes quadratiques a` coefficients entie`res, Me´m pre´sente´s par divers savants l’Acad 29 (2) (1887) 1–178 (reprinted in Gesammelte Abhandlungen von Hermann Minkowski, B. 1, Leipzig, Berlin, B.G. Teubner, 1911, pp. 3–144). [23] L.J. Mordell, On the representation of numbers as the sum of 2r squares, Quart. J. Pure and Appl. Math. Oxford 48 (1917) 93–104. [24] S. Ramanujan, Notebooks, 2 Vols, Tata Institute of Fundamental Research, Bombay, 1957. [25] S. Ramanujan, On certain arithmetical functions, Trans. Cambride Philos. Soc. 22 (1916) 159–184 (reprinted in Collected Papers of Srinivasa Ramanujan, AMS Chelsea, Providence, RI, 2000, pp. 136–162). [26] S. Ramanujan, Collected Papers of Srinivasa Ramanujan, AMS Chelsea, Providence, RI, 2000, pp. 179–199.
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H.F. Sandham, A square as the sum of 7 squares, Quart. J. Math. Oxford 4 (2) (1953) 230–236. H.F. Sandham, A square as the sum of 9, 11 and 13 squares, J. London Math. Soc. 29 (1954) 31–38. J.-P. Serre, Smith, Minkowski et l’Acade´mie des Sciences, Gaz. Math 56 (1993) 3–9. H.J.S. Smith, On the orders and genera of quadratic forms containing more than three indeterminates, Proc. Roy. Soc. 13 (1864) 199–203; H.J.S. Smith, On the orders and genera of quadratic forms containing more than three indeterminates, Proc. Roy. Soc. 16 (1867) 197–208 (reprinted in The Collected Mathematical Papers of H. J. S. Smith, Vol. 1, 1894, pp. 412–417, 510–523; reprinted by Chelsea, New York, 1965). [31] H.J.S. Smith, Report on the theory of numbers, Part VI, Report of the British Assn., 1865, pp. 322– 375. (reprinted in The Collected Mathematical Papers of H.J.S. Smith, Vol. 1, 1894, pp. 289–364, 308; reprinted by Chelsea, New York, 1965). [32] H.J.S. Smith, Me´moire sur la repre´sentation des nombres par des sommes de cinq carre´s, Me´moires pre´sente´s par divers savants a` l’Acad. 29 (1) (1887) 1–72 (reprinted in The Collected Mathematical Papers of H.J.S. Smith, Vol. 2, 1894, pp. 623–680; reprinted by Chelsea, New York, 1965). [33] T.J. Stieltjes, Sur le nombre de de´compositions d’un entier en cinq carre´s, Paris, C. R. Acad. Sci. 97 (1883) pp. 1545–1547; also in 0uvres Comple`tes de Thomas Jan Stieltjes, Tome 1, P. Noordhoff, Groningen, 1914, pp. 329–331.
Journal of Number Theory 103 (2003) 135–162
http://www.elsevier.com/locate/jnt
On the number of representations of certain integers as sums of 11 or 13 squares Shaun Cooper Institute of Information and Mathematical Sciences, Massey University Albany, Private Bag 102-904, North Shore Mail Centre, Auckland, New Zealand Received 22 August 2001; revised 22 May 2003 Communicated by D. Goss Dedicated to Sri Chinmoy on the occasion of his 70th birthday
Abstract Let rk ðnÞ denote the number of representations of an integer n as a sum of k squares. We prove that for odd primes p; ð p1Þ=2 4 9 p þ 352 r11 ð p2 Þ ¼ 330 31 ð p þ 1Þ 22ð1Þ 31 Hð pÞ;
where Hð pÞ is the coefficient of qp in the expansion of q
N N Y Y ð1 q2j Þ28 ð1 ðqÞ j Þ16 ð1 q2j Þ4 þ 32q2 : j 8 j¼1 j¼1 ð1 ðqÞ Þ
This result, together with the theory of modular forms of half integer weight is used to prove that 0 9Ilp =2m 29Il=2mþ9 1 Y p9Ilp =2mþ9 1 1 4 n p p ; r11 ðnÞ ¼ r11 ðn0 Þ p 29 1 p9 1 p9 1 p Q where n ¼ 2l p plp is the prime factorisation of n; n0 is the square-free part of n; and n0 is of the form 8k þ 7: The products here are taken over the odd primes p; and ðnpÞ is the Legendre symbol. We also prove that for odd primes p; 11 5 13936 r13 ð p2 Þ ¼ 4030 691 ð p þ 1Þ 26p þ 691 tð pÞ;
E-mail address: [email protected]. 0022-314X/$ - see front matter r 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.jnt.2003.06.001
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Q PN j 24 n where tðnÞ is Ramanujan’s t function, defined by q N j¼1 ð1 q Þ ¼ n¼1 tðnÞq : A conjectured formula for r2kþ1 ð p2 Þ is given, for general k and general odd primes p: r 2003 Elsevier Inc. All rights reserved. MSC: Primary 11E25; Secondary 05A15; 33D15; 33E05 Keywords: Hecke operator; Modular forms of half integer weight; Sums of squares
1. Introduction Let rk ðnÞ denote the number of representations of n as a sum of k squares. The generating function for rk ðnÞ is N X j¼N
!k q
j2
¼
N X
rk ðnÞqn :
n¼0
The study of rk ðnÞ has a long and interesting history. Generating functions which yield the value of rk ðnÞ for k ¼ 2; 4; 6 and 8 were found by Jacobi [18]. Glaisher [12] proved formulas for k ¼ 10; 12; 14; 16 and 18; and Ramanujan [25, Eqs. (145)–(147)] stated a general formula for rk ðnÞ for arbitrary even values of k: Ramanujan’s formula was proved by Mordell [23] and an elementary proof was given by Cooper [4]. The problem of determining rk ðnÞ for odd values of k is more difficult, and not so well known. The value of r3 ðnÞ was found by Gauss [11, Section 291]. Formulas for r5 ðnÞ and r7 ðnÞ for square-free values of n were stated without proof by Eisenstein [8,10]. These formulas were extended to all non-negative integers n (again without proof) by Smith [30]. The Paris Academy of Sciences, apparently unaware of Smith’s work [30], proposed as its Grand Prix des Sciences Mathe´matiques competition for 1882 the problem of completely determining the value of r5 ðnÞ: The prize was awarded jointly to Smith [32] and Minkowski [22], who both gave formulas as well as proofs. An interesting account of this competition and the controversy surrounding it has been given by Serre [29]. Others to have worked on this problem are (chronologically) Stieltjes [33], Hurwitz [16], Hardy [13,14], Lomadze [20,21] and Hirschhorn and Sellers [15]. More information can be found in [7, Chapters VI–IX]. Eisenstein [9,31] stated that the sequence of formulas for rk ðnÞ ceases for kX9: As a result of computer investigations [5], I found that there are in fact simple, closedform formulas for r9 ðnÞ; but only for certain values of n: The purpose of this article is to state and prove the corresponding results for r11 ðnÞ: These results were also initially discovered as a result of computer investigations. There appear to be no similar formulas for rk ðnÞ for k ¼ 13; 15; 17; y: Formulae can be given, but they all involve more complicated number theoretic functions.
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2. Summary of results Q Throughout this article p will always denote an odd prime, and p will always denote a product over all odd primes p: Accordingly, let us denote the prime factorisation of n by Y n ¼ 2l pl p ; p
where l and lp are all non-negative integers, only finitely many of which are nonzero. Let n0 denote the square-free part of n: Hirschhorn and Sellers [15] with the help of their referee proved that r3 ðnÞ ¼ r3 ðn0 Þ
YpIlp =2mþ1 1 p
p1
0 Ilp =2m n p 1 ; p1 p
ð2:1Þ
where n0 is the square-free part of n: The analogous results for r5 ðnÞ; r7 ðnÞ and r9 ðnÞ were proved by Cooper [5]. For sums of five squares we have 3Il=2mþ3 2 1 23Il=2m 1 þ e 23 1 23 1 Y p3Ilp =2mþ3 1 n0 p3Ilp =2m 1 p ; p3 1 p3 1 p p
r5 ðnÞ ¼ r5 ðn0 Þ
ð2:2Þ
where 8 > <0 e ¼ 4 > : 16=7
if n0 1 ðmod 8Þ; if n0 2 or 3 ðmod 4Þ;
ð2:3Þ
if n0 5 ðmod 8Þ:
The result for sums of seven squares is 5Il=2mþ5 2 1 25Il=2m 1 þe r7 ðnÞ ¼ r7 ðn Þ 25 1 25 1 5Ilp =2m 0 Y p5Ilp =2mþ5 1 1 2 n p p ; p5 1 p5 1 p p 0
ð2:4Þ
where 8 > <8 e¼ 0 > : 64=37
if n0 1 or 2 ðmod 4Þ; if n0 3 ðmod 8Þ; 0
if n 7 ðmod 8Þ:
ð2:5Þ
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For squares, these formulas reduce to lp Yplp þ1 1 ð p1Þ=2 p 1 2 ð1Þ r3 ðn Þ ¼ 6 ; p1 p1 p
r5 ðn2 Þ ¼ 10
3lþ3 2 1 Y p3lp þ3 1 p3lp 1 p ; 23 1 p3 1 p3 1 p
25lþ5 1 25l 1 þ 8 25 1 25 1 5lp Y p5lp þ5 1 1 ð p1Þ=2 2 p ð1Þ p : p5 1 p5 1 p
r7 ðn2 Þ ¼ 14
ð2:6Þ
ð2:7Þ
ð2:8Þ
Eqs. (2.6) and (2.7) seem to have been first explicitly stated by Hurwitz [17] and [16], respectively, and (2.8) is due to Sandham [27]. For sums of nine squares we have 0 7Ilp =2m 7Il=2mþ7 1 Y p7Ilp =2mþ7 1 1 0 2 3 n p p r9 ¼ r9 ðn Þ ð2:9Þ 27 1 p7 1 p7 1 p p in the case that n0 5 ðmod 8Þ: Results for when n0 c5 ðmod 8Þ were also given in [5], but these involve additional and more complicated number theoretic functions. The first purpose of this article is to prove that for odd primes p; ð p1Þ=2 4 9 r11 ð p2 Þ ¼ 330 p þ 352 31 ð p þ 1Þ 22ð1Þ 31 Hð pÞ;
where Hð pÞ is the coefficient of qp in the expansion of q
N N Y Y ð1 q2j Þ28 ð1 ðqÞ j Þ16 ð1 q2j Þ4 þ 32q2 : j 8 j¼1 j¼1 ð1 ðqÞ Þ
This result, together with the theory of modular forms of half integer weight, is then used to prove 0 9Ilp =2m 29Il=2mþ9 1 Y p9Ilp =2mþ9 1 1 4 n p p r11 ðnÞ ¼ r11 ðn0 Þ ð2:10Þ 29 1 p9 1 p9 1 p p in the case that n0 7 ðmod 8Þ: Some results for when n0 c7 ðmod 8Þ will also be given, but just as for sums of nine squares, these involve additional and more complicated number theoretic functions. A conjecture for the value of r11 ðn0 Þ for square-free values of n0 satisfying n0 7 ðmod 8Þ was given in [5, p. 471]. The second purpose of this article is to prove that for odd primes p; 11 5 13936 r13 ð p2 Þ ¼ 4030 691 ð p þ 1Þ 26p þ 691 tð pÞ;
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where tðnÞ is Ramanujan’s t function, defined by q
N N Y X ð1 q j Þ24 ¼ tðnÞqn : j¼1
ð2:11Þ
n¼1
We also compute the eigenvalues and eigenfunctions of the Hecke operators Tp2 : There does not appear to be a formula like (2.1), (2.2), (2.4), (2.9) or (2.10) that expresses r13 ðnÞ in terms of just r13 ðn0 Þ and the prime factorisation of n; for any special cases of n0 : Finally, we offer a conjectured formula for the value of r2kþ1 ð p2 Þ; for general k and a general odd prime p:
3. Definitions Let jqjo1 and set ða; qÞN ¼
N Y ð1 aq j1 Þ: j¼1
Let N X
fðqÞ ¼
2
qj ;
j¼N
cðqÞ ¼
N X
q jð jþ1Þ=2 :
j¼0
We will use the notation ½qn f ðqÞ :¼ the coefficient of qn in f ðqÞ: Let s and t be non-negative integers. We will always assume t is a multiple of 4: Put fs;t ðqÞ ¼ fðqÞs ð2q1=4 cðq2 ÞÞt ; and let us write fs;t ðqÞ ¼
N X
fs;t ðnÞqn ;
n¼0
so that fs;t ðnÞ is the coefficient of qn in the series expansion of fs;t ðqÞ: Geometrically, fs;t ðnÞ counts the number of lattice points, that is, points with integer coordinates, on the sphere s X j¼1
x2j þ
sþt X
ðxj 1=2Þ2 ¼ n:
j¼sþ1
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Let rs;t ðnÞ denotes the number of representations of n as a sum of s þ t squares, of which s are even and t are odd. Then rs;t ðnÞ has the generating function N X sþt sþt rs;t ðnÞqn ¼ ð2qÞt fðq4 Þs cðq8 Þt ¼ fs;t ðq4 Þ: t t n¼0 Let z ¼ fðqÞ2 ; x ¼ 16q
cðq2 Þ4 fðqÞ4
:
Then [3] or [24, Chapter 16, Entry 25(vii)] zðsþtÞ=2 xt=4 ¼ fs;t ðqÞ; 1x¼
fðqÞ4 fðqÞ4
:
By Jacobi’s triple product identity [3, p. 35], z¼
ðq; qÞ4N ðq2 ; q2 Þ2N
;
xð1 xÞ ¼ 16q
ðq2 ; q2 Þ24 N ðq; qÞ24 N
ð3:1Þ
:
Therefore z5 xð1 xÞ ¼ 16q
ðq2 ; q2 Þ14 N ðq; qÞ4N
ð3:2Þ
;
z6 xð1 xÞ ¼ 16qðq2 ; q2 Þ12 N; z10 x2 ð1 xÞ2 ¼ 256q2
ðq2 ; q2 Þ28 N ðq; qÞ8N
z12 xð1 xÞ ¼ 16qðq; qÞ24 N ¼ 16
ð3:3Þ ð3:4Þ
;
N X ð1Þnþ1 tðnÞqn ;
ð3:5Þ
n¼1
where tðnÞ is Ramanujan’s t-function defined in Eq. (2.11). Define HðqÞ and HðnÞ by 2 2 4 2 HðqÞ ¼ qðq; qÞ16 N ðq ; q ÞN þ 32q
ðq2 ; q2 Þ28 N ðq; qÞ8N
¼
N X n¼1
HðnÞqn :
ð3:6Þ
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141
Then by (3.1), (3.4) and (3.5), 1 10 z xð1 xÞð1 þ 2x 2x2 Þ: HðqÞ ¼ 16
ð3:7Þ
We will require some facts about Hecke operators on modular forms of half integer weight. All of the facts below can be found in [19]. * 0 ð4ÞÞ is the vector space consisting of all linear combinations of Fact 1. Mð2kþ1Þ=2 ðG f2kþ14j;4j ðqÞ; j ¼ 0; 1; y; Ik=2m [19, Proposition 4, p. 184]. Fact 2. The Hecke operators Tp2 ; where p is any prime, map Mð2kþ1Þ=2 ðG* 0 ð4ÞÞ into itself [19, p. 206]. That is, * 0 ð4ÞÞ ) Tp2 f AMð2kþ1Þ=2 ðG * 0 ð4ÞÞ: f AMð2kþ1Þ=2 ðG Fact 3. Suppose f ðqÞ ¼
N X
* 0 ð4ÞÞ: aðnÞqn AMð2kþ1Þ=2 ðG
n¼0
If p is an odd prime, then [19, Proposition 13, p. 207] implies ! N N N X X X ð1Þk n 2 n k1 Tp2 f ¼ að p nÞq þ p aðn=p2 Þqn ; aðnÞqn þ p2k1 p n¼0 n¼0 n¼0 while if p ¼ 2; the note in [19, p. 210] implies T22 f ¼
N X
að4nÞqn :
n¼0
4. Some preliminary lemmas Lemma 4.1. Let N X 1 5 z ð5 xÞð1 xÞ ¼ aðnÞqn ; 4 n¼0 N X 1 5 z ð4x þ x2 Þ ¼ bðnÞqn ; 64 n¼0 N X 1 5 z xð1 xÞ ¼ cðnÞqn ; 16 n¼0
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N X 1 z6 ð1 xÞð1 x þ x2 Þ ¼ dðnÞqn ; 8 n¼0 N X 1 6 z ð1 xÞð1 x2 Þ ¼ eðnÞqn : 8 n¼0
Q Let n ¼ 2l p plp be the prime factorisation of n: For any prime p 1 ðmod 4Þ; let xp and yp be the unique pair of positive integers satisfying x2p þ y2p ¼ p;
2jxp ;
and define yp by tan yp ¼
yp ; xp
p 0oyo : 2
Then að0Þ ¼ 5=4;
bð0Þ ¼ 0;
cð0Þ ¼ 0;
dð0Þ ¼ 1=8;
eð0Þ ¼ 1=8
and for nX1; " # 4ðlp þ1Þ Y ð1Þðlp þ1Þð p1Þ=2 lp ð p1Þ=2 p aðnÞ ¼ ð1Þ ; p4 ð1Þð p1Þ=2 p
bðnÞ ¼ 24l
Y p4ðlp þ1Þ ð1Þðlp þ1Þð p1Þ=2 p4 ð1Þð p1Þ=2
p
cðnÞ ¼
8 > < 2 > : n ð1Þ
; if lp is odd for any p 3 ðmod 4Þ;
0 l
Q p1 ðmod 4Þ
sin 4ð1 þ lp Þyp sin 4yp
8 > > > <
Q p5lp þ5 1 p5 1 p dðnÞ ¼ > 30 25l þ 63 Q p5lp þ5 1 > > : 25 1 p5 1 p 8 > > > <
Q p5lp þ5 1 p5 1 p eðnÞ ¼ 5lþ5 > 2 63 Q p5lp þ5 1 > > : 25 1 p5 1 p
otherwise;
if n is odd; if n is even;
if n is odd; if n is even:
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Proof. From [3] or [24, Chapter 17, Entry 17(viii)] we have N 1 5 5 X ð1Þk ð2k þ 1Þ4 q2kþ1 z ð5 xÞð1 xÞ ¼ þ : 4 4 k¼0 1 q2kþ1
Therefore að0Þ ¼ 5=4 and X ð1Þðd1Þ=2 d 4 aðnÞ ¼ djn d odd
" # 4ðlp þ1Þ Y ð1Þðlp þ1Þð p1Þ=2 lp ð p1Þ=2 p ¼ ð1Þ : p4 ð1Þð p1Þ=2 p Next, from [3] or [24, Chapter 17, Entry 17(iii)] we have N X 1 5 k 4 qk z ð4x þ x2 Þ ¼ : 64 1 þ q2k k¼1
Therefore bð0Þ ¼ 0 and bðnÞ ¼
X
ð1Þðd1Þ=2
djn d odd
¼ 24l
n 4 d
Y p4ðlp þ1Þ ð1Þðlp þ1Þð p1Þ=2 p
p4 ð1Þð p1Þ=2
:
For the coefficients cðnÞ; by (3.2) and one of the Macdonald identities for BC2 (for example, see [6]) we have 1 5 ðq2 ; q2 Þ14 1 N z xð1 xÞ ¼ q ¼ 4 16 ðq; qÞN 6
X
2
abða2 b2 Þqða þb
2
Þ=5
:
a2 ðmod 5Þ b1 ðmod 5Þ
Therefore cð0Þ ¼ 0 and cðnÞ ¼
¼
1 6
X
abða2 b2 Þ
a2 þb2 ¼5n a;b1 ðmod 5Þ
8 > <0
2 > : n ð1Þ
if lp is odd for any p 3 ðmod 4Þ; l
Q p1 ðmod 4Þ
sin4ð1 þ lp Þyp sin4yp
otherwise:
The second part of this formula expressing cðnÞ in terms of yp and lp was stated (without proof) by Ramanujan [25, Eq. (156)].
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Lastly, from [3] or [24, Chapter 17, Entries 14(iii) and (vi)] we have N 1 1 X ð1Þk k5 qk z6 ð1 xÞð1 x þ x2 Þ ¼ ; 8 8 k¼1 1 þ qk N 1 6 1 X ð1Þk k5 qk z ð1 xÞð1 x2 Þ ¼ : 8 8 k¼1 1 qk
Therefore, dðnÞ ¼
X
ð1Þðdþn=dÞ d 5
djn
8 Q p5lp þ5 1 > > > < p p5 1 ¼ > 30 25l þ 63 Q p5lp þ5 1 > > : 25 1 p5 1 p
eðnÞ ¼
X
if n is odd; if n is even;
ð1Þd d 5
djn
8 Q p5lp þ5 1 > > if n is odd; > < p p5 1 ¼ 5lþ5 > 2 63 Q p5lp þ5 1 > > if n is even: : 25 1 p5 1 p
&
Lemma 4.2. If p is an odd prime then að pÞ ¼ ð1Þð p1Þ=2 p4 þ 1; dð pÞ ¼ p5 þ 1; eð pÞ ¼ p5 þ 1:
Proof. These follow right away from Lemma 4.1.
&
Lemma 4.3. Each of aðnÞ; bðnÞ; cðnÞ; dðnÞ and eðnÞ are multiplicative. That is, aðmnÞ ¼ aðmÞaðnÞ; if m and n are relatively prime; and similarly for b; c; d and e: Proof. These follow right away from Lemma 4.1.
&
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145
Lemma 4.4. Under the change of variables z-zð1 xÞ1=2 ;
x-
x ; x1
the functions z10 ð5x5 þ 411x4 2118x3 þ 4262x2 4275x þ 1710Þ and z12 ð43870208x6 131830116x5 þ 162785487x4 105780950x3 þ 162565995x2 131610624x þ 43870208Þ are invariant and z6 xð1157 1157x þ 1008x2 Þ changes sign. Proof. This follows by straightforward calculation.
&
Lemma 4.5. The q-expansions of z10 ð5x5 þ 411x4 2118x3 þ 4262x2 4275x þ 1710Þ and z12 ð43870208x6 131830116x5 þ 162785487x4 105780950x3 þ 162565995x2 131610624x þ 43870208Þ contain only even powers of q; and the q-expansion of z6 xð1157 1157x þ 1008x2 Þ contains only odd powers. Proof. This follows from the Change of Sign Principle [3, p. 126] or [17, Chapter 24, Entry 14(xii)]. & Lemma 4.6. ½qp z10 ð1 þ xÞð1 þ 14x þ x2 Þð1 34x þ x2 Þ ¼ 264ð p9 þ 1Þ; ½qp z12 ½441ð1 þ 14x þ x2 Þ3 þ 250ð1 þ xÞ2 ð1 34x þ x2 Þ2 ¼ 65520ð p11 þ 1Þ:
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Proof. From [3] or [24, Chapter 17, Entries 13(iii) and (iv)], followed by [2] or [24, Chapter 15, Entry 12(iii)] we have z10 ð1 þ xÞð1 þ 14x þ x2 Þð1 34x þ x2 Þ ¼ MðqÞNðqÞ ¼ 1 264
N X k 9 qk : 1 qk k¼1
The coefficient of qp in this is readily seen to be 264ð p9 þ 1Þ: Also, from [3] or [24, Chapter 17, Entries 13(iii) and (iv)] followed by [2] or [24, Chapter 15, Entry 13(i)] we have 441z12 ð1 þ 14x þ x3 Þ3 þ 250z12 ð1 þ xÞ2 ð1 34x þ x2 Þ2 ¼ 441MðqÞ3 þ 250NðqÞ2 ¼ 691 þ 65520
N X j 11 q j : 1qj j¼1
The coefficient of qp in this is readily seen to be 65520ð p11 þ 1Þ: & 5. The value of r11 ð p2 Þ for odd primes p Theorem 5.1. If p is an odd prime and Hð pÞ is defined by Eq. (3.6), then ð p1Þ=2 4 9 r11 ð p2 Þ ¼ 330 p þ 352 31 ð p þ 1Þ 22ð1Þ 31 Hð pÞ:
Proof. Let us write p2 ¼
11 X
x2i :
i¼1
There are three possibilities: either one, five or nine of the xi are odd, and the others are even. Accordingly, we have 22 X 22 X r6;4 ð p2 j 2 Þ þ r2;8 ð p2 j 2 Þ 5 9 j odd j odd j odd X 2 2 10 22 16q4 fðq4 Þ6 cðq8 Þ4 ¼ ½q p j 22fðq4 Þ10 þ 5 4 j odd 10 22 2 8 4 2 8 8 16 q fðq Þ cðq Þ þ 9 8 X 11 11 2 2 ¼2 ½qð p j Þ=4 fðqÞ10 þ 16qfðqÞ6 cðq2 Þ4 1 5 j odd
r11 ð p2 Þ ¼ 22
X
r10;0 ð p2 j 2 Þ þ
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þ
11
9
162 q2 fðqÞ2 cðq2 Þ8
147
p X 11 11 11 ½q jð pjÞ z5 þ xþ x2 1 5 9 j¼0
¼
p X ½q jð pjÞ z5 ð11 þ 462x þ 55x2 Þ:
¼
j¼0
In general, we can also prove that p X r2kþ1 ð p Þ ¼ ½q jð pjÞ zk 2
j¼0
Ik=2m X i¼0
2k þ 1 4i þ 1
! xi :
ð5:2Þ
Now 5 2 528 5 264 5 z5 ð11 þ 462x þ 55x2 Þ ¼ 11 5 z ð5 xÞð1 xÞ þ 5 z ð4x þ x Þ þ 5 z xð1 xÞ:
Using this together with Lemmas 4.1–4.3 gives r11 ð p2 Þ ¼
p X
44 5 að jð p
4224 jÞÞ þ 33792 5 bð jð p jÞÞ þ 5 cð jð p jÞÞ
j¼0 88 44 ¼ 88 5 að0Þ 5 að0Það pÞ þ 5
p X
að jÞað p jÞ
j¼0
þ 33792 5
p X j¼0
bð jÞbð p jÞ þ 4224 5
p X
cð jÞcð p jÞ
j¼0
2 p 1 5 ¼ 22ð1Þð p1Þ=2 p4 þ 44 5 ½q 4z ð5 xÞð1 xÞ 2 p 1 5 2 2 4224 p 1 5 þ 33792 5 ½q 64 z ð4x þ x Þ þ 5 ½q 16z xð1 xÞ p 10 4 2 ¼ 22ð1Þð p1Þ=2 p4 þ 11 4 ½q z ð2x þ 20x 12x þ 5Þ:
Now 4 11 4 ð2x
5 þ 20x2 12x þ 5Þ ¼ 124 p1 ðxÞ þ 22 31 p2 ðxÞ þ p3 ðxÞ
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where p1 ðxÞ ¼ ð1 þ xÞð1 þ 14x þ x2 Þð1 34x þ x2 Þ; p2 ðxÞ ¼ xð1 xÞð1 þ 2x 2x2 Þ; 1 ð5x5 þ 411x4 2118x3 þ 4262x2 4275x þ 1710Þ: p3 ðxÞ ¼ 124
This, together with Lemmas 4.5, 4.6 and Eq. (3.7) gives 5 ð264Þð p9 þ 1Þ þ 22 r11 ð p2 Þ ¼ 22ð1Þð p1Þ=2 p4 124 31 16Hð pÞ þ 0 ð p1Þ=2 4 9 ¼ 330 p þ 352 31 ð p þ 1Þ 22ð1Þ 31 Hð pÞ:
&
Remark 5.3. The coefficients HðnÞ have some interesting properties. 1. H is multiplicative. That is, HðmnÞ ¼ HðmÞHðnÞ for any pair of relatively prime positive integers m and n: 2. jHðnÞjpn9=2 dðnÞ; where dðnÞ is the number of divisors of n: In particular, jHð pÞjp2p9=2
ð5:4Þ
for primes p: Ramanujan [25, Section 28] made some conjectures concerning the orders of some similar functions. See his Eqs. (157), (160) and (163). See also Berndt’s commentary [26, p. 367] for references and more information.
6. The Hecke operator Tp2 when p ¼ 2 for eleven squares Theorem 6.1. T4 f11;0 ¼ f11;0 þ 330f7;4 þ 165f3;8 ; T4 f7;4 ¼ 336f7;4 þ 176f3;8 ; T4 f3;8 ¼ 320f7;4 þ 192f3;8 : That is, f11;0 ð4nÞ ¼ f11;0 ðnÞ þ 330f7;4 ðnÞ þ 165f3;8 ðnÞ; f7;4 ð4nÞ ¼ 336f7;4 ðnÞ þ 176f3;8 ðnÞ; f3;8 ð4nÞ ¼ 320f7;4 ðnÞ þ 192f3;8 ðnÞ:
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Proof. Let us consider f11;0 first. By Facts 1 and 2, we have T4 f11;0 ¼ a1 f11;0 þ a2 f7;4 þ a3 f3;8 ; for some constants a1 ; a2 ; a3 : By Fact 3, we have
T4 f11;0 ¼
N X
f11;0 ð4nÞqn :
n¼0
Therefore N X n¼0
f11;0 ð4nÞqn ¼ a1
N X
f11;0 ðnÞqn þ a2
n¼0
N X
f7;4 ðnÞqn þ a3
n¼0
N X
f3;8 ðnÞqn :
n¼0
Equating coefficients of 1; q and q2 gives a1 ¼ 1; 22a1 þ 16a2 ¼ 5302; 220a1 þ 224a2 þ 256a3 ¼ 116380 and so a1 ¼ 1;
a2 ¼ 330;
a3 ¼ 165:
This proves the first part of the theorem. The results for T4 f7;4 and T4 f3;8 follow similarly. & Corollary 6.2. The eigenfunctions and eigenvalues of T4 are w11;1 :¼ 511f11;0 682f7;4 þ 187f3;8 ; w11;2 :¼ 20f7;4 þ 11f3;8 ; w11;3 :¼ f7;4 f3;8 ;
l1 ¼ 1;
l2 ¼ 512;
l3 ¼ 16:
These coefficients satisfy many interesting properties. Some of these are summarised in
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Theorem 6.3. 32w11;1 ðnÞ 33w11;2 ðnÞ ¼ 0
if n 1 or 2 ðmod 4Þ;
512w11;1 ðnÞ 17w11;2 ðnÞ ¼ 0 15872w11;1 ðnÞ þ 495w11;2 ðnÞ ¼ 0
if n 3 ðmod 8Þ; if n 7 ðmod 8Þ;
w11;3 ðnÞ ¼ 0
if n 7 ðmod 8Þ:
Proof. A proof of Theorem 6.3 has been given by Barrucand and Hirschhorn [1]. & As a consequence of the previous results we have Theorem 6.4. r11 ð4l ð8k þ 7ÞÞ ¼
29lþ9 1 r11 ð8k þ 7Þ: 29 1
Proof. r11 ð4l ð8k þ 7ÞÞ ¼ f11;0 ð4l ð8k þ 7ÞÞ 1 ð31w11;1 ð4l ð8k þ 7ÞÞ þ 495w11;2 ð4l ð8k þ 7ÞÞ þ 11242w11;3 ð4l ð8k þ 7ÞÞÞ ¼ 31511 1 ð31w11;1 ð8k þ 7Þ þ 495 512l w11;2 ð8k þ 7Þ þ 11242 ð16Þl w11;3 ð8k þ 7ÞÞ; ¼ 31511
where we have applied Corollary 6.2 in the last two equalities. Applying Theorem 6.3 gives 1 r11 ð4l ð8k þ 7ÞÞ ¼ 31511 ð31w11;1 ð8k þ 7Þ 15872 512l w11;1 ð8k þ 7Þ þ 0Þ 1 ð512lþ1 1Þw11;1 ð8k þ 7Þ: ¼ 511
Taking l ¼ 0 in this gives r11 ð8k þ 7Þ ¼ w11;1 ð8k þ 7Þ and so 1 r11 ð4l ð8k þ 7ÞÞ ¼ 511 ð512lþ1 1Þr11 ð8k þ 7Þ ¼
as required.
&
29lþ9 1 r11 ð8k þ 7Þ; 29 1
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7. The Hecke operator Tp2 when p is an odd prime for sums of eleven squares By Facts 2 and 3 we have Tp2 f11;0 ¼
N X
f11;0 ð p2 nÞqn þ p4
n¼0
N N X X n f11;0 ðn=p2 Þqn f11;0 ðnÞqn þ p9 p n¼0 n¼0
¼ c1 f11;0 ðqÞ þ c2 f7;4 ðqÞ þ c3 f3;8 ðqÞ;
Tp2 f7;4 ¼
N X
f7;4 ð p2 nÞqn þ p4
n¼0
N N X X n f7;4 ðn=p2 Þqn f7;4 ðnÞqn þ p9 p n¼0 n¼0
¼ d1 f11;0 ðqÞ þ d2 f7;4 ðqÞ þ d3 f3;8 ðqÞ;
Tp2 f3;8 ¼
N X
f3;8 ð p2 nÞqn þ p4
n¼0
ð7:1Þ
ð7:2Þ
N N X X n f3;8 ðn=p2 Þqn f3;8 ðnÞqn þ p9 p n¼0 n¼0
¼ e1 f11;0 ðqÞ þ e2 f7;4 ðqÞ þ e3 f3;8 ðqÞ;
ð7:3Þ
for some constants c1 ; c2 ; c3 ; d1 ; d2 ; d3 ; e1 ; e2 ; e3 : Equating the constant terms in each of the three equations above gives c1 ¼ p9 þ 1; d1 ¼ 0; e1 ¼ 0: Equating the coefficients of q in (7.1)–(7.3) gives 22c1 þ 16c2 ¼ f11;0 ð p2 Þ þ 22ð1Þð p1Þ=2 p4 ; 22d1 þ 16d2 ¼ f7;4 ð p2 Þ þ 16ð1Þð p1Þ=2 p4 ; 22e1 þ 16e2 ¼ f3;8 ð p2 Þ: By Theorem 5.1 we have ð p1Þ=2 4 9 f11;0 ð p2 Þ ¼ 330 p þ 352 31 ð p þ 1Þ 22ð1Þ 31 Hð pÞ:
ð7:4Þ
By the same methods it can be shown that ð p1Þ=2 4 9 f7;4 ð p2 Þ ¼ 320 p þ 176 31 ð p þ 1Þ 16ð1Þ 31 Hð pÞ;
ð7:5Þ
9 320 f3;8 ð p2 Þ ¼ 320 31 ð p þ 1Þ 31 Hð pÞ:
ð7:6Þ
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It follows that 9 22 c2 ¼ 22 31 ð p þ 1Þ þ 31 Hð pÞ; 9 11 d2 ¼ 20 31 ð p þ 1Þ þ 31 Hð pÞ; 9 20 e2 ¼ 20 31 ð p þ 1Þ 31 Hð pÞ:
Next, equating coefficients of q4 in (7.1)–(7.3) gives f11;0 ð4p2 Þ þ 5302ð1Þð p1Þ=2 p4 ¼ 5302c1 þ 5376c2 þ 5120c3 ; f7;4 ð4p2 Þ þ 5376ð1Þð p1Þ=2 p4 ¼ 5302d1 þ 5376d2 þ 5120d3 ; f3;8 ð4p2 Þ þ 5120ð1Þð p1Þ=2 p4 ¼ 5302e1 þ 5376e2 þ 5120e3 : By Theorem 6.1 and Eqs. (7.4)–(7.6) we have f11;0 ð4p2 Þ ¼ f11;0 ð p2 Þ þ 330f7;4 ð p2 Þ þ 165f3;8 ð p2 Þ ð p1Þ=2 4 9 ¼ 158730 p þ 5632 31 ð p þ 1Þ 5302ð1Þ 31 Hð pÞ:
Similarly, ð p1Þ=2 4 9 f7;4 ð4p2 Þ ¼ 163840 p þ 2816 31 ð p þ 1Þ 5376ð1Þ 17 Hð pÞ ð p1Þ=2 4 9 f3;8 ð4p2 Þ ¼ 163840 p 5120 31 ð p þ 1Þ 5120ð1Þ 17 Hð pÞ:
It follows that 9 22 c3 ¼ 22 31 ð p þ 1Þ 31Hð pÞ; 9 11 d3 ¼ 11 31 ð p þ 1Þ 31Hð pÞ; 9 20 e3 ¼ 11 31 ð p þ 1Þ þ 31Hð pÞ:
The above results may be summarised as Theorem 7.7. Let a ¼ p9 þ 1;
b ¼ Hð pÞ:
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Then 0 1 Tp2 f11;0 31a 1B B T 2f C @ p 7;4 A ¼ @ 0 31 Tp2 f3;8 0 0
22a þ 22b 20a þ 11b 20a 20b
10 1 f11;0 22a 22b CB C 11a 11b A@ f7;4 A: f3;8 11a þ 20b
Corollary 7.8. The eigenfunctions and eigenvalues of Tp2 are z11;1 :¼ f11;0 2f7;4 ;
l1 ¼ a;
z11;2 :¼ 20f7;4 þ 11f3;8 ¼ w11;2 ; z11;3 :¼ f7;4 f3;8 ¼ w11;3 ;
l2 ¼ a;
l3 ¼ b:
That is, Tp2 z11;1 ¼ ð p9 þ 1Þz11;1 ; Tp2 z11;2 ¼ ð p9 þ 1Þz11;2 ; Tp2 z11;3 ¼ Hð pÞz11;3 :
Using Fact 3, together with Corollary 7.8, we have 2 4 n z11;1 ð p nÞ þ p z11;1 ðnÞ þ p9 z11;1 ðn=p2 Þ ¼ ð p9 þ 1Þz11;1 ðnÞ; p
2
z11;2 ð p nÞ þ p
4
z11;3 ð p2 nÞ þ p4
ð7:9Þ
n z11;2 ðnÞ þ p9 z11;2 ðn=p2 Þ ¼ ð p9 þ 1Þz11;2 ðnÞ; p
ð7:10Þ
n z11;3 ðnÞ þ p9 z11;3 ðn=p2 Þ ¼ Hð pÞz11;3 ðnÞ: p
ð7:11Þ
Iterating (7.9)–(7.11) we obtain Theorem 7.12. If p2 [n; then 9m 9mþ9 p 1 4 n p 1 z11;1 ð p nÞ ¼ p z ðnÞ; p9 1 p p9 1 11;1 9mþ9 9m p 1 4 n p 1 z11;2 ð p2m nÞ ¼ p z ðnÞ; p9 1 p p9 1 11;2 2m
z11;3 ð p2m nÞ ¼ ½c1 ðnÞrm1 þ c2 ðnÞrm2 z11;3 ðnÞ;
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where pffiffiffiffiffiffiffiffiffiffiffi Hð pÞ þ Dð pÞ ; r1 ¼ 2 pffiffiffiffiffiffiffiffiffiffiffi Hð pÞ Dð pÞ ; r2 ¼ 2 c1 ðnÞ ¼
4 n 1 Hð pÞ 2p ð p Þ pffiffiffiffiffiffiffiffiffiffiffi þ ; 2 2 Dð pÞ
c2 ðnÞ ¼
4 n 1 Hð pÞ 2p ð p Þ pffiffiffiffiffiffiffiffiffiffiffi ; 2 2 Dð pÞ
Dð pÞ ¼ Hð pÞ2 4p9 : Remark 7.13. Eq. (5.4) implies Dð pÞo0: It is possible to express r11 ðnÞ in terms of r11 ðno Þ and z11;3 ðno Þ; where no is the oddsquare free part of n; that is, no is the greatest divisor of n which is not divisible by an odd square. Let gp;m;m ¼
9m p9mþ9 1 4 m p 1 p ; p9 1 p p9 1
hp;m;m ¼ c1 ðmÞrm1 þ c2 ðmÞrm2 ; where c1 ; c2 ; r1 ; r2 are as for Theorem 7.12. If p2 [m then 1 r11 ð p2m mÞ ¼ 31 ½31z11;1 ð p2m mÞ þ 2z11;2 ð p2m mÞ þ 22z11;3 ð p2m mÞ 1 ½31gp;m;m z11;1 ðmÞ þ 2gp;m;m z11;2 ðmÞ þ 22hp;m;m z11;3 ðmÞ ¼ 31
¼
1 ½31gp;m;m z11;1 ðmÞ þ 2gp;m;m z11;2 ðmÞ þ 22gp;m;m z11;3 ðmÞ 31 þ 22 31 ½hp;m;m gp;m;m z11;3 ðmÞ
¼ gp;m;m r11 ðmÞ þ 22 31 ½hp;m;m gp;m;m z11;3 ðmÞ: By induction on j; it follows that if p1 ; p2 ; y; pj are distinct odd primes, and p2i [m for 1pipj; then 2m
2m
2m
r11 ð p1 1 p2 2 ?pj j mÞ ! " ! j j Y 22 Y ¼ gpi ;mi ;m r11 ðmÞ þ hp m ;m 31 i¼1 i i i¼1
j Y i¼1
!# gpi ;mi ;m
z11;3 ðmÞ:
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Consequently, if no is the odd-square free part of n and either p2mp jjn or p2mp þ1 jjn; then o 9m n p p 1 r11 ðnÞ ¼ r11 ðn Þ p p9 1 p9 1 p p " o 9m # Y Yp9mp þ9 1 22 p p 1 o 4 n p hp;mp ;no : þ z11;3 ðn Þ 31 p9 1 p9 1 p p p o
Yp9mp þ9 1
4
If no ¼ 4m ð8k þ 7Þ for some nonnegative integers m and k; then Theorem 6.3 implies z11;3 ðno Þ ¼ 0: In this case we have r11 ðnÞ ¼ r11 ðno Þ
Yp9mp þ9 1 p
p9 1
p4
o 9m n p p 1 : p9 1 p
This, together with Theorem 6.4 implies Q Theorem 7.14. Let n ¼ 2l p plp be the prime factorisation of n and let n0 be the square-free part of n: If n0 is of the form 8k þ 7 then r11 ðnÞ ¼ r11 ðn0 Þ
0 9Ilp =2m 29Il=2mþ9 1 Y p9Ilp =2mþ9 1 1 4 n p p : 29 1 p9 1 p9 1 p p
8. The value of r13 ð p2 Þ for odd primes p Theorem 8.1. If p is an odd prime and tð pÞ is defined by Eq. (2.11), then 11 5 13936 r13 ð p2 Þ ¼ 4030 691 ð p þ 1Þ 26p þ 691 tð pÞ;
Proof. We proceed in a similar way to Section 5. Taking k ¼ 6 in (5.2) gives r13 ð p2 Þ ¼
p X
½q jð pjÞ z6 f13 þ 1287x þ 715x2 þ x3 g:
ð8:2Þ
j¼0
Now z6 ð13 þ 1287x þ 715x2 þ x3 Þ ¼ 1014z6 ð1 xÞð1 x þ x2 Þ 1001z6 ð1 xÞð1 x2 Þ þ 2z6 xð1157 1157x þ 1008x2 Þ:
ð8:3Þ
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Therefore Lemmas 4.1–4.3 give r13 ð p2 Þ ¼ 8112
p X
dð jð p jÞÞ 8008
j¼0
p X
eð jð p jÞÞ þ 0
j¼0
¼ 16224dð0Þ þ 16224dð0Þdð pÞ 8112
p X
dð jÞdð p jÞ
j¼0
16016eð0Þ þ 16016eð0Þeð pÞ 8008
p X
eð jÞeð p jÞ
j¼0
1 8
¼ 26p5 8112½qp z6 ð1 xÞð1 x þ x2 Þ
2
8008½qp
2 1 6 z ð1 xÞð1 x2 Þ 8
1 ¼ 26p5 ½q p z12 ð1014ð1 xÞ2 ð1 x þ x2 Þ2 þ 1001ð1 xÞ2 ð1 x2 Þ2 Þ: 8 Now 18 ð1014ð1 xÞ2 ð1 x þ x2 Þ2 þ 1001ð1 xÞ2 ð1 x2 Þ2 Þ 31 p1 ðxÞ þ 871 ¼ 504691 691 p2 ðxÞ þ p3 ðxÞ;
ð8:4Þ
where p1 ðxÞ ¼ 441ð1 þ 14x þ x2 Þ3 þ 250ð1 þ xÞ2 ð1 34x þ x2 Þ2 ; p2 ðxÞ ¼ xð1 xÞ; 1 ð43870208x6 131830116x5 þ 162785487x4 105780950x3 p3 ðxÞ ¼ 174132
þ 162565995x2 131610624x þ 43870208Þ: Therefore Lemmas 4.5, 4.6 and Eq. (3.5) give 31 65520ð p11 þ 1Þ þ 871 r13 ð p2 Þ ¼ 26p5 þ 504691 691 16tð pÞ þ 0 11 5 13936 ¼ 4030 691 ð p þ 1Þ 26p þ 691 tð pÞ;
as required.
&
Remark 8.5. The coefficients tðnÞ satisfy some well-known interesting properties. 1. t is multiplicative. That is, tðmnÞ ¼ tðmÞtðnÞ for any pair of relatively prime positive integers m and n:
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2. jtðnÞjpn11=2 dðnÞ; where dðnÞ is the number of divisors of n: In particular, jtð pÞjp2p11=2 for primes p: Both properties were conjectured by Ramanujan [25, Eqs. (103), (104) and (105)]. The first was proved by Mordell and the second by Deligne. See Berndt’s commentary [26, p. 367] for references and more information.
9. The eigenvalues and eigenfunctions for the Hecke operators Tp2 for 13 squares In this section we give the analogues of the results in Sections 6.1 and 7 for the Hecke operator Tp2 for thirteen squares. The methods of proof are identical with those in Sections 6.1 and 7, and therefore we omit them. Theorem 9.1. T4 f13;0 ¼ f13;0 þ 715f9;4 þ 1287f5;8 þ 13f1;12 ; T4 f9;4 ¼ 744f9;4 þ 1296f5;8 þ 8f1;12 ; T4 f5;8 ¼ 768f9;4 þ 1280f5;8 ; T4 f1;8 ¼ 512f9;4 þ 1536f5;8 :
Corollary 9.2. The eigenfunctions of T4 are w13;1 :¼ 1414477f13;0 2298777f9;4 þ 908427f5;8 2015f1;12 ; w13;2 :¼ 256f9;4 þ 434f5;8 þ f1;12 ; w13;3 :¼ of9;4 ðo þ 2Þf5;8 þ 2f1;12 ; w13;4 :¼ of % 9;4 ðo % þ 2Þf5;8 þ 2f1;12 and the corresponding eigenvalues are l1 ¼ 1; l2 ¼ 2048; l3 ¼ 4o; l4 ¼ 4o; % where pffiffiffiffiffiffiffiffiffiffiffi o ¼ 3 þ 119: The coefficients in the expansions of f13;0 ; f9;4 ; f5;8 and f1;12 satisfy some interesting properties: Some of these are summarised in
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Theorem 9.3. If n 2 or 3 ðmod 4Þ then 8f9;4 ðnÞ 9f5;8 ðnÞ þ f1;12 ðnÞ ¼ 0; 64f13;0 ðnÞ 104f9;4 ðnÞ þ 39f5;8 ðnÞ ¼ 0: If n 1 ðmod 8Þ then f5;8 ðnÞ f1;12 ðnÞ ¼ 0; 2048f13;0 ðnÞ 3328f9;4 ðnÞ þ 1313f5;8 ðnÞ ¼ 0: If n 5 ðmod 8Þ then 64f9;4 ðnÞ 57f5;8 ðnÞ 7f1;12 ðnÞ ¼ 0; 14336f13;0 ðnÞ 23488f9;4 ðnÞ þ 9369f5;8 ðnÞ ¼ 0:
Proof. A proof of Theorem 9.3 has been given by Barrucand and Hirschhorn [1]. & For odd primes p we have Theorem 9.4. Let a ¼ p11 þ 1; Then 0 0 1 Tp2 f13;0 691a B B T 2f C 1 B 0 B p 9;4 C B B C¼ @ Tp2 f5;8 A 691@ 0 Tp2 f1;12
0
b ¼ tð pÞ:
871a þ 871b 871a 871b 256a þ 435b 256a 256b 256a 256b
0
z13;1 :¼ 691f13;0 871f9;4 þ 871f5;8 ;
l1 ¼ a;
z13;2 :¼ 256f9;4 þ 434f5;8 þ f1;12 ¼ w13;2 ;
z13;4 :¼ f5;8 f1;12 ¼ w13;4 ;
f13;0
1
B C ab C CB f9;4 C CB C: 434a þ 257b a b A@ f5;8 A f1;12 434a 434b a þ 690b
434a 434b
Corollary 9.5. The eigenfunctions and eigenvalues of Tp2 are
z13;3 :¼ f9;4 f5;8 ¼ w13;3 ;
10
l3 ¼ b; l4 ¼ b:
l2 ¼ a;
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That is, Tp2 z13;1 ¼ ð p11 þ 1Þz13;1 ; Tp2 z13;2 ¼ ð p11 þ 1Þz13;2 ; Tp2 z13;3 ¼ tð pÞz13;3 ; Tp2 z13;4 ¼ tð pÞz13;4 : 10. Sums of an odd number of squares It is interesting to compare the values of rk ð p2 Þ for various k: We have: r1 ð p2 Þ ¼ 2;
ð10:1Þ
r2 ð p2 Þ ¼ 4ð2 þ ð1Þð p1Þ=2 Þ;
ð10:2Þ
r3 ð p2 Þ ¼ 6ð p þ 1 ð1Þð p1Þ=2 Þ;
ð10:3Þ
r4 ð p2 Þ ¼ 8ð p2 þ p þ 1Þ;
ð10:4Þ
r5 ð p2 Þ ¼ 10ð p3 p þ 1Þ;
ð10:5Þ
r6 ð p2 Þ ¼ 12ð p4 þ ð1Þð p1Þ=2 p2 þ 1Þ;
ð10:6Þ
r7 ð p2 Þ ¼ 14ð p5 ð1Þð p1Þ=2 p2 þ 1Þ;
ð10:7Þ
r8 ð p2 Þ ¼ 16ð p6 þ p3 þ 1Þ;
ð10:8Þ
7 3 32 r9 ð p2 Þ ¼ 274 17 ð p þ 1Þ 18p þ 17 yð pÞ;
ð10:9Þ
ð p1Þ=2 4 8 2 p þ 1Þ þ 32 r10 ð p2 Þ ¼ 68 5 ð p þ ð1Þ 5 cð p Þ;
ð10:10Þ
ð p1Þ=2 4 9 p þ 352 r11 ð p2 Þ ¼ 330 31 ð p þ 1Þ 22ð1Þ 31 Hð pÞ;
ð10:11Þ
r12 ð p2 Þ ¼ 8ð p10 þ p5 þ 1Þ þ 16Oð p2 Þ;
ð10:12Þ
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11 5 13936 r13 ð p2 Þ ¼ 4030 691 ð p þ 1Þ 26p þ 691 tð pÞ:
ð10:13Þ
where N X
1 8 yðnÞqn ¼ qðq; qÞ8N ðq2 ; q2 Þ8N ¼ 16 z xð1 xÞ;
n¼1 N X
cðnÞqn 11 ¼ q
n¼1 N X
ðq2 ; q2 Þ14 N ðq; qÞ4N
1 5 ¼ 16 z xð1 xÞ;
2 2 4 2 HðnÞqn ¼ qðq; qÞ16 N ðq ; q ÞN þ 32q
n¼1
ðq2 ; q2 Þ28 N ðq; qÞ8N
1 10 ¼ 16 z xð1 xÞð1 þ 2x 2x2 Þ; N X
1 6 OðnÞqn ¼ qðq2 ; q2 Þ12 N ¼ 16z xð1 xÞ;
n¼1 N X
1 12 ð1Þnþ1 tðnÞqn ¼ qðq; qÞ24 N ¼ 16z xð1 xÞ:
n¼1
Observe that formulas (10.9)–(10.13) are significantly more complicated than (10.1)–(10.8), as anticipated by Eisenstein [8] and Smith [31]. Eq. (10.1) is trivial. Eqs. (10.2), (10.4), (10.6), (10.8), (10.10) and (10.12) follow readily from the well known formulas for r2 ðnÞ; r4 ðnÞ; r6 ðnÞ; r8 ðnÞ; r10 ðnÞ and r12 ðnÞ; see, for example, [4]. Eqs. (10.3), (10.5) and (10.7) follow right away from (2.6)–(2.8). Direct proofs of (10.5) and (10.7) are given in [5], and it is possible to prove (10.3) in the same way. Eqs. (10.11) and (10.13) are Theorems 5.1 and 8.1, respectively. Formulas for r9 ðn2 Þ; r11 ðn2 Þ and r13 ðn2 Þ for any integer n were given by Sandham [28]. Results for r2k ð p2 Þ can be written down immediately from the general formulas for r2k ðnÞ in, for example, [4]. For sums of an odd number of squares, we have Conjecture 10.14. r2kþ1 ð p2 Þ ¼ Ak ð p2k1 þ 1Þ ð4k þ 2Þð1Þkð p1Þ=2 pk1 þ ½qp
Ik=2m X
cj;k q j
j¼1
ðq; qÞ8k24j N ðq2 ; q2 Þ4k24j N
where Ak and cj;k are rational numbers and Ak þ c1;k ¼ 4k þ 2:
;
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