AN ODD SQUARE AS A SUM OF AN ODD NUMBER

AN ODD SQUARE AS A SUM OF AN ODD NUMBER OF ODD SQUARES HENG HUAT CHAN, SHAUN COOPER AND WEN-CHIN LIAW Abstract. Let sk (n) be the number of representations of n as a sum of k odd squares. We study the function s8k+1 (n2 ), where n is any odd positive integer. Special cases of our results are 1 1 1 s9 (p2 ) = s16 (8p) and s17 (p2 ) = s32 (8p), 16 17 32 where p is any odd prime.

1. Introduction Let sk (n) be the number of representations of n as sum of k odd squares. The generating function for sk (n) is  k ∞ ∞ X X 2 Sk (q) := sk (n)q n =  q (2j+1)  , n=0

j=0

and clearly sk (n) = 0 if n 6≡ k (mod 8). The goal of this article is to prove that for any odd positive integer n and every positive integer k,   8n (24k − 1)B4k X µ(d)d4k−1 s16k + O(n6k−1 ), (1.1) s8k+1 (n2 ) = − 8k d d|n

where Bk is the kth Bernoulli number 1 given by ∞ X xk x = B , k ex − 1 k! k=0

and   1 µ(n) = (−1)j   0

if n = 1, if n = p1 p2 · · · pj , for distinct primes p1 , p2 , . . . , pj , otherwise.

When k = 1 or 2, the error term in (1.1) is zero, and we obtain the identities   1 X 8n 2 3 s9 (n ) = µ(d)d s16 16 d d|n

2000 Mathematics Subject Classification. 11F03, 11F11, 11F20. Key words and phrases. sums of odd squares, modular forms, Eisenstein series. 1Note that B < 0. 4k 1

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HENG HUAT CHAN, SHAUN COOPER AND WEN-CHIN LIAW

and 17 X s17 (n ) = µ(d)d7 s32 32 2



d|n

8n d

 .

When n = p is prime, these simplify further to 1 s16 (8p) 16

s9 (p2 ) =

(1.2) and

1 1 s17 (p2 ) = s32 (8p). 17 32 The method we shall use in proving (1.1) is motivated by the work of A. Hurwitz [6]. In Section 2, we illustrate the main idea by proving (1.1) in the case when k = 1 and deducing (1.2). In Section 3, we prove (1.1) and give a precise formula for the error term. We also give an asymptotic formula for s8k+1 (n2 ) in terms of the divisors of n. (1.3)

2. Proof of (1.1) for k = 1 Let Tk (q) =

∞ X

 k ∞ X 2 tk (n)q n =  q (2j+1) /8  .

n=0

j=0

Note that Tk (q 8 ) = Sk (q) and for all positive integers n, (2.1)

t8k (n) = s8k (8n).

Next, recall that for |q| < 1, (2.2)

∞ X n=0

∞ X n3 q n . t8 (n)q = 1 − q 2n n

n=1

Identity (2.2) is the classical sum of eight triangular numbers formula of Legendre [8, p. 133] and Jacobi [7, p. 170]. It was rediscovered by Ramanujan [10, p. 191], and many proofs have since been given. For example, see [3, eq. (3.71)]. Equating coefficients of q n on both sides of (2.2), we deduce that  n 3 X , (2.3) t8 (n) = (d) d d|n

where (2.4)

( 0 (n) = 1

if n is even, if n is odd.

AN ODD SQUARE AS A SUM OF AN ODD NUMBER OF ODD SQUARES

3

From (2.3), we see that the corresponding Dirichlet series for T8 (q) (for Re s > 3) is ∞ X t8 (n) ζT8 (s) := = ζ(s − 3)L(s), ns n=1

where ζ(s) is the Riemann zeta function, and L(s) =

∞ X (n) n=1

Hence ζT8 (s) has the Euler product Y (2.5) ζT8 (s) = p 1−

ns

.

1 ((p)+p3 ) ps

+

(p) p2s−3

.

For positive integers m, k and some arithmetical function χ, let Tm,χ be the operator on a power series A(q) =

∞ X

a(n)q n

n=0

defined by Tm,χ (A(q)) =

∞ X

b(n)q n ,

n=0

where b(n) =

X

χ(d)dk−1 a

d|gcd(m,n)

Let

 Γ0 (N ) =

 A ∈ SL(2, Z) | A ≡

 mn  d2

 1 ∗ 0 1

.

 (mod N ) .

Let Mk (N ) = M (Γ0 (N ), k, 1) be the space of weight k modular forms on Γ0 (N ) with multiplier 1. When χ =  with  given by (2.4), Tm, are the Hecke operators on Mk (2). It is known that if A(q) ∈ Mk (2), with a(1) = 1, and the corresponding Dirichlet series for A(q) has an Euler product, then (see for example [9, Theorem 4.5.16]) Tm, (A(q)) = a(m)A(q). Hence  mn  X . (2.6) a(m)a(n) = (d)dk−1 a d2 d|gcd(m,n)

Using (2.5) and the fact that T8k ∈ M4k (2) [11, p. 222, Theorem 7.1.4] for q = e2πiτ , we deduce from (2.6) that  mn  X . (2.7) t8 (m)t8 (n) = (d)d3 t8 d2 d|gcd(m,n)

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HENG HUAT CHAN, SHAUN COOPER AND WEN-CHIN LIAW

When m = 2 and n is any positive integer, we deduce from (2.7) that (2.8)

t8 (2)t8 (n) = t8 (2n).

We now state and prove a simple lemma. Lemma 2.1. If f satisfies (2.9)

X

f (m)f (n) =

g(d)f

 mn  d2

d|gcd(m,n)

,

where g is a completely multiplicative function, then (2.10) m n X m n X f (mn) = µ(d)g(d)f f = µ(d)g(d)f f , d d d d d

d|gcd(m,n)

where we use the convention that the coefficient f (n/d) is zero if d is not a divisor of n. Proof. Let N be any positive integer. Let m and n be any positive integers satisfying N = gcd(m, n) and write m = N s and n = N t, with gcd(s, t) = 1. From (2.9), we have   X stN 2 f (N s)f (N t) = . g(d)f d2 d|N

Since X

h(d) =

d|N

X

h(N/d),

d|N

we conclude that X N 
f std2 , f (N s)f (N t) = g d d|N

or


f (N s)f (N t) X 1 = f std2 . g(N ) g(d) d|N

Applying the M¨obius inversion formula, we find that
X  N  f (ds)f (dt) f stN 2 = µ . g(N ) d g(d) d|N

Simplifying the above, we obtain the first equality of (2.10). The second equality of (2.10) follows immediately.  Applying Lemma 2.1 with f = t8 and using equation (2.7), we conclude that m n X (2.11) t8 (mn) = (d)µ(d)d3 t8 t8 . d d d

We are now ready to prove the main result of this section.

AN ODD SQUARE AS A SUM OF AN ODD NUMBER OF ODD SQUARES

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Theorem 2.2. For all odd positive integers n, s9 (n2 ) =

1 X µ(d)d3 s16 16



d|n

8n d

 .

Proof. Observe that s9 (n2 ) =

X

s8 (n2 − i2 )

0