Res. Lett. Inf. Math. Sci., (2002) 3, 37?58
Available online at http://www.massey.ac.nz/?wwiims/research/letters/
On the number of representations of certain integers as sums of eleven
or thirteen squares
Shaun Cooper
I.I.M.S., Massey University Albany Campus, Auckland, New Zealand
s.cooper@massey.ac.nz
Dedicated to Sri Chinmoy on the occasion of his 70th birthday.
Abstract
Let rk(n) denote the number of representations of an integer n as a sum of k squares. We prove
that for odd primes p,
r11(p2) =
330
31
(p9 + 1) ? 22(?1)(p?1)/2p4 + 352
31
H(p),
where H(p) is the coefficient of qp in the expansion of
q
?
?
j=1
(1 ? (?q)j)16(1 ? q2j)4 + 32q2
?
?
j=1
(1 ? q2j)28
(1 ? (?q)j)8 .
This result, together with the theory of modular forms of half integer weight is used to prove that
r11(n) = r11(n?)
29?/2+9 ? 1
29 ? 1
?
p
[
p9?p/2+9 ? 1
p9 ? 1 ? p
4
(?n?
p
)
p9?p/2 ? 1
p9 ? 1
]
,
where n = 2?
?
p p
?p is the prime factorisation of n and n? is the square-free part of n, in the case
that n? is of the form 8k + 7. The products here are taken over the odd primes p, and
(
n
p
)
is the
Legendre symbol.
We also prove that for odd primes p,
r13(p2) =
4030
691
(p11 + 1) ? 26p5 + 13936
691
?(p),
where ?(n) is Ramanujan?s ? function, defined by q
?
?
j=1
(1 ? qj)24 =
?
?
n=1
?(n)qn. A conjectured
formula for r2k+1(p2) is given, for general k and general odd primes p.
1 Introduction
Let rk(n) denote the number of representations of n as a sum of k squares. The generating function
for rk(n) is
?
?
?
?
j=??
qj
2
?
?
k
=
?
?
n=0
rk(n)qn.
The study of rk(n) has a long and interesting history. Generating functions which yield the value
of rk(n) for k = 2, 4, 6 and 8 were found by Jacobi [18]. Glaisher [12] proved formulas for
k = 10, 12, 14, 16 and 18, and Ramanujan [25, eqs. (145)?(147)] stated a general formula for
38 R.L.I.M.S. Vol. 3, April, 2002
rk(n) for arbitrary even values of k. Ramanujan?s formula was proved by Mordell [23] and a simple
proof was given by Cooper [4].
The problem of determining rk(n) for odd values of k is more difficult, and not so well known.
The value of r3(n) was found by Gauss [11, Section 291]. Formulas for r5(n) and r7(n) for square-
free values of n were stated without proof by Eisenstein [9], [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(n). 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], Sandham [27], [28] and Hirschhorn and Sellers [15]. More information
can be found in [7, Chapters VI?IX].
Eisenstein [8], [31] stated that the sequence of formulas for rk(n) ceases for k ? 9. As a result
of computer investigations [5], I found that there are in fact simple, closed form formulas for r9(n),
but only for certain values of n. The purpose of this article is to state and prove the corresponding
results for r11(n). These results were also initially discovered as a result of computer investigations.
There appear to be no similar formulas for rk(n) for k = 13, 15, 17, ? ? ?. Formulae can be
given, but they all involve more complicated number theoretic functions.
2 Summary of results
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
n = 2?
?
p
p?p ,
where ? and ?p are all nonnegative integers, only finitely many of which are non-zero. Let n?
denote the square-free part of n.
Hirschhorn and Sellers [15] proved that
r3(n) = r3(n?)
?
p
[
p?p/2+1 ? 1
p ? 1 ?
(?n?
p
)
p?p/2 ? 1
p ? 1
]
(2.1)
where n? is the square-free part of n. The analogous results for r5(n), r7(n) and r9(n) were proved
by Cooper [5]. For sums of five squares we have
r5(n) = r5(n?)
[
23?/2+3 ? 1
23 ? 1 + 
23?/2 ? 1
23 ? 1
]
?
?
p
[
p3?p/2+3 ? 1
p3 ? 1 ? p
(
n?
p
)
p3?p/2 ? 1
p3 ? 1
]
, (2.2)
where
 =
?
?
?
0 if n? ? 1 (mod 8)
?4 if n? ? 2 or 3 (mod 4)
?16/7 if n? ? 5 (mod 8).
(2.3)
The result for sums of seven squares is
r7(n) = r7(n?)
[
25?/2+5 ? 1
25 ? 1 + 
25?/2 ? 1
25 ? 1
]
?
?
p
[
p5?p/2+5 ? 1
p5 ? 1 ? p
2
(?n?
p
)
p5?p/2 ? 1
p5 ? 1
]
, (2.4)
S. Cooper, On the Number of Representations of Certain Integers 39
where
 =
?
?
?
8 if n? ? 1 or 2 (mod 4)
0 if n? ? 3 (mod 8)
?64/37 if n? ? 7 (mod 8).
(2.5)
For squares, these formulas reduce to
r3(n2) = 6
?
p
[
p?p+1 ? 1
p ? 1 ? (?1)
(p?1)/2 p
?p ? 1
p ? 1
]
, (2.6)
r5(n2) = 10
[
23?+3 ? 1
23 ? 1
]
?
p
[
p3?p+3 ? 1
p3 ? 1 ? p
p3?p ? 1
p3 ? 1
]
, (2.7)
r7(n2) = 14
[
25?+5 ? 1
25 ? 1 + 8
25? ? 1
25 ? 1
]
?
?
p
[
p5?p+5 ? 1
p5 ? 1 ? (?1)
(p?1)/2p2
p5?p ? 1
p5 ? 1
]
. (2.8)
Equations (2.6) and (2.7) seem to have been first been explicitly stated by Hurwitz [17] and [16],
respectively, and (2.8) is due to Sandham [27].
For sums of nine squares we have
r9(n) = r9(n?)
27?/2+7 ? 1
27 ? 1
?
p
[
p7?p/2+7 ? 1
p7 ? 1 ? p
3
(
n?
p
)
p7?p/2 ? 1
p7 ? 1
]
, (2.9)
in the case that n? ? 5 (mod 8). Results for when n? ? 5 (mod 8) 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,
r11(p2) =
330
31
(p9 + 1) ? 22(?1)(p?1)/2p4 + 352
31
H(p),
where H(p) is the coefficient of qp in the expansion of
q
?
?
j=1
(1 ? (?q)j)16(1 ? q2j)4 + 32q2
?
?
j=1
(1 ? q2j)28
(1 ? (?q)j)8 .
This result, together with the theory of modular forms of half integer weight, is then used to prove
r11(n) = r11(n?)
29?/2+9 ? 1
29 ? 1
?
p
[
p9?p/2+9 ? 1
p9 ? 1 ? p
4
(?n?
p
)
p9?p/2 ? 1
p9 ? 1
]
. (2.10)
in the case that n? ? 7 (mod 8). Some results for when n? ? 7 (mod 8) will also be given, but
just as for sums of nine squares, these involve additional and more complicated number theoretic
functions.
The value of r11(n?) for square-free n? is given by
r11(n?) =
31680
31
(n??1)/2
?
j=1
(
j
n?
)
j3(j ? n?), if n? ? 7 (mod 8). (2.11)
The second purpose of this article is to prove that for odd primes p,
r13(p2) =
4030
691
(p11 + 1) ? 26p5 + 13936
691
?(p),
40 R.L.I.M.S. Vol. 3, April, 2002
where ?(n) is Ramanujan?s ? function, defined by
q
?
?
j=1
(1 ? qj)24 =
?
?
n=1
?(n)qn. (2.12)
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(n) in terms of just
r13(n?) and the prime factorisation of n, for any special cases of n?.
Finally, we offer a conjectured formula for the value of r2k+1(p2), for general k and a general
odd prime p.
3 Definitions
Let |q| < 1 and set
(a; q)? =
?
?
j=1
(1 ? aqj?1).
Let
?(q) =
?
?
j=??
qj
2
,
?(q) =
?
?
j=0
qj(j+1)/2.
Let s and t be non-negative integers. We will always assume t is a multiple of 4. Put
?s,t(q) = ?(q)s
(
2q1/4?(q2)
)t
,
and let us write
?s,t(q) =
?
?
n=0
?s,t(n)qn,
so that ?s,t(n) is the coefficient of qn in the series expansion of ?s,t(q). Geometrically, ?s,t(n)
counts the number of lattice points, that is, points with integer coordinates, on the sphere
s
?
j=1
x2j +
s+t
?
j=s+1
(xj ? 1/2)2 = n.
Let rs,t(n) denotes the number of representations of n as a sum of s + t squares, of which s are
even and t are odd. Then rs,t(n) has the generating function
?
?
n=0
rs,t(n)qn =
(
s + t
t
)
(2q)t?(q4)s?(q8)t =
(
s + t
t
)
?s,t(q4).
Let
z = ?(q)2
x = 16q
?(q2)4
?(q)4
.
S. Cooper, On the Number of Representations of Certain Integers 41
Then [3] or [24, Ch. 16, Entry 25 (vii)]
z(s+t)/2xt/4 = ?s,t(q)
1 ? x = ?(?q)
4
?(q)4
.
By Jacobi?s triple product identity,
z =
(q2; q2)10?
(q; q)4?(q4; q4)4?
=
(?q;?q)4?
(q2; q2)2?
x(1 ? x) = 16q (q
2; q2)24?
(?q;?q)24?
.
Therefore
z5x(1 ? x) = 16q (q
2; q2)14?
(?q;?q)4?
(3.1)
z6x(1 ? x) = 16q(q2; q2)? (3.2)
z10x2(1 ? x)2 = 256q2 (q
2; q2)28?
(?q;?q)8?
(3.3)
z12x(1 ? x) = 16q(?q;?q)24? = 16
?
?
n=1
(?1)n+1?(n)qn, (3.4)
where ?(n) is Ramanujan?s ? -function defined in equation (2.12).
Let H(q) and H(n) be defined by
H(q) = q(?q;?q)16?(q2; q2)4? + 32q2
(q2; q2)28?
(?q;?q)8?
=
?
?
n=1
H(n)qn. (3.5)
Then
H(q) =
1
16
z10x(1 ? x)(1 + 2x ? 2x2). (3.6)
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].
Fact 1
M(2k+1)/2(??0(4)) is the vector space consisting of all linear combinations of ?2k+1?4j,4j(q), j =
0, 1, ? ? ? , k/2 [19, p. 184, Prop. 4].
Fact 2
The Hecke operators Tp2 , where p is any prime, map M(2k+1)/2(??0(4)) into itself [19, p. 206]. That
is,
f ? M(2k+1)/2(??0(4)) ? Tp2f ? M(2k+1)/2(??0(4)).
42 R.L.I.M.S. Vol. 3, April, 2002
Fact 3
Suppose
f(q) =
?
?
n=0
a(n)qn ? M(2k+1)/2(??0(4)).
If p is an odd prime, then [19, p. 207, Prop. 13] implies
Tp2f =
?
?
n=0
a(p2n)qn + pk?1
?
?
n=0
(
(?1)kn
p
)
a(n)qn + p2k?1
?
?
n=0
a(n/p2)qn,
while if p = 2, the note in [19, page 210] implies
T22f =
?
?
n=0
a(4n)qn.
4 Some preliminary lemmas
Lemma 4.1 Let
1
4
z5(5 ? x)(1 ? x) =
?
?
n=0
a(n)qn
1
64
z5(4x + x2) =
?
?
n=0
b(n)qn
1
16
z5x(1 ? x) =
?
?
n=0
c(n)qn
?1
8
z6(1 ? x)(1 ? x + x2) =
?
?
n=0
d(n)qn
1
8
z6(1 ? x)(1 ? x2) =
?
?
n=0
e(n)qn.
Let n = 2?
?
p p
?p be the prime factorisation of n. For any prime p ? 1 (mod 4), let xp and yp
be the unique pair of positive integers satisfying
x2p + y
2
p = p, 2|xp,
and define ?p by
tan ?p =
yp
xp
, 0 < ? <
?
2
.
Then
a(0) = 5/4, b(0) = 0, c(0) = 0, d(0) = ?1/8, e(0) = 1/8
and for n ? 1,
a(n) =
?
p
(?1)?p(p?1)/2
[
p4(?p+1) ? (?1)(?p+1)(p?1)/2
p4 ? (?1)(p?1)/2
]
b(n) = 24?
?
p
p4(?p+1) ? (?1)(?p+1)(p?1)/2
p4 ? (?1)(p?1)/2
S. Cooper, On the Number of Representations of Certain Integers 43
c(n) =
?
?
?
?
?
0, if ?p is odd for any p ? 3 (mod 4)
n2(?1)?
?
p?1 (mod 4)
sin 4(1 + ?p)?p
sin 4?p
, otherwise
d(n) =
?
?
?
?
?
?
?
?
?
?
p
p5?p+5 ? 1
p5 ? 1 if n is odd
?
(
30 ? 25? + 63
25 ? 1
)
?
p
p5?p+5 ? 1
p5 ? 1 if n is even
e(n) =
?
?
?
?
?
?
?
?
?
?
p
p5?p+5 ? 1
p5 ? 1 if n is odd
?
(
25?+5 ? 63
25 ? 1
)
?
p
p5?p+5 ? 1
p5 ? 1 if n is even.
Proof.
From [3] or [24, Ch. 17, Entry 17 (viii)] we have
1
4
z5(5 ? x)(1 ? x) = 5
4
+
?
?
k=0
(?1)k(2k + 1)4q2k+1
1 ? q2k+1 .
Therefore a(0) = 5/4 and
a(n) =
?
d|n
d odd
(?1)(d?1)/2d4
=
?
p
(?1)?p(p?1)/2
[
p4(?p+1) ? (?1)(?p+1)(p?1)/2
p4 ? (?1)(p?1)/2
]
.
Next, from [3] or [24, Ch. 17, Entry17 (iii)] we have
1
64
z5(4x + x2) =
?
?
k=1
k4qk
1 + q2k
.
Therefore b(0) = 0 and
b(n) =
?
d|n
d odd
(?1)(d?1)/2
(n
d
)4
= 24?
?
p
p4(?p+1) ? (?1)(?p+1)(p?1)/2
p4 ? (?1)(p?1)/2 .
For the coefficients c(n), by (3.1) and one of the Macdonald identities for BC2 (for example, see
[6]) we have
1
16
z5x(1 ? x) = q (q
2; q2)14?
(?q;?q)4?
=
1
6
?
??2 (mod 5)
??1 (mod 5)
??(?2 ? ?2)q(?2+?2)/5.
Therefore c(0) = 0 and
c(n) =
1
6
?
?2+?2=5n
??2,??1 (mod 5)
??(?2 ? ?2)
44 R.L.I.M.S. Vol. 3, April, 2002
=
?
?
?
?
?
0, if ?p is odd for any p ? 3 (mod 4)
n2(?1)?
?
p?1 (mod 4)
sin 4(1 + ?p)?p
sin 4?p
, otherwise.
The second part of this formula expressing c(n) in terms of ?p and ?p was stated (without proof)
by Ramanujan [25, eq. (156)].
Lastly, from [3] or [24, Ch. 17, Entries 14 (iii) & (vi)] we have
?1
8
z6(1 ? x)(1 ? x + x2) = ?1
8
?
?
?
k=1
(?1)kk5qk
1 + qk
1
8
z6(1 ? x)(1 ? x2) = 1
8
?
?
?
k=1
(?1)kk5qk
1 ? qk .
Therefore
d(n) =
?
d|n
(?1)(d+n/d)d5
=
?
?
?
?
?
?
?
?
?
?
p
p5?p+5 ? 1
p5 ? 1 if n is odd
?
(
30 ? 25? + 63
25 ? 1
)
?
p
p5?p+5 ? 1
p5 ? 1 if n is even
e(n) = ?
?
d|n
(?1)dd5
=
?
?
?
?
?
?
?
?
?
?
p
p5?p+5 ? 1
p5 ? 1 if n is odd
?
(
25?+5 ? 63
25 ? 1
)
?
p
p5?p+5 ? 1
p5 ? 1 if n is even.


Lemma 4.2 If p is an odd prime then
a(p) = (?1)(p?1)/2p4 + 1
d(p) = p5 + 1
e(p) = p5 + 1.
Proof.
These follow right away from Lemma 4.1. 

Lemma 4.3 Each of a(n), b(n), c(n), d(n) and e(n) are multiplicative. That is, a(mn) =
a(m)a(n), if m and n are relatively prime; and similarly for b, c, d and e.
Proof.
These follow right away from Lemma 4.1. 

Lemma 4.4 Under the change of variables
z ? z(1 ? x)1/2, x ? x
x ? 1
S. Cooper, On the Number of Representations of Certain Integers 45
the functions
z10(5x5 + 411x4 ? 2118x3 + 4262x2 ? 4275x + 1710) and
z12(43870208x6 ? 131830116x5 + 162785487x4 ? 105780950x3
+162565995x2 ? 131610624x + 43870208)
are invariant and
z6x(1157 ? 1157x + 1008x2)
changes sign.
Proof.
This follows by straightforward calculation. 

Lemma 4.5 The q-expansions of
z10(5x5 + 411x4 ? 2118x3 + 4262x2 ? 4275x + 1710) and
z12(43870208x6 ? 131830116x5 + 162785487x4 ? 105780950x3
+162565995x2 ? 131610624x + 43870208)
contain only even powers of q, and the q-expansion of
z6x(1157 ? 1157x + 1008x2)
contains only odd powers.
Proof.
This follows from the Change of Sign Principle [3, p. 126] or [24, Ch. 17, Entry 14 (xii)]. 

Lemma 4.6
[qp]z10(1 + x)(1 + 14x + x2)(1 ? 34x + x2) = ?264(p9 + 1)
[qp] z12
[
441(1 + 14x + x2)3 + 250(1 + x)2(1 ? 34x + x2)2
]
= 65520(p11 + 1).
Proof.
From [3] or [24, Ch. 17, Entries 13 (iii) and (iv)], followed by [2] or [24, Ch. 15, Entry 12 (iii)] we
have
z10(1 + x)(1 + 14x + x2)(1 ? 34x + x2) = M(q)N(q) = 1 ? 264
?
?
k=1
k9qk
1 ? qk .
The coefficient of qp in this is readily seen to be ?264(p9 + 1).
Also, from [3] or [24, Ch. 17, Entries 13 (iii) and (iv)] followed by [2] or [24, Ch. 15, Entry
13 (i)] we have
441z12(1 + 14x + x3)3 + 250z12(1 + x)2(1 ? 34x + x2)2
= 441M(q)3 + 250N(q)2 = 691 + 65520
?
?
j=1
j11qj
1 ? qj .
The coefficient of qp in this is readily seen to be 65520(p11 + 1). 

46 R.L.I.M.S. Vol. 3, April, 2002
5 The value of r11(p2) for odd primes p
Theorem 5.1 If p is an odd prime and H(p) is defined by equation (3.5), then
r11(p2) =
330
31
(p9 + 1) ? 22(?1)(p?1)/2p4 + 352
31
H(p).
Proof.
Let us write
p2 =
11
?
i=1
x2i .
There are three possibilities: either one, five or nine of the xi are odd, and the others are even.
Accordingly, we have
r11(p2) = 22
?
j odd
r10,0(p2 ? j2) +
22
5
?
j odd
r6,4(p2 ? j2) +
22
9
?
j odd
r2,8(p2 ? j2)
=
?
j odd
[
qp
2?j2
]
{
22?(q4)10 +
22
5
?
(
10
4
)
? 16q4?(q4)6?(q8)4
+
22
9
?
(
10
8
)
? 162q8?(q4)2?(q8)8
}
= 2
?
j odd
[
q(p
2?j2)/4
]
{(
11
1
)
?(q)10 +
(
11
5
)
? 16q?(q)6?(q2)4
+
(
11
9
)
? 162q2?(q)2?(q2)8
}
=
p
?
j=0
[
qj(p?j)
]
z5
{(
11
1
)
+
(
11
5
)
x +
(
11
9
)
x2
}
=
p
?
j=0
[
qj(p?j)
]
z5(11 + 462x + 55x2).
In general we have
r2k+1(p2) =
p
?
j=0
[qj(p?j)]zk
?
?
k/2
?
i=0
(
2k + 1
4i + 1
)
xi
?
? . (5.2)
Now
z5(11 + 462x + 55x2) =
11
5
z5(5 ? x)(1 ? x) + 528
5
z5(4x + x2) +
264
5
z5x(1 ? x).
Using this together with Lemmas 4.1, 4.2 and 4.3 gives
r11(p2) =
p
?
j=0
(
44
5
a(j(p ? j)) + 33792
5
b(j(p ? j)) + 4224
5
c(j(p ? j))
)
=
88
5
a(0) ? 88
5
a(0)a(p) +
44
5
p
?
j=0
a(j)a(p ? j)
+
33792
5
p
?
j=0
b(j)b(p ? j) + 4224
5
p
?
j=0
c(j)c(p ? j)
S. Cooper, On the Number of Representations of Certain Integers 47
= ?22(?1)(p?1)/2p4 + 44
5
[qp]
(
1
4
z5(5 ? x)(1 ? x)
)2
+
33792
5
[qp]
(
1
64
z5(4x + x2)
)2
+
4224
5
[qp]
(
1
16
z5x(1 ? x)
)2
= ?22(?1)(p?1)/2p4 + 11
4
[qp]z10(2x4 + 20x2 ? 12x + 5).
Now
11
4
(2x4 + 20x2 ? 12x + 5) = ? 5
124
p1(x) +
22
31
p2(x) + p3(x)
where
p1(x) = (1 + x)(1 + 14x + x2)(1 ? 34x + x2)
p2(x) = x(1 ? x)(1 + 2x ? 2x2)
p3(x) =
1
124
(5x5 + 411x4 ? 2118x3 + 4262x2 ? 4275x + 1710).
This, together with Lemmas 4.5, 4.6 and Equation (3.6) gives
r11(p2) = ?22(?1)(p?1)/2p4 ?
5
124
? (?264)(p9 + 1) + 22
31
? 16H(p) + 0
=
330
31
(p9 + 1) ? 22(?1)(p?1)/2p4 + 352
31
H(p).


Remark 5.3 The coefficients H(n) have some interesting properties.
1. H is multiplicative. That is, H(mn) = H(m)H(n) for any pair of relatively prime positive
integers m and n.
2. |H(n)| ? n9/2d(n), where d(n) is the number of divisors of n. In particular,
|H(p)| ? 2p9/2 (5.4)
for primes p. Ramanujan [25, ?28] made some conjectures concerning the orders of some
similar functions. See his equations (157), (160) and (163). See also Berndt?s commentary
[26, p. 367] for references and more information.
6 The Hecke operator Tp2 when p = 2 for eleven squares
Theorem 6.1
T4?11,0 = ?11,0 + 330?7,4 + 165?3,8
T4?7,4 = 336?7,4 + 176?3,8
T4?3,8 = 320?7,4 + 192?3,8.
That is,
?11,0(4n) = ?11,0(n) + 330?7,4(n) + 165?3,8(n)
?7,4(4n) = 336?7,4(n) + 176?3,8(n)
?3,8(4n) = 320?7,4(n) 192?3,8(n).
Proof.
Let us consider ?11,0 first. By Facts 1 and 2, we have
T4?11,0 = ?1?11,0 + ?2?7,4 + ?3?3,8,
48 R.L.I.M.S. Vol. 3, April, 2002
for some constants ?1, ?2, ?3. By Fact 3, we have
T4?11,0 =
?
?
n=0
?11,0(4n)qn.
Therefore
?
?
n=0
?11,0(4n)qn = ?1
?
?
n=0
?11,0(n)qn + ?2
?
?
n=0
?7,4(n)qn + ?3
?
?
n=0
?3,8(n)qn.
Equating coefficients of 1, q and q2 gives
?1 = 1
22?1 + 16?2 = 5302
220?1 + 224?2 + 256?3 = 116380
and so
?1 = 1, ?2 = 330, ?3 = 165.
This proves the first part of the Theorem. The results for T4?7,4 and T4?3,8 follow similarly. 

Lemma 6.2 The eigenfunctions and eigenvalues of T4 are
?11,1 := 511?11,0 ? 682?7,4 + 187?3,8, ?1 = 1,
?11,2 := 20?7,4 + 11?3,8, ?2 = 512,
?11,3 := ?7,4 ? ?3,8, ?3 = 16.
These coefficients satisfy many interesting properties. Some of these are summarised in
Theorem 6.3
32?11,1(n) ? 33?11,2(n) = 0 if n ? 1 or 2 (mod 4)
512?11,1(n) ? 17?11,2(n) = 0 if n ? 3 (mod 8)
15872?11,1(n) + 495?11,2(n) = 0 if n ? 7 (mod 8)
?11,3(n) = 0 if n ? 7 (mod 8).
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(4?(8k + 7)) =
29?+9 ? 1
29 ? 1 r11(8k + 7).
Proof.
r11(4?(8k + 7))
= ?11,0(4?(8k + 7))
=
1
31 ? 511
(
31?11,1(4?(8k + 7)) + 495?11,2(4?(8k + 7)) + 11242?11,3(4?(8k + 7))
)
=
1
31 ? 511
(
31?11,1(8k + 7) + 495 ? 512??11,2(8k + 7) + 11242 ? (16)??11,3(8k + 7)
)
.
S. Cooper, On the Number of Representations of Certain Integers 49
Applying Theorem 6.3 gives
r11(4?(8k + 7)) =
1
31 ? 511
(
31?11,1(8k + 7) ? 15872 ? 512??11,1(8k + 7) + 0
)
=
?1
511
(512?+1 ? 1)?11,1(8k + 7).
Taking ? = 0 in this gives
r11(8k + 7) = ??11,1(8k + 7)
and so
r11(4?(8k + 7)) =
1
511
(512?+1 ? 1)r11(8k + 7) =
29?+9 ? 1
29 ? 1 r11(8k + 7),
as required. 

7 The Hecke operator Tp2 when p is an odd prime for sums of eleven
squares
By Facts 2 and 3 we have
Tp2?11,0 =
?
?
n=0
?11,0(p2n)qn + p4
?
?
n=0
(?n
p
)
?11,0(n)qn + p9
?
?
n=0
?11,0(n/p2)qn
= c1?11,0(q) + c2?7,4(q) + c3?3,8(q), (7.1)
Tp2?7,4 =
?
?
n=0
?7,4(p2n)qn + p4
?
?
n=0
(?n
p
)
?7,4(n)qn + p9
?
?
n=0
?7,4(n/p2)qn
= d1?11,0(q) + d2?7,4(q) + d3?3,8(q), (7.2)
Tp2?3,8 =
?
?
n=0
?3,8(p2n)qn + p4
?
?
n=0
(?n
p
)
?3,8(n)qn + p9
?
?
n=0
?3,8(n/p2)qn
= e1?11,0(q) + e2?7,4(q) + e3?3,8(q), (7.3)
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 = p9 + 1
d1 = 0
e1 = 0.
Equating the coefficients of q in (7.1)?(7.3) gives
22c1 + 16c2 = ?11,0(p2) + 22(?1)(p?1)/2p4
22d1 + 16d2 = ?7,4(p2) + 16(?1)(p?1)/2p4
22e1 + 16e2 = ?3,8(p2).
By Theorem 5.1 we have
?11,0(p2) =
330
31
(p9 + 1) ? 22(?1)(p?1)/2p4 + 352
31
H(p). (7.4)
By the same methods it can be shown that
?7,4(p2) =
320
31
(p9 + 1) ? 16(?1)(p?1)/2p4 + 176
31
H(p) (7.5)
?3,8(p2) =
320
31
(p9 + 1) ? 320
31
H(p). (7.6)
50 R.L.I.M.S. Vol. 3, April, 2002
It follows that
c2 = ?
22
31
(p9 + 1) +
22
31
H(p)
d2 =
20
31
(p9 + 1) +
11
31
H(p)
e2 =
20
31
(p9 + 1) ? 20
31
H(p).
Next, equating coefficients of q4 in (7.1)?(7.3) gives
?11,0(4p2) + 5302(?1)(p?1)/2p4 = 5302c1 + 5376c2 + 5120c3
?7,4(4p2) + 5376(?1)(p?1)/2p4 = 5302d1 + 5376d2 + 5120d3
?3,8(4p2) + 5120(?1)(p?1)/2p4 = 5302e1 + 5376e2 + 5120e3.
By Theorem 6.1 and equations (7.4)?(7.6) we have
?11,0(4p2) = ?11,0(p2) + 330?7,4(p2) + 165?3,8(p2)
=
158730
31
(p9 + 1) ? 5302(?1)(p?1)/2p4 + 5632
31
H(p).
Similarly,
?7,4(4p2) =
163840
31
(p9 + 1) ? 5376(?1)(p?1)/2p4 + 2816
17
H(p)
?3,8(4p2) =
163840
31
(p9 + 1) ? 5120(?1)(p?1)/2p4 ? 5120
17
H(p).
It follows that
c3 =
22
31
(p9 + 1) ? 22
31
H(p)
d3 =
11
31
(p9 + 1) ? 11
31
H(p)
e3 =
11
31
(p9 + 1) +
20
31
H(p).
The above results may be summarised as
Theorem 7.7 Let
? = p9 + 1, ? = H(p).
Then
?
?
Tp2?11,0
Tp2?7,4
Tp2?3,8
?
? =
1
31
?
?
31? ?22? + 22? 22? ? 22?
0 20? + 11? 11? ? 11?
0 20? ? 20? 11? + 20?
?
?
?
?
?11,0
?7,4
?3,8
?
? .
Corollary 7.8 The eigenfunctions and eigenvalues of Tp2 are
?11,1 := ?11,0 ? 2?7,4, ?1 = ?
?11,2 := 20?7,4 + 11?3,8 = ?11,2, ?2 = ?
?11,3 := ?7,4 ? ?3,8 = ?11,3, ?3 = ?.
That is,
Tp2?11,1 = (p
9 + 1)?11,1
Tp2?11,2 = (p
9 + 1)?11,2
Tp2?11,3 = H(p)?11,3.
S. Cooper, On the Number of Representations of Certain Integers 51
Using Fact 3, together with Corollary 7.8, we have
?11,1(p2n) + p4
(?n
p
)
?11,1(n) + p9?11,1(n/p2) = (p9 + 1)?11,1(n) (7.9)
?11,2(p2n) + p4
(?n
p
)
?11,2(n) + p9?11,2(n/p2) = (p9 + 1)?11,2(n) (7.10)
?11,3(p2n) + p4
(?n
p
)
?11,3(n) + p9?11,3(n/p2) = H(p)?11,3(n). (7.11)
Iterating (7.9)?(7.11) we obtain
Theorem 7.12 If p2  | n, then
?11,1(p2?n) =
[
p9?+9 ? 1
p9 ? 1 ? p
4
(?n
p
)
p9? ? 1
p9 ? 1
]
?11,1(n)
?11,2(p2?n) =
[
p9?+9 ? 1
p9 ? 1 ? p
4
(?n
p
)
p9? ? 1
p9 ? 1
]
?11,2(n)
?11,3(p2?n) = [c1(n)r
?
1 + c2(n)r
?
2 ] ?11,3(n),
where
r1 =
H(p) +
?
?(p)
2
r2 =
H(p) ?
?
?(p)
2
c1(n) =
1
2
+
H(p) ? 2p4
(
?n
p
)
2
?
?(p)
c2(n) =
1
2
?
H(p) ? 2p4
(
?n
p
)
2
?
?(p)
?(p) = H(p)2 ? 4p9.
Remark 7.13 Equation 5.4 implies ?(p) ? 0.
It is possible to express r11(n) in terms of r11(no) and ?11,3(no), where no is the odd-square free
part of n, that is, no is the greatest divisor of n which is not divisible by an odd square. Let
gp,?,m =
p9?+9 ? 1
p9 ? 1 ? p
4
(?m
p
)
p9? ? 1
p9 ? 1
hp,?,m = c1(m)r
?
1 + c2(m)r
?
2 ,
where c1, c2, r1, r2 are as for Theorem 7.12. If p2  | m then
r11(p2?m) =
1
31
[
31?11,1(p2?m) + 2?11,2(p2?m) + 22?11,3(p2?m)
]
=
1
31
[31gp,?,m?11,1(m) + 2gp,?,m?11,2(m) + 22hp,?,m?11,3(m)]
=
1
31
[31gp,?,m?11,1(m) + 2gp,?,m?11,2(m) + 22gp,?,m?11,3(m)]
+
22
31
[hp,?,m ? gp,?,m] ?11,3(m)
= gp,?,mr11(m) +
22
31
[hp,?,m ? gp,?,m] ?11,3(m).
52 R.L.I.M.S. Vol. 3, April, 2002
By induction on j, it follows that if p1, p2, ? ? ? , pj are distinct odd primes, and p2i  | m for 1 ? i ? j,
then
r11(p
2?1
1 p
2?2
2 ? ? ? p
2?j
j m)
=
(
j
?
i=1
gpi,?i,m
)
r11(m) +
22
31
[(
j
?
i=1
hpi,?i,m
)
?
(
j
?
i=1
gpi,?i,m
)]
?11,3(m).
Consequently, if no is the odd-square free part of n and either p2?p || n or p2?p+1 || n, then
r11(n) = r11(no)
?
p
[
p9?p+9 ? 1
p9 ? 1 ? p
4
(?no
p
)
p9?p ? 1
p9 ? 1
]
+
22
31
?11,3(no)
[
?
p
hp,?p,no ?
?
p
(
p9?p+9 ? 1
p9 ? 1 ? p
4
(?no
p
)
p9?p ? 1
p9 ? 1
)
]
.
If no = 4?(8k + 7) for some nonnegative integers ? and k, then Theorem 6.3 implies ?11,3(no) = 0.
In this case we have
r11(n) = r11(no)
?
p
[
p9?p+9 ? 1
p9 ? 1 ? p
4
(?no
p
)
p9?p ? 1
p9 ? 1
]
.
This, together with Theorem 6.4 implies
Theorem 7.14 Let n = 2?
?
p p
?p be the prime factorisation of n and let n? be the square-free
part of n. If n? is of the form 8k + 7 then
r11(n) = r11(n?)
29?/2+9 ? 1
29 ? 1
?
p
[
p9?p/2+9 ? 1
p9 ? 1 ? p
4
(?n?
p
)
p9?p/2 ? 1
p9 ? 1
]
.
8 The value of r13(p2) for odd primes p
Theorem 8.1 If p is an odd prime and ?(p) is defined by equation (2.12), then
r13(p2) =
4030
691
(p11 + 1) ? 26p5 + 13936
691
?(p).
Proof.
We proceed in a similar way to Section 5. Taking k = 6 in (5.2) gives
r13(p2) =
p
?
j=0
[qj(p?j)]z6
{
13 + 1287x + 715x2 + x3
}
. (8.2)
Now
z6(13 + 1287x + 715x2 + x3)
= 1014z6(1 ? x)(1 ? x + x2) ? 1001z6(1 ? x)(1 ? x2)
+2z6x(1157 ? 1157x + 1008x2). (8.3)
Therefore Lemmas 4.1, 4.2 and 4.3 give
r13(p2)
= ?8112
p
?
j=0
d(j(p ? j)) ? 8008
p
?
j=0
e(j(p ? j)) + 0
S. Cooper, On the Number of Representations of Certain Integers 53
= ?16224d(0) + 16224d(0)d(p) ? 8112
p
?
j=0
d(j)d(p ? j)
?16016e(0) + 16016e(0)e(p) ? 8008
p
?
j=0
e(j)e(p ? j)
= ?26p5 ? 8112[qp]
(
?1
8
z6(1 ? x)(1 ? x + x2)
)2
? 8008[qp]
(
1
8
z6(1 ? x)(1 ? x2)
)2
= ?26p5 ? 1
8
[qp]z12
(
1014(1 ? x)2(1 ? x + x2)2 + 1001(1 ? x)2(1 ? x2)2
)
.
Now
?1
8
(
1014(1 ? x)2(1 ? x + x2)2 + 1001(1 ? x)2(1 ? x2)2
)
=
31
504 ? 691p1(x) +
871
691
p2(x) + p3(x), (8.4)
where
p1(x) = 441(1 + 14x + x2)3 + 250(1 + x)2(1 ? 34x + x2)2,
p2(x) = x(1 ? x),
p3(x) =
?1
174132
(
43870208x6 ? 131830116x5 + 162785487x4 ? 105780950x3
+ 162565995x2 ? 131610624x + 43870208
)
.
Therefore Lemmas 4.5, 4.6 and Equation (3.4) give
r13(p2) = ?26p5 +
31
504 ? 691 ? 65520(p
11 + 1) +
871
691
? 16?(p) + 0
=
4030
691
(p11 + 1) ? 26p5 + 13936
691
?(p),
as required. 

Remark 8.5 The coefficients ?(n) satisfy some well-known interesting properties.
1. ? is multiplicative. That is, ?(mn) = ?(m)?(n) for any pair of relatively prime positive
integers m and n.
2. |?(n)| ? n11/2d(n), where d(n) is the number of divisors of n. In particular,
|?(p)| ? 2p11/2
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
thirteen 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.
54 R.L.I.M.S. Vol. 3, April, 2002
Theorem 9.1
T4?13,0 = ?13,0 + 715?9,4 + 1287?5,8 + 13?1,12
T4?9,4 = 744?9,4 + 1296?5,8 + 8?1,12
T4?5,8 = 768?9,4 + 1280?5,8
T4?1,8 = 512?9,4 + 1536?5,8.
Corollary 9.2 The eigenfunctions of T4 are
?13,1 := 1414477?13,0 ? 2298777?9,4 + 908427?5,8 ? 2015?1,12
?13,2 := 256?9,4 + 434?5,8 + ?1,12
?13,3 := ??9,4 ? (? + 2)?5,8 + 2?1,12
?13,4 := ??9,4 ? (? + 2)?5,8 + 2?1,12
and the corresponding eigenvalues are ?1 = 1, ?2 = 2048, ?3 = 4?, ?4 = 4?, where ? = ?3 +?
?119.
The coefficients in the expansions of ?13,0, ?9,4, ?5,8 and ?1,12 satisfy some interesting properties:
Some of these are summarised in
Theorem 9.3
If n ? 2 or 3 (mod 4) then
8?9,4(n) ? 9?5,8(n) + ?1,12(n) = 0
64?13,0(n) ? 104?9,4(n) + 39?5,8(n) = 0.
If n ? 1 (mod 8) then
?5,8(n) ? ?1,12(n) = 0
2048?13,0(n) ? 3328?9,4(n) + 1313?5,8(n) = 0.
If n ? 5 (mod 8) then
64?9,4(n) ? 57?5,8(n) ? 7?1,12(n) = 0
14336?13,0(n) ? 23488?9,4(n) + 9369?5,8(n) = 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
? = p11 + 1, ? = ?(p).
Then
?
?
?
?
Tp2?13,0
Tp2?9,4
Tp2?5,8
Tp2?1,12
?
?
?
?
=
1
691
?
?
?
?
691? ?871? + 871? 871? ? 871? 0
0 256? + 435? 434? ? 434? ? ? ?
0 256? ? 256? 434? + 257? ? ? ?
0 256? ? 256? 434? ? 434? ? + 690?
?
?
?
?
?
?
?
?
?13,0
?9,4
?5,8
?1,12
?
?
?
?
.
Corollary 9.5 The eigenfunctions and eigenvalues of Tp2 are
?13,1 := 691?13,0 ? 871?9,4 + 871?5,8, ?1 = ?
?13,2 := 256?9,4 + 434?5,8 + ?1,12 = ?13,2, ?2 = ?
?13,3 := ?9,4 ? ?5,8 = ?13,3, ?3 = ?
?13,4 := ?5,8 ? ?1,12 = ?13,4, ?4 = ?.
S. Cooper, On the Number of Representations of Certain Integers 55
That is,
Tp2?13,1 = (p
11 + 1)?13,1
Tp2?13,2 = (p
11 + 1)?13,2
Tp2?13,3 = ?(p)?13,3
Tp2?13,4 = ?(p)?13,4.
10 Sums of an odd number of squares
It is interesting to compare the values of rk(p2) for various k. We have:
r1(p2) = 2 (10.1)
r2(p2) = 4(2 + (?1)(p?1)/2) (10.2)
r3(p2) = 6(p + 1 ? (?1)(p?1)/2) (10.3)
r4(p2) = 8(p2 + p + 1) (10.4)
r5(p2) = 10(p3 ? p + 1) (10.5)
r6(p2) = 12(p4 + (?1)(p?1)/2p2 + 1) (10.6)
r7(p2) = 14(p5 ? (?1)(p?1)/2p2 + 1) (10.7)
r8(p2) = 16(p6 + p3 + 1) (10.8)
r9(p2) =
274
17
(p7 + 1) ? 18p3 + 32
17
?(p) (10.9)
r10(p2) =
68
5
(p8 + (?1)(p?1)/2p4 + 1) + 32
5
c(p2) (10.10)
r11(p2) =
330
31
(p9 + 1) ? 22(?1)(p?1)/2p4 + 352
31
H(p) (10.11)
r12(p2) = 8(p10 + p5 + 1) + 16?(p2) (10.12)
r13(p2) =
4030
691
(p11 + 1) ? 26p5 + 13936
691
?(p). (10.13)
where
?
?
n=1
?(n)qn = q(?q;?q)8?(q2; q2)8? =
1
16
z8x(1 ? x)
?
?
n=1
c(n)qn = q
(q2; q2)14?
(?q;?q)4?
=
1
16
z5x(1 ? x)
?
?
n=1
H(n)qn = q(?q;?q)16?(q2; q2)4? + 32q2
(q2; q2)28?
(?q;?q)8?
=
1
16
z10x(1 ? x)(1 + 2x ? 2x2)
?
?
n=1
?(n)qn = q(q2; q2)12? =
1
16
z6x(1 ? x)
?
?
n=1
(?1)n+1?(n)qn = q(?q;?q)24? =
1
16
z12x(1 ? x).
Observe that formulas (10.9)?(10.13) are significantly more complicated in nature than (10.1)?
(10.8). This is a vivid illustration of Eisenstein?s remark [8], [31].
Equation (10.1) is trivial. Equations (10.2), (10.4), (10.6), (10.8), (10.10) and (10.12) follow
readily from the well known formulas for r2(n), r4(n), r6(n), r8(n), r10(n) and r12(n); see, for
56 R.L.I.M.S. Vol. 3, April, 2002
example, [4]. Equations (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. Equations
(10.11) and (10.13) are Theorems 5.1 and 8.1, respectively. Equations (10.7), (10.9), (10.11) and
(10.13) are originally due to Sandham [27], [28]. Slightly different proofs of (10.7) and (10.9) were
given in [5], and the proofs we have given here of (10.11) and (10.13) are different from Sandham?s.
Results for r2k(p2) can be written down immediately from the general formulas for r2k(n) in, for
example, [4].
For sums of an odd number of squares, we have
Conjecture 10.14
r2k+1(p2) = Ak(p2k?1 + 1) ? (4k + 2)(?1)k(p?1)/2pk?1
+[qp]
k/2
?
j=1
cj,kqj
(?q;?q)8k?24j?
(q2; q2)4k?24j?
,
where Ak and cj,k are rational numbers and
Ak + c1,k = 4k + 2.
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