THETA FUNCTION IDENTITIES AND REPRESENTATIONS BY

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International Journal of Number Theory Vol. 4, No. 2 (2008) 219–239 c World Scientific Publishing Company °

THETA FUNCTION IDENTITIES AND REPRESENTATIONS BY CERTAIN QUATERNARY QUADRATIC FORMS

AYS ¸ E ALACA∗ , S ¸ ABAN ALACA† , MATHIEU F. LEMIRE‡ and KENNETH S. WILLIAMS§ Centre for Research in Algebra and Number Theory School of Mathematics and Statistics Carleton University, Ottawa Ontario, K1S 5B6, Canada ∗[email protected] †[email protected] ‡[email protected] §[email protected] Received 7 December 2006 Accepted 19 January 2007 Some new theta function identities are proved and used to determine the number of representations of a positive integer n by certain quaternary quadratic forms. Keywords: Quaternary quadratic forms; theta functions. Mathematics Subject Classification 2000: 11E20, 11E25

1. Introduction Let N, N0 , Z, Q, R and C denote the sets of positive integers, nonnegative integers, integers, rational numbers, real numbers and complex numbers respectively. For q ∈ C with |q| < 1 Jacobi’s one-dimensional theta function ϕ(q) is defined by X 2 qn , ϕ(q) := (1.1) n∈Z

and the Borweins’ two-dimensional theta function a(q) is defined by X 2 2 a(q) := q m +mn+n .

(1.2)

(m,n)∈Z2

For a, b, c, d ∈ N and n ∈ N0 we define N (a, b, c, d; n) = card{(x, y, z, t) ∈ Z4 | n = ax2 + by 2 + cz 2 + dt2 }

(1.3)

M (a, b; n) = card{(x, y, z, t) ∈ Z4 | n = a(x2 + y 2 ) + b(z 2 + zt + t2 )}.

(1.4)

and

219

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Clearly (1.5)

N (a, b, c, d; 0) = M (a, b; 0) = 1. Also ∞ X

N (a, b, c, d; n)q n = ϕ(q a )ϕ(q b )ϕ(q c )ϕ(q d )

(1.6)

n=0

and ∞ X

M (a, b; n)q n = ϕ2 (q a )a(q b ).

(1.7)

n=0

In this paper, using some recent results of Alaca, Alaca and Williams [1, 2], we prove some new theta function identities, see Theorems 2.6 and 2.7. For example we show that 3 1 1 ϕ(q)ϕ(q 2 )ϕ(q 6 ) = ϕ3 (q 3 ) + ϕ2 (q)ϕ(q 3 ) − ϕ(q)ϕ(−q)ϕ(−q 3 ) 4 4 2 1 3 − ϕ(q 3 )ϕ2 (−q) + ϕ(q 3 )ϕ2 (−q 3 ), 4 4 see Theorem 2.6(a). Then, using these identities in conjunction with some results of Petr [17], we determine the number of representations of n ∈ N by each of the following quaternary quadratic forms: x2 + y 2 + z 2 + 3t2 x2 + y 2 + 2z 2 + 6t2 x2 + 2y 2 + 2z 2 + 3t2 x2 + 2y 2 + 4z 2 + 6t2 x2 + 3y 2 + 3z 2 + 3t2 x2 + 3y 2 + 6z 2 + 6t2 2x2 + 3y 2 + 3z 2 + 6t2 x2 + y 2 + z 2 + zt + t2 x2 + y 2 + 2(z 2 + zt + t2 ) x2 + y 2 + 4(z 2 + zt + t2 ) 3(x2 + y 2 ) + (z 2 + zt + t2 ) 3(x2 + y 2 ) + 2(z 2 + zt + t2 ) 3(x2 + y 2 ) + 4(z 2 + zt + t2 )

(Sec. (Sec. (Sec. (Sec. (Sec. (Sec. (Sec. (Sec. (Sec. (Sec. (Sec. (Sec. (Sec.

4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16)

2. Theta Function Identities As in [2, pp. 32–33] we set p = p(q) =

ϕ2 (q) − ϕ2 (q 3 ) 2ϕ2 (q 3 )

(2.1)

ϕ3 (q 3 ) . ϕ(q)

(2.2)

and k = k(q) =

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221

Next we recall the duplication, triplication and change of sign principles for p and k, which were proved in [1, Theorems 9–11]. Theorem 2.1 (Duplication Principle). p(q 2 ) =

1 + p − p2 − ((1 − p)(1 + p)(1 + 2p))1/2 , p2

k(q 2 ) =

(1 + p − p2 + ((1 − p)(1 + p)(1 + 2p))1/2 )k . 2

Theorem 2.2 (Triplication Principle). p(q 3 ) = 3−1 ((−4 − 3p + 6p2 + 4p3 ) + 22/3 (1 − 2p − 2p2 )((1 − p)(1 + 2p)(2 + p))1/3 + 21/3 (1 + 2p)((1 − p)(1 + 2p)(2 + p))2/3 ), k(q 3 ) = 3−2 (3 + 22/3 (1 + 2p)((1 − p)(1 + 2p)(2 + p))1/3 + 24/3 ((1 − p)(1 + 2p)(2 + p))2/3 )k. Theorem 2.3 (Change of Sign Principle). p(−q) =

−p , 1+p

k(−q) = (1 + p)2 k.

From (2.1) and (2.2) we obtain ϕ(q) = (1 + 2p)3/4 k 1/2 ,

ϕ(q 3 ) = (1 + 2p)1/4 k 1/2 .

(2.3)

Applying the change of sign principle (Theorem 2.3) to (2.3), we obtain ϕ(−q) = (1 − p)3/4 (1 + p)1/4 k 1/2 ,

ϕ(−q 3 ) = (1 − p)1/4 (1 + p)3/4 k 1/2 .

(2.4)

Now ϕ(q) + ϕ(−q) =

∞ X

2

∞ X

2

(1 + (−1)n )q n = 2

n=−∞

2

q 4n = 2ϕ(q 4 ),

(2.5)

n=−∞

so appealing to (2.3)–(2.5), we obtain ϕ(q 4 ) =

1 ((1 + 2p)3/4 + (1 − p)3/4 (1 + p)1/4 )k 1/2 . 2

(2.6)

Replacing q by q 3 in (2.5), we have ϕ(q 3 ) + ϕ(−q 3 ) = 2ϕ(q 12 ).

(2.7)

Then, appealing to (2.3), (2.4) and (2.7), we deduce ϕ(q 12 ) =

1 ((1 + 2p)1/4 + (1 − p)1/4 (1 + p)3/4 )k 1/2 . 2

(2.8)

Next let r(n) = card{(x, y) ∈ Z2 | n = x2 + y 2 },

n ∈ N0 .

(2.9)

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Then ϕ2 (q) =

∞ X

∞ X

r(n)q n =

n=0

∞ X

r(n)q n +

n=0 n even

r(n)q n

(2.10)

n=0 n odd

and ϕ2 (−q) =

∞ X

r(n)(−q)n =

n=0

∞ X

r(n)q n −

n=0 n even

∞ X

r(n)q n .

(2.11)

n=0 n odd

Adding (2.10) and (2.11), and recalling that r(2n) = r(n) (n ∈ N0 ), we obtain ϕ2 (q) + ϕ2 (−q) = 2

∞ X

r(n)q n = 2

n=0 n even

∞ X

r(2n)q 2n = 2

n=0

∞ X

r(n)q 2n ,

n=0

that is ϕ2 (q) + ϕ2 (−q) = 2ϕ2 (q 2 ).

(2.12)

Appealing to (2.3), (2.4) and (2.12), we obtain 1 ϕ(q 2 ) = √ ((1 + 2p)3/2 + (1 − p)3/2 (1 + p)1/2 )1/2 k 1/2 . 2

(2.13)

Replacing q by q 3 in (2.12), we have ϕ2 (q 3 ) + ϕ2 (−q 3 ) = 2ϕ2 (q 6 ).

(2.14)

Then, appealing to (2.3), (2.4) and (2.14), we deduce 1 ϕ(q 6 ) = √ ((1 + 2p)1/2 + (1 − p)1/2 (1 + p)3/2 )1/2 k 1/2 . 2 We have proved the following result. Theorem 2.4. Define p and k by (2.1) and (2.2), respectively. Then (a) ϕ(q) = (1 + 2p)3/4 k 1/2 , 1 (b) ϕ(q 2 ) = √ ((1 + 2p)3/2 + (1 − p)3/2 (1 + p)1/2 )1/2 k 1/2 , 2 (c) ϕ(q 3 ) = (1 + 2p)1/4 k 1/2 , 1 (d) ϕ(q 4 ) = ((1 + 2p)3/4 + (1 − p)3/4 (1 + p)1/4 )k 1/2 , 2 1 6 (e) ϕ(q ) = √ ((1 + 2p)1/2 + (1 − p)1/2 (1 + p)3/2 )1/2 k 1/2 , 2 1 (f) ϕ(q 12 ) = ((1 + 2p)1/4 + (1 − p)1/4 (1 + p)3/4 )k 1/2 , 2

(2.15)

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(g) ϕ(−q) = (1 − p)3/4 (1 + p)1/4 k 1/2 , (h) ϕ(−q 3 ) = (1 − p)1/4 (1 + p)3/4 k 1/2 . We remark that parts (b)–(f) can also be obtained from part (a) by applying Theorems 2.1 and 2.2. From Theorem 2.4(b), (e), we obtain the following result. Theorem 2.5. ϕ(q 2 )ϕ(q 6 ) =

1 (1 + p + p2 + (1 − p)1/2 (1 + p)1/2 (1 + 2p)1/2 )k. 2

Proof. From parts (b) and (e) of Theorem 2.4, we have ϕ(q 2 )ϕ(q 6 ) =

1 (((1 + 2p)3/2 + (1 − p)3/2 (1 + p)1/2 ) 2 × ((1 + 2p)1/2 + (1 − p)1/2 (1 + p)3/2 ))1/2 k

=

1 (2 + 4p + 2p2 + p4 2 + (2 + 2p + 2p2 )(1 − p)1/2 (1 + p)1/2 (1 + 2p)1/2 )1/2 k

=

1 (1 + p + p2 + (1 − p)1/2 (1 + p)1/2 (1 + 2p)1/2 )k. 2

The theta function identities given in the next theorem follow from Theorems 2.4 and 2.5. Theorem 2.6. 3 3 3 1 1 ϕ (q ) + ϕ2 (q)ϕ(q 3 ) − ϕ(q)ϕ(−q)ϕ(−q 3 ) 4 4 2 1 3 − ϕ(q 3 )ϕ2 (−q) + ϕ(q 3 )ϕ2 (−q 3 ). 4 4 1 1 1 (b) ϕ(q 2 )ϕ(q 3 )ϕ(q 6 ) = − ϕ(q)ϕ2 (q 3 ) + ϕ3 (q) + ϕ(q 3 )ϕ(−q)ϕ(−q 3 ) 4 4 2 1 2 1 2 + ϕ (−q)ϕ(q) + ϕ (−q 3 )ϕ(q). 4 4 1 1 3 (c) ϕ(q)ϕ(q 2 )ϕ(q 4 )ϕ(q 6 ) = ϕ(q)ϕ3 (q 3 ) + ϕ3 (q)ϕ(q 3 ) − ϕ2 (q)ϕ(−q)ϕ(−q 3 ) 8 8 8 1 3 2 3 − ϕ (q )ϕ(−q)ϕ(−q 3 ) − ϕ2 (q 4 )ϕ(−q 4 )ϕ(−q 12 ) 8 2 3 2 12 + ϕ (q )ϕ(−q 4 )ϕ(−q 12 ). 2 (a) ϕ(q)ϕ(q 2 )ϕ(q 6 ) =

Proof. We just give the proof of (a) as the remaining identities can be proved in a similar manner.

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(a) By Theorem 2.4(a), (c), (g), (h) and Theorem 2.5, we have 3ϕ3 (q 3 ) + ϕ2 (q)ϕ(q 3 ) − 2ϕ(q)ϕ(−q)ϕ(−q 3 ) − ϕ(q 3 )ϕ2 (−q) + 3ϕ(q 3 )ϕ2 (−q 3 ) = 3(1 + 2p)3/4 k 3/2 + (1 + 2p)7/4 k 3/2 − 2(1 + 2p)3/4 (1 − p)(1 + p)k 3/2 − (1 + 2p)1/4 (1 − p)3/2 (1 + p)1/2 k 3/2 + 3(1 + 2p)1/4 (1 − p)1/2 (1 + p)3/2 k 3/2 = (1 + 2p)1/4 k 3/2 ((2 + 2p + 2p2 )(1 + 2p)1/2 + (2 + 4p)(1 − p)1/2 (1 + p)1/2 ) = 2(1 + 2p)3/4 k 3/2 (1 + p + p2 + (1 − p)1/2 (1 + p)1/2 (1 + 2p)1/2 ) = 4ϕ(q)ϕ(q 2 )ϕ(q 6 ). In [1, p. 178] Alaca, Alaca and Williams showed that a(q) = (1 + 4p + p2 )k,

(2.16)

a(q 2 ) = (1 + p + p2 )k, µ ¶ 1 a(q 4 ) = 1 + p − p2 k. 2

(2.17) (2.18)

Making use of Theorem 2.4 and (2.16)–(2.18), we obtain the theta function identities stated in the next theorem. Theorem 2.7. (a) a(q) = 2ϕ(q)ϕ(q 3 ) − ϕ(−q)ϕ(−q 3 ). 1 1 (b) a(q 4 ) = ϕ(q)ϕ(q 3 ) + ϕ(−q)ϕ(−q 3 ). 2 2 3 1 (c) ϕ(q)a(q 2 ) = ϕ3 (q 3 ) + ϕ2 (q)ϕ(q 3 ) − ϕ(q)ϕ(−q)ϕ(−q 3 ). 2 2 1 1 (d) ϕ(q 3 )a(q 2 ) = − ϕ(q)ϕ2 (q 3 ) + ϕ3 (q) + ϕ(q 3 )ϕ(−q)ϕ(−q 3 ). 2 2 Proof. (a) By Theorem 2.4(a), (c), (g), (h) and (2.16), we have 2ϕ(q)ϕ(q 3 ) − ϕ(−q)ϕ(−q 3 ) = (1 + 4p + p2 )k = a(q). (b) By Theorem 2.4(a), (c), (g), (h) and (2.18), we have µ ¶ 1 1 1 ϕ(q)ϕ(q 3 ) + ϕ(−q)ϕ(−q 3 ) = 1 + p − p2 k = a(q 4 ). 2 2 2 (c) By Theorem 2.4(a), (c), (g), (h) and (2.17), we have 1 3 3 3 ϕ (q ) + ϕ2 (q)ϕ(q 3 ) − ϕ(q)ϕ(−q)ϕ(−q 3 ) 2 2 3 1 = (1 + 2p)3/4 k 3/2 + (1 + 2p)7/4 k 3/2 − (1 − p2 )(1 + 2p)3/4 k 3/2 2 2

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= (1 + 2p)3/4 (1 + p + p2 )k 3/2 = ϕ(q)a(q 2 ). (d) By Theorem 2.4(a), (c), (g), (h) and (2.17), we have 1 1 − ϕ(q)ϕ2 (q 3 ) + ϕ3 (q) + ϕ(q 3 )ϕ(−q)ϕ(−q 3 ) 2 2 1 1 = − (1 + 2p)5/4 k 3/2 + (1 + 2p)9/4 k 3/2 + (1 − p2 )(1 + 2p)1/4 k 3/2 2 2 = (1 + 2p)1/4 (1 + p + p2 )k 3/2 = ϕ(q 3 )a(q 2 ). Parts (a) and (b) are due to Borwein, Borwein and Garvan [4, Lemma 2.1(i), (a), (b), p. 36]. 3. Some Results of Petr A nonsquare integer ∆ satisfying ∆ ≡ 0 or 1 (mod 4) is called a discriminant. If ∆ is a discriminant the Legendre–Jacobi–Kronecker symbol corresponding to the ¡ ¢ discriminant ∆ is denoted by ∆ . We begin with a definition. ∗ Definition 3.1. For n ∈ N, we set X µ 12 ¶ X n µ 12 ¶ (a) A(n) := d = , n/d d d d|n d|n X µ −3 ¶ µ −4 ¶ X n µ −3 ¶ µ −4 ¶ (b) B(n) := d = , d n/d d n/d d d|n d|n X µ −3 ¶ µ −4 ¶ X n µ −3 ¶ µ −4 ¶ = , (c) C(n) := d n/d d d d n/d d|n d|n X µ 12 ¶ X n µ 12 ¶ (d) D(n) := d = . d d n/d d|n

d|n

The following result is easily proved so the proof is omitted. Theorem 3.1. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then (a) A(n) = 2α 3β A(N ), (b) B(n) = (−1)α+β 2α

µ

N 3

¶ A(N ),

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(c) C(n) = (−1)α+β+(N −1)/2 3β A(N ), µ ¶ µ ¶ N 3 (N −1)/2 (d) D(n) = (−1) A(N ) = A(N ). 3 N ¡ ¢ ¡3¢ ¡3¢ Here N3 and are Legendre–Jacobi symbols, which are related by = N N ¡ ¢ (N −1)/2 N (−1) by the Law of Quadratic Reciprocity. 3 We now turn to the results of Petr [17] that we need. Petr’s notation is as follows: Θ1 = 2

∞ X

q (2n+1)

2

/4

,

Θ1 (0, 3τ ) = 2

n=0

Θ2 = 1 + 2

∞ X

/4

,

(3.1)

n=0 ∞ X

2

(−1)n q n ,

Θ2 (0, 3τ ) = 1 + 2

n=1

Θ3 = 1 + 2

2

q 3(2n+1)

∞ X

∞ X

2

(−1)n q 3n ,

(3.2)

n=1 2

qn ,

Θ3 (0, 3τ ) = 1 + 2

n=1

∞ X

2

q 3n .

(3.3)

n=1

Petr [17, Eq. (23), p. 12] notes the following classical relation of Jacobi Θ3 Θ3 (0, 3τ ) = Θ1 Θ1 (0, 3τ ) + Θ2 Θ2 (0, 3τ )

(3.4)

and proves the identities   X µ 12 ¶ q n ,  (Θ23 + 3Θ23 (0, 3τ ))Θ2 Θ2 (0, 3τ ) = 4 − 4 d d n=1 ∞ X

(3.5)

d|n

 X µ −3 ¶ µ −4 ¶  q n , (Θ23 − Θ23 (0, 3τ ))Θ2 Θ2 (0, 3τ ) = 4 d n/d d n=1 ∞ X



(3.6)

d|n

(3Θ23 (0, 3τ )

−

Θ23 )Θ1 Θ1 (0, 3τ )

 X µ −3 ¶ µ −4 ¶  q n , =8 d d n/d n=1 ∞ X



(3.7)

d|n

 X µ 12 ¶  q n , (Θ23 (0, 3τ ) + Θ23 )Θ1 Θ1 (0, 3τ ) = 8 d n/d n=1 ∞ X



(3.8)

d|n

see [17, p. 15, 5th Eq. (30)], [17, p. 15, 6th Eq. (30)], [17, p. 15, 1st Eq. (30)], [17, p. 15, 2nd Eq. (30)], respectively. From (1.1), (3.2) and (3.3), we obtain Θ2 = ϕ(−q), Θ3 = ϕ(q),

Θ2 (0, 3τ ) = ϕ(−q 3 ), Θ3 (0, 3τ ) = ϕ(q 3 ).

(3.9) (3.10)

From (3.4), (3.9) and (3.10), we deduce Θ1 Θ1 (0, 3τ ) = ϕ(q)ϕ(q 3 ) − ϕ(−q)ϕ(−q 3 ).

(3.11)

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Using (3.9)–(3.11) in (3.5)–(3.8), and recalling Definition 3.1, we have the following result. Theorem 3.2. For |q| < 1, ∞ X

A(n)q n =

n=1

1 1 ϕ(q)ϕ3 (q 3 ) + ϕ3 (q)ϕ(q 3 ) 8 8 1 1 − ϕ2 (q)ϕ(−q)ϕ(−q 3 ) − ϕ2 (q 3 )ϕ(−q)ϕ(−q 3 ), 8 8

∞ X

B(n)q n =

n=1

1 3 ϕ(q)ϕ3 (q 3 ) − ϕ3 (q)ϕ(q 3 ) 8 8 1 3 + ϕ2 (q)ϕ(−q)ϕ(−q 3 ) − ϕ2 (q 3 )ϕ(−q)ϕ(−q 3 ), 8 8

∞ X

C(n)q n =

n=1

1 2 1 ϕ (q)ϕ(−q)ϕ(−q 3 ) − ϕ2 (q 3 )ϕ(−q)ϕ(−q 3 ), 4 4

∞ X

1 3 D(n)q n = 1 − ϕ2 (q)ϕ(−q)ϕ(−q 3 ) − ϕ2 (q 3 )ϕ(−q)ϕ(−q 3 ). 4 4 n=1 Solving for the quantities ϕ(q)ϕ3 (q 3 ), ϕ3 (q)ϕ(q 3 ), ϕ2 (q)ϕ(−q)ϕ(−q 3 ) and ϕ2 (q 3 )ϕ(−q)ϕ(−q 3 ) in Theorem 3.2, we obtain the following theorem. Theorem 3.3. For |q| < 1, (a) ϕ(q)ϕ3 (q 3 ) = 1 +

∞ X

(2A(n) + 2B(n) − C(n) − D(n))q n ,

n=1

(b) ϕ3 (q)ϕ(q 3 ) = 1 +

∞ X

(6A(n) − 2B(n) + 3C(n) − D(n))q n ,

n=1

(c) ϕ2 (q)ϕ(−q)ϕ(−q 3 ) = 1 +

∞ X

(3C(n) − D(n))q n ,

n=1

(d) ϕ2 (q 3 )ϕ(−q)ϕ(−q 3 ) = 1 −

∞ X

(C(n) + D(n))q n .

n=1

Replacing q by −q in (a)–(d) of Theorem 3.2, we obtain the expansions of ϕ(−q)ϕ3 (−q 3 ), ϕ3 (−q)ϕ(−q 3 ), ϕ(q)ϕ(q 3 )ϕ2 (−q) and ϕ(q)ϕ(q 3 )ϕ2 (−q 3 ) respectively in powers of q.

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4. The Form x2 + y 2 + z 2 + 3t2 By (1.3), (1.5), (1.6) and Theorem 3.3(b), we have 1+

∞ X

N (1, 1, 1, 3; n)q n =

n=1

∞ X

N (1, 1, 1, 3; n)q n

n=0

= ϕ3 (q)ϕ(q 3 ) = 1+

∞ X

(6A(n) − 2B(n) + 3C(n) − D(n))q n .

n=1 n

Equating the coefficients of q (n ∈ N), we obtain N (1, 1, 1, 3; n) = 6A(n) − 2B(n) + 3C(n) − D(n). Appealing to Theorem 3.1, we have the following result. Theorem 4.1. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then µ µ ¶¶ N α+1 α+β+ N 2−1 β+1 α+β N (1, 1, 1, 3; n) = (2 + (−1) ) 3 − (−1) A(N ). 3 This theorem was first stated but not proved by Liouville [6]. Proofs have been given by Benz [3, pp. 175–192], Demuth [5, pp. 243–245] and Petr [17, p. 16] and [18, p. 186]. 5. The Form x2 + y 2 + 2z 2 + 6t2 By (1.3), (1.5), (1.6), Theorem 2.6(a) and Theorem 3.3, we have 1+

∞ X

N (1, 1, 2, 6; n)q n = ϕ2 (q)ϕ(q 2 )ϕ(q 6 )

n=1

1 1 3 ϕ(q)ϕ3 (q 3 ) + ϕ3 (q)ϕ(q 3 ) − ϕ2 (q)ϕ(−q)ϕ(−q 3 ) 4 4 2 1 3 − ϕ(q)ϕ(q 3 )ϕ2 (−q) + ϕ(q)ϕ(q 3 )ϕ2 (−q 3 ) 4 4 ∞ X = 1+ (3A(n) + B(n)

=

n=1

3 1 − (1 + (−1)n )C(n) − (1 + (−1)n )D(n))q n . 2 2 Equating the coefficients of q n (n ∈ N), we have 3 1 N (1, 1, 2, 6; n) = 3A(n) + B(n) − (1 + (−1)n )C(n) − (1 + (−1)n )D(n). 2 2 Appealing to Theorem 3.1, we deduce the following result.

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Theorem 5.1. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then µ ¶ N −1 1 N (1, 1, 2, 6; n) = 2α − (1 + (−1)n )(−1)α+β+ 2 2 µ µ ¶¶ N × 3β+1 + (−1)α+β A(N ). 3 This theorem was first stated but not proved by Liouville [9]. The only proof of this result that the authors have found in the literature is due to Benz [3, pp. 175–192]. Our proof is considerably simpler and more direct than that of Benz. 6. The Form x2 + 2y 2 + 2z 2 + 3t2 By (1.3), (1.5), (1.6), (2.12) and Theorem 3.3(b), (c), we have 1+

∞ X

N (1, 2, 2, 3; n)q n = ϕ(q)ϕ2 (q 2 )ϕ(q 3 )

n=1

1 ϕ(q)(ϕ2 (q) + ϕ2 (−q))ϕ(q 3 ) 2 1 1 = ϕ3 (q)ϕ(q 3 ) + ϕ(q)ϕ(q 3 )ϕ2 (−q) 2 2 ∞ X 3 = 1+ (3A(n) − B(n) + (1 + (−1)n )C(n) 2 n=1 =

1 − (1 + (−1)n )D(n))q n . 2 Equating coefficients of q n (n ∈ N), we obtain 3 1 N (1, 2, 2, 3; n) = 3A(n) − B(n) + (1 + (−1)n )C(n) − (1 + (−1)n )D(n). 2 2 Appealing to Theorem 3.1, we deduce the following result. Theorem 6.1. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then ¶ µ 1 n α+β+ N 2−1 α N (1, 2, 2, 3; n) = 2 + (1 + (−1) )(−1) 2 µ µ ¶¶ N β+1 α+β × 3 − (−1) A(N ). 3 This result was stated but not proved by Liouville [10]. The authors have not found a proof of Theorem 6.1 in the literature.

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7. The Form x2 + 2y 2 + 4z 2 + 6t2 Making use of (1.3), (1.5), (1.6), Theorem 2.6(c) and Theorem 3.3, we obtain 1+

∞ X

N (1, 2, 4, 6; n)q n = ϕ(q)ϕ(q 2 )ϕ(q 4 )ϕ(q 6 )

n=1

3 1 1 ϕ(q)ϕ3 (q 3 ) + ϕ3 (q)ϕ(q 3 ) − ϕ2 (q)ϕ(−q)ϕ(−q 3 ) 8 8 8 1 3 2 3 − ϕ (q )ϕ(−q)ϕ(−q 3 ) − ϕ2 (q 4 )ϕ(−q 4 )ϕ(−q 12 ) 8 2 3 + ϕ2 (q 12 )ϕ(−q 4 )ϕ(−q 12 ) 2 ! Ã ∞ X 3 n 1+ (2A(n) + 2B(n) − C(n) − D(n))q = 8 n=1 Ã ! ∞ X 1 n + 1+ (6A(n) − 2B(n) + 3C(n) − D(n))q 8 n=1 Ã ! ∞ X 1 1+ − (3C(n) − D(n))q n 8 n=1 ! Ã ∞ X 3 n 1− (C(n) + D(n))q − 8 n=1 ! Ã ∞ X 1 4n − 1+ (3C(n) − D(n))q 2 n=1 Ã ! ∞ X 3 4n + 1− (C(n) + D(n))q 2 n=1 =

= 1+

∞ µ X 3 n=1

+

∞ X

¶ 1 A(n) + B(n) q n 2 2

(−3C(n) − D(n))q 4n .

n=1

Equating coefficients of q n (n ∈ N), we obtain  3    A(n) + 2 N (1, 2, 4, 6; n) =  3   A(n) + 2

1 B(n), if n 6≡ 0 2 ³n´ ³n´ 1 B(n) − 3C −D , if n ≡ 0 2 4 4

Appealing to Theorem 3.1, we deduce the following result.

(mod 4), (mod 4).

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Theorem 7.1. Let n ∈ N. Set n = 2α 3β N, where α, β gcd(N, 6) = 1. Then  µ µ ¶¶ N 1  β+1 β  3 + (−1) A(N ),    2 3   ¶¶ µ µ    N   3β+1 − (−1)β A(N ), 3 N (1, 2, 4, 6; n) = ³ ´  N −1   2α−1 − (−1)α+β+ 2     µ ¶¶ µ   N   A(N ), × 3β+1 + (−1)α+β  3

231

∈ N0 , N ∈ N and

if n ≡ 1

(mod 2),

if n ≡ 2

(mod 4),

if n ≡ 0

(mod 4).

This result was stated but not proved by Liouville [11]. The authors have not found a proof in the literature. 8. The Form x2 + 3y 2 + 3z 2 + 3t2 By (1.3), (1.5), (1.6) and Theorem 3.3(a), we have 1+

∞ X

N (1, 3, 3, 3; n)q n = ϕ(q)ϕ3 (q 3 )

n=1

= 1+

∞ X

(2A(n) + 2B(n) − C(n) − D(n))q n .

n=1 n

Equating the coefficients of q (n ∈ N), we have N (1, 3, 3, 3; n) = 2A(n) + 2B(n) − C(n) − D(n).

(8.1)

Appealing to Theorem 3.1, we deduce the following result. Theorem 8.1. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then µ ¶¶ ´µ ³ N −1 N 3β + (−1)α+β N (1, 3, 3, 3; n) = 2α+1 − (−1)α+β+ 2 A(N ). 3 A less compact form of Theorem 8.1 was given without proof by Liouville [14]. A proof of Theorem 8.1 have been given by Benz [3, pp. 175–192]. 9. The Form x2 + 3y 2 + 6z 2 + 6t2 By (1.3), (1.5), (1.6), (2.14) and Theorem 3.3(a)(d), we have 1+

∞ X

N (1, 3, 6, 6; n)q n = ϕ(q)ϕ(q 3 )ϕ2 (q 6 )

n=1

=

1 ϕ(q)ϕ(q 3 )(ϕ2 (q 3 ) + ϕ2 (−q 3 )) 2

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1 1 ϕ(q)ϕ3 (q 3 ) + ϕ(q)ϕ(q 3 )ϕ2 (−q 3 ) 2 2 ∞ µ X 1 = 1+ A(n) + B(n) − (1 + (−1)n )C(n) 2 n=1 ¶ 1 − (1 + (−1)n )D(n) q n . 2

=

Equating the coefficients of q n (n ∈ N), we obtain 1 1 N (1, 3, 6, 6; n) = A(n) + B(n) − (1 + (−1)n )C(n) − (1 + (−1)n )D(n). 2 2

(9.1)

Appealing to Theorem 3.1, we deduce the following result. Theorem 9.1. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then µ ¶ N −1 1 N (1, 3, 6, 6; n) = 2α − (1 + (−1)n )(−1)α+β+ 2 2 µ µ ¶¶ N A(N ). × 3β + (−1)α+β 3 A complicated result equivalent to Theorem 9.1 was stated but not proved by Liouville [12]. The authors have not found a proof of Theorem 9.1 in the literature. 10. The Form 2x2 + 3y 2 + 3z 2 + 6t2 By (1.3), (1.5), (1.6), Theorem 2.6(b) and Theorem 3.3, we have 1+

∞ X

N (2, 3, 3, 6; n)q n = ϕ(q 2 )ϕ2 (q 3 )ϕ(q 6 )

n=1

1 1 1 = − ϕ(q)ϕ3 (q 3 ) + ϕ3 (q)ϕ(q 3 ) + ϕ2 (q 3 )ϕ(−q)ϕ(−q 3 ) 4 4 2 1 2 1 + ϕ (−q)ϕ(q)ϕ(q 3 ) + ϕ2 (−q 3 )ϕ(q)ϕ(q 3 ) 4 4 ∞ µ X 1 = 1+ A(n) − B(n) + (1 + (−1)n )C(n) 2 n=1 ¶ 1 n − (1 + (−1) )D(n) q n . 2 Equating the coefficients of q n (n ∈ N), we obtain 1 1 N (2, 3, 3, 6; n) = A(n) − B(n) + (1 + (−1)n )C(n) − (1 + (−1)n )D(n). 2 2 Appealing to Theorem 3.1, we deduce the following result.

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Theorem 10.1. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then ¶ µ 1 n α+β+ N 2−1 α N (2, 3, 3, 6; n) = 2 + (1 + (−1) )(−1) 2 µ µ ¶¶ N × 3β − (−1)α+β A(N ). 3 This result was stated by Liouville in [13] without proof. A proof has been given by Benz [3, pp. 175–192]. 11. The Form x2 + y 2 + z 2 + zt + t2 By (1.4), (1.5), (1.7), Theorem 2.7(a) and Theorem 3.3(b), (c), we have 1+

∞ X n=1

M (1, 1; n)q n =

∞ X

M (1, 1; n)q n

n=0

= ϕ2 (q)a(q) = 2ϕ3 (q)ϕ(q 3 ) − ϕ2 (q)ϕ(−q)ϕ(−q 3 ) Ã ! ∞ X n = 2 1+ (6A(n) − 2B(n) + 3C(n) − D(n))q Ã

n=1

− 1+

∞ X

! (3C(n) − D(n))q

n

n=1

= 1+

∞ X

(12A(n) − 4B(n) + 3C(n) − D(n))q n .

n=1

Equating the coefficients of q n (n ∈ N), we obtain M (1, 1; n) = 12A(n) − 4B(n) + 3C(n) − D(n). Appealing to Theorem 3.1, we have the following result. Theorem 11.1. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then µ µ ¶¶ ¡ N −1 ¢ N A(N ). 3β+1 − (−1)α+β M (1, 1; n) = 2α+2 + (−1)α+β+ 2 3 This result was stated without proof by Liouville in [8]. A proof was given by Benz in [3, pp. 192–194]. Our proof is considerably shorter than that of Benz.

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12. The Form x2 + y 2 + 2(z 2 + zt + t2 ) By (1.4), (1.5), (1.7), Theorem 2.7(c) and Theorem 3.3(a)–(c), we obtain 1+

∞ X

M (1, 2; n)q n = ϕ2 (q)a(q 2 )

n=1

1 3 ϕ(q)ϕ3 (q 3 ) + ϕ3 (q)ϕ(q 3 ) − ϕ2 (q)ϕ(−q)ϕ(−q 3 ) 2 2 ∞ X = 1+ (6A(n) + 2B(n) − 3C(n) − D(n))q n .

=

n=1 n

Equating the coefficients of q (n ∈ N), we obtain M (1, 2; n) = 6A(n) + 2B(n) − 3C(n) − D(n). Appealing to Theorem 3.1, we deduce the following result. Theorem 12.1. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then µ µ ¶¶ ¢ β+1 ¡ α+1 N α+β α+β+ N 2−1 3 + (−1) A(N ). M (1, 2; n) = 2 − (−1) 3 This theorem was stated but not proved by Liouville in [7]. Benz considered the evaluation of M (1, 2; n) in [3, pp. 194–198]. 13. The Form x2 + y 2 + 4(z 2 + zt + t2 ) By (1.4), (1.5), (1.7), Theorem 2.7(b) and Theorem 3.3(b), (c), we obtain 1+

∞ X

M (1, 4; n)q n = ϕ2 (q)a(q 4 )

n=1

1 1 3 ϕ (q)ϕ(q 3 ) + ϕ2 (q)ϕ(−q)ϕ(−q 3 ) 2 2 ∞ X = 1+ (3A(n) − B(n) + 3C(n) − D(n))q n .

=

n=1 n

Equating the coefficients of q (n ∈ N), we deduce M (1, 4; n) = 3A(n) − B(n) + 3C(n) − D(n).

(13.1)

Appealing to Theorem 3.1, we obtain the following result. Theorem 13.1. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then µ µ ¶¶ ¡ α ¢ β+1 N α+β+ N 2−1 α+β M (1, 4; n) = 2 + (−1) 3 − (−1) A(N ). 3

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This result was not stated by Liouville and appears to be new. We observe that M (1, 4; n) = card{(x, y, z, t) ∈ Z4 | n = x2 + y 2 + z 2 + 3t2 , z ≡ t (mod 2)}. (13.2) Hence, by Theorems 4.1 and 13.1, we have the following result. Theorem 13.2. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then card{(x, y, z, t) ∈ Z4 | n = x2 + y 2 + z 2 + 3t2 , z 6≡ t(mod 2)} µ µ ¶¶ N α β α+β = 2 3 − (−1) A(N ). 3 14. The Form 3(x2 + y 2 ) + z 2 + zt + t2 By (1.4), (1.5), (1.7), Theorem 2.7(a) and Theorem 3.3(a), (d), we have 1+

∞ X n=1

M (3, 1; n)q n = ϕ2 (q 3 )a(q) = ϕ2 (q 3 )(2ϕ(q)ϕ(q 3 ) − ϕ(−q)ϕ(−q 3 )) = 2ϕ(q)ϕ3 (q 3 ) − ϕ2 (q 3 )ϕ(−q)ϕ(−q 3 ) = 1+

∞ X

(4A(n) + 4B(n) − C(n) − D(n))q n .

n=1

Equating the coefficients of q n (n ∈ N), we obtain M (3, 1; n) = 4A(n) + 4B(n) − C(n) − D(n). Appealing to Theorem 3.1, we have the following result. Theorem 14.1. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then µ ¶¶ ´µ ³ N −1 N 3β + (−1)α+β A(N ). M (3, 1; n) = 2α+2 − (−1)α+β+ 2 3 This result was stated without proof by Liouville in [16]. No proof seems to exist in the literature.

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15. The Form 3(x2 + y 2 ) + 2(z 2 + zt + t2 ) By (1.4), (1.5), (1.7), Theorem 2.7(d) and Theorem 3.3(a), (b), (d), we obtain 1+

∞ X

M (3, 2; n)q n = ϕ2 (q 3 )a(q 2 ) n=1 ¶ µ 1 3 1 2 3 3 3 3 = ϕ(q ) − ϕ(q)ϕ (q ) + ϕ (q) + ϕ(q )ϕ(−q)ϕ(−q ) 2 2 = 1+

∞ X

(2A(n) − 2B(n) + C(n) − D(n))q n .

n=1 n

Equating the coefficients of q (n ∈ N), we deduce M (3, 2; n) = 2A(n) − 2B(n) + C(n) − D(n). Appealing to Theorem 3.3, we have the following result. Theorem 15.1. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then µ ¶¶ ³ ´µ N α+1 α+β+ N 2−1 β α+β M (3, 2; n) = 2 + (−1) A(N ). 3 − (−1) 3 This result was stated without proof by Liouville in [15]. No proof seems to exist in the literature. 16. The Form 3(x2 + y 2 ) + 4(z 2 + zt + t2 ) By (1.4), (1.5), (1.7), Theorem 2.7(b) and Theorem 3.3(a), (d), we obtain 1+

∞ X

M (3, 4; n)q n = ϕ2 (q 3 )a(q 4 ) n=1 ¶ µ 1 1 ϕ(q)ϕ(q 3 ) + ϕ(−q)ϕ(−q 3 ) = ϕ2 (q 3 ) 2 2 1 1 ϕ(q)ϕ3 (q 3 ) + ϕ2 (q 3 )ϕ(−q)ϕ(−q 3 ) 2 2 Ã ! ∞ X 1 n 1+ (2A(n) + 2B(n) − C(n) − D(n))q = 2 n=1 Ã ! ∞ X 1 n 1− (C(n) + D(n))q + 2 n=1

=

= 1+

∞ X

(A(n) + B(n) − C(n) − D(n))q n .

n=1

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Equating the coefficients of q n (n ∈ N), we obtain M (3, 4; n) = A(n) + B(n) − C(n) − D(n). Appealing to Theorem 3.1, we deduce the following result. Theorem 16.1. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then µ ¶¶ ³ ´µ N −1 N 3β + (−1)α+β A(N ). M (3, 4; n) = 2α − (−1)α+β+ 2 3 This result was not stated by Liouville and appears to be new. We observe that M (3, 4; n) = card{(x, y, z, t) ∈ Z4 | n = x2 + 3y 2 + 3z 2 + 3t2 , z ≡ t (mod 2)}. From Theorems 8.1 and 16.1 we deduce our final result of this paper. Theorem 16.2. Let n ∈ N. Set n = 2α 3β N, where α, β ∈ N0 , N ∈ N and gcd(N, 6) = 1. Then card{(x, y, z, t) ∈ Z4 | n = x2 + 3y 2 + 3z 2 + 3t2 , z 6≡ t (mod 2)} µ µ ¶¶ N = 2α 3β + (−1)α+β A(N ). 3 Note added in proof. The authors recently noticed that the methods of their paper also apply to the form 2x2 + 3y 2 + 6z 2 + 12t2 . From (2.12) with q replaced by q 2 and Theorem 2.4(b), (d), we obtain ϕ(−q 2 ) in terms of p and k. Applying the duplication principle to ϕ(−q 2 ), we obtain ϕ(−q 4 ) in terms of p and k. Applying the triplication principle to ϕ(−q 4 ), we obtain ϕ(−q 12 ) in terms of p and k. Then, appealing to Theorem 2.4, we can verify the identity 1 1 ϕ(q 2 )ϕ(q 3 )ϕ(q 6 )ϕ(q 12 ) = − ϕ(q)ϕ3 (q 3 ) + ϕ3 (q)ϕ(q 3 ) 8 8 1 1 + ϕ(−q)ϕ2 (q 3 )ϕ(−q 3 ) − ϕ2 (q)ϕ(−q)ϕ(−q 3 ) 8 8 1 1 + ϕ2 (q 4 )ϕ(−q 4 )ϕ(−q 12 ) + ϕ2 (q 12 )ϕ(−q 4 )ϕ(−q 12 ). 2 2 From this identity and Theorem 3.3, we deduce 1 1    2 A(n) − 2 B(n),  N (2, 3, 6, 12; n) = µ ¶ µ ¶   1 1 n n   A(n) − B(n) + C −D , 2 2 4 4

if n 6≡ 0 (mod 4), if n ≡ 0 (mod 4).

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Finally, with n = 2α 3β N , we deduce from Theorem 3.1

N (2, 3, 6, 12; n) =

µ ¶¶  µ 1 β β N   3 − (−1) A(N ),   2 3     µ µ ¶¶    β β N   3 A(N ), + (−1)   3 µ ¶    α−1 α+β+ N 2−1  2 + (−1)       µ µ ¶¶    N   × 3β − (−1)α+β A(N ), 3

if n ≡ 1 (mod 2), if n ≡ 2 (mod 4),

if n ≡ 0 (mod 4).

Acknowledgment The fourth author was supported by the research grant A-7233 from the Natural Sciences and Engineering Research Council of Canada.

References [1] A. Alaca, S ¸ . Alaca and K. S. Williams, On the two-dimensional theta functions of the Borweins, Acta Arith. 124 (2006) 177–195. [2] A. ¸ . Alaca and PK. S. Williams, Evaluation of the convolution sums P Alaca, S 3l+4m=n σ(l)σ(m), Adv. Theoret. Appl. Math. 1 (2006) l+12m=n σ(l)σ(m) and 27–48. ¨ [3] E. Benz, Uber die Anzahl Darstellungen einer Zahlen n durch gewisse quatern¨ are quadratische Formen: Beweise, welche auf Identit¨ aten aus dem Gebiete der Thetafunktionen basieren, Dissertation, Z¨ urich, 1964, in Studien zur Theorie der quadratischen Formen, eds. B. L. van der Waerden and H. Gross (Stuttgart, 1968), pp. 165–198. [4] J. M. Borwein, P. B. Borwein and F. G. Garvan, Some cubic modular identities of Ramanujan, Trans. Amer. Math. Soc. 343 (1994) 35–47. [5] P. Demuth, Die Zahl der Darstellungen einer nat¨ urlichen Zahl durch spezielle quatern¨ are quadratishe Formen aufgrund der Siegelschen Massformel, in Studien zur Theorie der quadratischen Formen, eds. B. L. van der Waerden and H. Gross (Stuttgart, 1968), pp. 224–254. [6] J. Liouville, Sur la forme x2 + y 2 + z 2 + 3t2 , J. Math. Pures Appl. 8 (1863) 105–114. [7] J. Liouville, Sur la forme x2 + y 2 + 2z 2 + 2zt + 2t2 , J. Math. Pures Appl. 8 (1863) 115–119. [8] J. Liouville, Sur la forme x2 +y 2 +z 2 +zt+t2 , J. Math. Pures Appl. 8 (1863) 120–123. [9] J. Liouville, Sur la forme x2 + y 2 + 2z 2 + 6t2 , J. Math. Pures Appl. 8 (1863) 124–128. [10] J. Liouville, Sur la forme x2 +2y 2 +2z 2 +3t2 , J. Math. Pures Appl. 8 (1863) 129–133. [11] J. Liouville, Sur la forme x2 +2y 2 +4z 2 +6t2 , J. Math. Pures Appl. 8 (1863) 134–136. [12] J. Liouville, Sur la forme x2 +3y 2 +6z 2 +6t2 , J. Math. Pures Appl. 8 (1863) 209–213. [13] J. Liouville, Sur la forme 2x2 +3y 2 +3z 2 +6t2 , J. Math. Pures Appl. 8 (1863) 214–218. [14] J. Liouville, Sur la forme x2 + 3(y 2 + z 2 + t2 ), J. Math. Pures Appl. 8 (1863) 219–224. [15] J. Liouville, Sur la forme 2x2 + 2xy + 2y 2 + 3(z 2 + t2 ), J. Math. Pures Appl. 8 (1863) 225–226.

April 1, 2008

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[16] J. Liouville, Sur la forme x2 + xy + y 2 + 3(z 2 + t2 ), J. Math. Pures Appl. 8 (1863) 227–228. [17] K. Petr, O poˇctu tˇrid forem kvadratick´ ych z´ aporného diskriminantu, Rozpravy Ceské Akademie C´ısare Frantiska Josefa I 10 (1901) 1–22. ¨ [18] K. Petr, Uber die Klassenzahl der quadratischen Formen mit negativer Discriminante, Bull. Internat. de l’Acad. des Sciences de Bohème (Prague) 7 (1903) 180–187.