REPRESENTATIONS OF SQUARES BY CERTAIN SEPTENARY QUADRATIC FORMS

For any positive integer n and for certain fixed positive integers a1, a2, . . . , a7, we study the number of solutions in integers of a1x 2 1 + a2x 2 2 + a3x 2 3 + a4x 2 4 + a5x 2 5 + a6x 2 6 + a7x 2 7 = n . When a1 = a2 = · · · = a7 = 1, this reduces to the classical formula for the number of representations of a square as a sum of seven squares. A further eighteen analogous results will be given.


Introduction
Let n be a positive integer and let its prime factorization be given by n = p p λp .Let r k (n) denote the number of solutions in integers of Three classical results are given by Hurwitz (see [6] and [5]) and Sandham (see [9]), respectively: and where the values of the Legendre symbol are given by −1 p = 1 if p ≡ 1 (mod 4), −1 if p ≡ −1 (mod 4).
In recent work [3], the number of solutions in integers of was investigated for certain values of b and c.When b = c = 1, the number of solutions of (1.4) is given by (1.1).The number of solutions of (1.4) in the case b = 1, c = 2 is given by Theorem 1.1.[3, Theorem 1.2]The number of (x 1 , x 2 , x 3 ) ∈ Z 3 such that is given by ) and the values of the Legendre symbol are given by −8 p = 1 if p ≡ 1 or 3 (mod 8), −1 if p ≡ 5 or 7 (mod 8).
Further three-variable analogues of Theorem 1.1 were given in [3] and five-variable analogues were analyzed in [4].In this work we study seven-variable analogues of Theorem 1.1.That is, we investigate the number of solutions in integers of for certain fixed positive integers a 1 , a 2 , a 3 , a 4 , a 5 , a 6 and a 7 .The number of solutions in the case a 1 = a 2 = • • • = a 7 = 1 is given by Sandham's identity (1.3).This work is organized as follows.In Section 2 we define some notation and list all the results by grouping them into three theorems.The results in the first theorem are different from the others and they are treated in Section 3. Proofs of results in the second theorem are given in Section 4. Finally, results in the third theorem can be deduced from the results in the second theorem.A proof of one of them is given as an illustration in Section 5.

Proof of Theorem 2.1
In this section we will outline the proof of Theorem 2.1.Lemma 3.1.Fix an odd integer m.For any nonnegative integer k, let Then and Hence, on solving the recurrence relation, we have: for k ≥ 1 Proof.These can all be deduced by the methods in [3, Section 4].

It remains to determine the values of r
) in the case that m is odd.
Proposition 3.2.Let m be a positive odd number and let its prime factorization be given by m = p≥3 p λp .
To prove Proposition 3.2, we will need: Lemma 3.3.Let f 1 , f 2 and f 3 be defined by And let their series expansions be given by and Then for any nonnegative integer n and any prime p we have where χ is the completely multiplicative function defined on the positive integers by and −8 r is the Kronecker symbol, and a j (x) is defined to be zero if x is not an integer.We may note that

Proof
and The next result is due to Hurwitz.
Lemma 3.4.Suppose that a(n) is a function, defined for all non-negative integers n, that satisfies the property for all primes p, where χ is a completely multiplicative function.Then the coefficient of where µ is the Möbius function, A(n) is defined by and A(x) is defined to be 0 if x is not a non-negative integer.
Before starting the next lemma, let us define [q k ]f (q) to be the coefficient of q k in the Taylor expansion of f (q).Lemma 3.5.Let m be a positive odd number and let its prime factorization be given by m = p≥3 p λp .
We are now ready for Proof of Proposition 3.2.We will deal with (3.1) first.The proofs of (3.2) and (3.3) will be similar.By (3.4) and (3.8), By Lemmas 3.3 and 3.4 and (3.11) this is equivalent to where χ(r) is the completely multiplicative function defined by (3.7).Since χ(r) = 0 if r is even, the last sum in the above is over odd values of r only.Moreover, since m is odd, we may apply Lemma 3.5 to deduce that Similarly, we can deduce: and

Proof of Theorem 2.2
In this section, we will outline proofs of results in Theorem 2.2.The proof of (2.10) depends on: Lemma 4.1.Fix an odd integer m.For any nonnegative integer k let Hence, on solving the recurrence relation, we have Proof.These may be deduced by the methods in [3, Section 4].

INTEGERS: 13 (2013)
13 Now we will outline the proof of (2.11).This will be achieved in two steps according to whether n is even or odd.We begin with the case when n is even.Lemma 4.2.Fix an odd integer m.For any nonnegative integer k let ). Then Hence, on solving the recurrence relation, we have: for k ≥ 1 and thus where n is even.
Proof.These may be deduced by the methods in [ It remains to deal with the case when n is odd.We will need two lemmas.Lemma 4.3.Let g 1 , g 2 and g 3 be defined by ϕ 2 (q)ϕ 4 (−q), and g 3 (q) = qψ 2 (q 4 )ϕ 4 (−q 2 ); and let their series expansions be given by Then for any nonnegative integer n and any prime p we have where χ is the completely multiplicative function defined on the positive integers by and −1 r is the Kronecker symbol, and a j (x) is defined to be zero if x is not an integer.
Proof.By (4.2), (4.3) and (4.5), By Lemmas 3.4 and 4.3, and (4.6), this is equivalent to where χ(r) is the completely multiplicative function defined by (4.4).Since χ(r) = 0 if r is even, the last sum in the above is over odd values of r only.Moreover, since m is odd, we may apply Lemma 4.4 to deduce that

Proof of Theorem 2.3
In this section, we will give the proof of (2.12) and regard it as an illustration.Proofs of the remaining results are all similar: for n is even, the value can be deduced from the value of r (1,1,1,1,1,1,1) (n 2 ) and for n is odd, the value can be deduced from Theorem 2.2.Lemma 5.1.Fix an odd integer m.For any nonnegative integer k let ). Then Hence, on solving the recurrence relation, we have: for k ≥ 1 for the values of a, c a (λ 2 ), d a and e a given in the  (1, 1, 1, 1, 2, 2, 4) (1, 1, 1, 2, 2, 4, 4) Table 1: Data for (2.13)-(2.24)

Concluding Remarks
It is natural to ask if Theorem 2.1 can be extended by allowing some of the a j to be equal to 4, that is, to consider the case 1 = a 1 ≤ a 2 ≤ • • • ≤ a 7 = 4 for which the product a 1 a 2 • • • a 7 is an odd power of 2. For example, consider the case r (1,1,1,1,1,2,4) (n 2 ).The methods in [3,Section 4] can be used to find a formula in the case that n is even.For odd values of n, it would be necessary to study the sextenary form r (1,1,1,1,2,4) (n) and be able to express ϕ 4 (q)ϕ(q 2 )ϕ(q 4 ) as a linear combination of functions whose coefficients satisfy (3.6).Such a formula is not known.This could be the subject of further investigation.

8 )
Proof.Equation (4.7) may be deduced by the methods in [3, Section 4], and (4.8) follows from the result of [q 2m