Ayşe Alaca, Şaban Alaca and Kenneth S. Williams School of Mathematics and Statistics, Carleton University, Ottawa, Ontario, Canada. Abstract.

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1 Journal of Cobinatorics and Nuber Theory Volue 6, Nuber, ISSN: c Nova Science Publishers, Inc. DOUBLE GAUSS SUMS Ayşe Alaca, Şaban Alaca and Kenneth S. Willias School of Matheatics and Statistics, Carleton University, Ottawa, Ontario, Canada We evaluate the double Gauss su Ga, b, c; ; n : Abstract x,y0 e πiax +bxy+cy / n, where a, b, c are integers such that gcda, b, c 1 and b 4ac 0, is a rie, n is a ositive integer, and is an integer not divisible by. We aly the evaluation of this su to the deterination of the nuber of solutions of certain congruences of the for ax +bxy +cy +dz +ezt+ft k od n, where k is a nonzero integer. 010 Matheatics Subject Classification: 11L0, 11L05, 11T, 11E16, 11E5, 11D79 Keywords and hrases: Gauss sus, double Gauss sus, exonential sus, binary quadratic fors, sus of two binary quadratic fors, congruences 1. Introduction Let N denote the set of ositive integers and Z the set of all integers. Set N 0 N {0}. For a rie, n N and Z, the quadratic Gauss su G; n is defined by G; n : x0 e πix / n. 1.1 The value of this su is well-known and goes back to Gauss. If 0 od n then G; n x0 1 n. The research of the first and second authors was suorted by research grants fro the Natural Sciences and Engineering Research Council of Canada. E-ail address: aalaca@ath.carleton.ca E-ail address: salaca@ath.carleton.ca E-ail address: kwillia@connect.carleton.ca

2 18 Ayşe Alaca, Şaban Alaca and Kenneth S. Willias On the other hand if 0 od n then for soe k N 0 with k < n we have k so that x0 e πix / n x0 n k 1 e πi 1x / n k k x0 where 1 / k Z and 1. Hence in this case we have G; n k G 1 ; n k. e πi 1x / n k, Thus we ay suose that. For and we have ni G; n n/, 1. where is the Legendre sybol, see for exale [, Theore 1.5.,. 6]. For and we have G; n { 0 if n 1, n 1 + i n/ if n, 1. where { 1 if 1, 7 od 8, 1 if, 5 od 8, see for exale [, Theore 1.5.1, Proosition 1.5.,. 6]. We note that for odd we have 1 if 1 od 4, i or od 4, n 0 od, 1.4 i if od 4, n 1 od. In this aer we consider the exonential su fored by relacing x in 1.1 by a riitive binary quadratic for ax + bxy + cy Z[x, y] with a nonzero discriinant, that is, we consider the su where Ga, b, c;; n x,y0 e πiax +bxy+cy / n, od, gcda, b, c 1, b 4ac 0. We call such a su a double quadratic Gauss su. We first evaluate the su 1.5 when is an odd rie. We rove the following theore in Section, where it is used to deterine soe double Gauss sus exlicitly Corollary

3 Double Gauss Sus The underlying idea in the roof of Theore 1.1 is the deterination of integers r, s, t and u such that [ ] T [ ] [ ] r s a b/ r s t u b/ c t u [ ar + brt + ct ] ars + bru + st + ctu/ ars + bru + st + ctu/ as + bsu + cu is a diagonal atrix with integral entries and deterinant not divisible by in order to exress the double Gauss su Ga, b, c;; n as the roduct of two single Gauss sus. The roof is given in Section. Theore 1.1. Let be an odd rie. Let ax + bxy + cy Z[x, y] be such that and gcda, b, c b 4ac Let A be any integer with A 0 od reresented by the binary quadratic for ax + bxy + cy. For exale we can take a if a as a a 1 + b c 0, A c if a, c as c a 0 + b c 1, a + b + c if a, c as a + b + c a 1 + b c 1. If a and c the condition 1.6 ensures that b so a + b + c. Define l N 0 by and set h : l 4ac b 1.8 4ac b l Z \ {0} 1.9 so h. Let n N. Let Z be such that 0 od. Then A n n 1 Ga, b, c;; n n 1/ n if l n, l+1n A l h l+n n 1 l 1/ l if l n. We reark that when l n the two exressions for Ga, b, c;; n in Theore 1.1 agree as they should as n+1n A n h n n 1 n 1.10 A n n 1 n.

4 10 Ayşe Alaca, Şaban Alaca and Kenneth S. Willias We also note that the evaluation of Ga, b, c; ; n given in Theore 1.1 is as it should be indeendent of the choice of the integer A. This follows fro the theory of binary quadratic fors. We give a sile direct roof. It suffices to show that if A and A are two integers reresented by ax +bxy+cy with A 0 od and A 0 od then A A when l If l 0 the indeendence is clear as A l ±1 0 1, A l A ±1 0 l A 1, so l. We now rove As A is reresented by the binary quadratic for ax +bxy+cy there are integers u and v such that au + buv + cv A. Suose first that a. Then as 4ac b since l 1 we have au + bv au + bv + 4ac b v 4aau + buv + cv 4aA od 4aA 4aA A a so 0 or 1. But 4aA so 1. Thus. Siilarly A a A A and hence. Next suose that a and c. Then, by the arguent just given, we find that A c A. Finally suose that a and c. As gcda, b, c 1 we have b and so a + b + c. Let z u Z and w v u Z. Then a + b + cz + b + czw + cw au + buv + cv A. Hence, A is reresented by the binary quadratic for a + b + cx + b + cxy + cy and, by the revious arguent, we find A a + b + c A. This coletes the roof of We will aly Theore 1.1 to soe secific binary quadratic fors in Corollary.1. Regarding double Gauss sus 1.5 for, we begin by treating the case when b is even. We rove the following result in Section using the sae ethod of roof as for Theore 1.1.

5 Theore 1.. Let ax + bxy + cy Z[x, y] be such that and Double Gauss Sus 11 gcda, b, c 1, 1.1 b 4ac 0, 1.1 b 0 od Fro 1.1 and 1.14 we see that at least one of a and c is odd. As Ga, b, c;; n Gc, b, a;; n we ay suose without loss of generality that a is odd. Define l N 0 by and set h : l ac b/ 1.15 ac b/ l Z \ {0} 1.16 so h 0 od. Let n N. Let Z be such that 0 od. Then 0 if n 1, n 1 + i a n/ if n l, Ga, b, c;; n a 0 if n l + 1, n+l 1 + i al a 1 + i ah n+l/ if n l +. h Soe nuerical exales illustrating Theore 1. are given in Corollary.1. We now turn to the case when and b is odd. In this case we cannot use the ethod, which was used to rove Theores [ 1.1 and ] 1., to evaluate Ga, b, c;; n since r s there are no integers r, s, t and u with det 0 od such that t u [ r s t u ] T [ a b/ b/ c ][ ] r s t u is a diagonal atrix. This is clear as ru st 1 od ilies ars+bru+st+ctu is odd and thus nonzero. Hence we ust roceed differently. If a c 0 we have Ga, b, c;; n x0 y0 e πibxy/n x 0 n bx n x 0 n x n n. Thus we ay suose that a, c 0, 0. As Ga, b, c;; n Gc, b, a;; n we ay assue without loss of generality that a 0. Our aroach involves ultilying both the nuerator and the denoinator in the exonent of e in Ga, b, c;; n by a suitable ower f so that we can colete the square in f ax + bxy + cy odulo n+f. We rove the following result in Section.

6 1 Ayşe Alaca, Şaban Alaca and Kenneth S. Willias Theore 1.. Let ax + bxy + cy Z[x, y] be such that gcda, b, c 1, 1.17 and Note that 1.18 ensures that Let n N. Let Z be such that 0 od. Then b 1 od 1.18 a b 4ac Ga, b, c;; n 1 acn n. Soe nuerical exales illustrating Theore 1. are given in Corollary.. Let N na, b, c, d,e, f;k denote the nuber of solutions of the quadratic congruence ax + bxy + cy + dz + ezw + fw k od n, where a, b, c, d,e, f are integers, k a nonzero integer, a rie and n a ositive integer. We use the results given in Corollary.1, Corollary.1 and Corollary. to deterine exlicit forulae for N na, b, c,d,e,f; k for certain quaternary quadratic fors ax + bxy + cy + dz + ezw + fw when n α + 1, where α k. These forulae will be used in a forthcoing aer of the authors to deterine forulae for the nuber of reresentations of a ositive integer k by each of these fors using the local densities ethod described in [1]. We rove the following theores Theores in Section 4. Iortant Note. In Theores , k is a nonzero integer, is a rie, α is the largest ower of dividing k, K k/ α and n is a ositive integer with n α + 1. For convenience, we also define A,α by α+1 A,α n α α+1, which is clearly an integer. Theore 1.4. N n1, 0, 1,,,5;k + 1 n 1 n α if,, k n + n 1 if, α 0, 4 n 1 n α 1 if, α 1, n if, α 0, n α if, α 1.

7 Double Gauss Sus 1 Theore 1.5. N n1, 0,,,,;k A,α 15 if,, 5, K 5 n + 1 α 5 n α 1 if 5, 5 K n + n α 1 if, n n α 1 1 K 1 if. Theore 1.6. N n1, 0, 1, 5,, 5; k A,α 6 if,, K n + n α 1 if, 8 if, α 0, n 1, K 1 if, α 0, n, 1 4 n + + if, α 0, n, K K K n 1 if, α 1, n, n 1 if, α, n,,, 4,, 4, n α+1 n α 1 K if, α n, n 5, n 1 if, α n, n 1, n 5. Theore 1.7. N n1, 0, 5,,,;k Theore 1.8. N n1, 1, 1,,,1;k + 1 n 1 n α if, 5, n if, α 0, n+1 n α if, α 1, 6 5 n α 1 if 5. A,α 5 if, 5, K 1 n + if, α 0, 1 n 1 α 5 n α if, α 1, K 5 n + 5 n α 1 if 5. 5

8 14 Ayşe Alaca, Şaban Alaca and Kenneth S. Willias Theore 1.9. N n, 1,,,,;k A,α 5 if, 5, n K 1 if, α 0, 1 n 1 α 5 n α if, α 1, K 5 n + 5 n α 1 if 5. 5 Theore N n1, 0,,,,5;k A,α if,, n 1 if, α 0, K 5 n 1 1 α n α if, α 1, n K+1 α+ 1 n α 1 if. Theore N n1, 0,,,, 5; k A,α 7 if, 7, K 7 n + 1 α 7 n α 1 7 if 7, n if, α 0, n n α 1 K 1 if, α 1. Theore 1.1. N n1, 0,, 6, 4,6;k + 1 n 1 n α if, 1 n + n 1 K K if, α 0, n if, α 1, n n 1 1 K 1 if, α, n 1 if, α, n 1 if, α 4, n α+ if, α 5. Theore 1.1. N n1, 1, 4, 5,5,5;k A,α 5 if, 5, 1 n+1 1 α 5 n α 1 if, K 5 n + 5 n α if 5. 5

9 Double Gauss Sus 15 Theore N n, 1,,5,5,5;k A,α 5 if, 5, n α 5 α 1 if, K 5 n 5 n α if 5. 5 Theore N n,, 5,, 0,;k + 1 n 1 n α if,, n K 1 if, α 0, 4 n α if, α 1, n if, α 0, n+1 n α if, α 1. We conclude this introduction by recording a few easily-roved eleentary identities that we will use in this aer. Let n 1, n N. Let be an odd rie. Then 1 n 1 1 n 1/4 Taking n 1 n n N in 1.1 we deduce that Let k, l be odd integers. Then 1 /4 n 1 n. 1.1 n. 1. i k 1/ +l 1/ 1 k 1l 1/4 i kl 1/ 1. and { 1 + i k 1 + i l i if k l od 4, k if k l od Proof of Theore 1.1 In this section we rove Theore 1.1, which gives the evaluation of the double Gauss su Ga, b, c;; n when is an odd rie. Proof of Theore 1.1. Let A be any integer not divisible by which is reresented by the binary quadratic for ax + bxy + cy. Then there exist integers r and t such that A ar + brt + ct 0 od..1

10 16 Ayşe Alaca, Şaban Alaca and Kenneth S. Willias Define integers s and u by Then we have and s : br ct, u : ar + bt.. as + bsu + cu 4ac b ar + brt + ct l ha,. ars + bru + st + ctu 0,.4 ru st ar + brt + ct A 0 od..5 Let Z n denote the ring of residue classes odulo n. We write k for the residue class od n containing the integer k. The aing λ : Z n Z n Z n Z n given by λ x, ȳ rx + sy, tx + uy is well-defined. It is both injective and surjective in view of.5. Hence it is a bijection. As e πix/n is a eriodic function of x Z with eriod n, we have by. and.4 Ga, b, c; ; n x,y0 x,y0 x0 e πiarx+sy +brx+sytx+uy+ctx+uy / n e πiax + l hay / n n 1 e πiax / n y0 e πiahy / n l. If l n we obtain by 1. n 1 Ga, b, c;; n n x0 n A n A If l < n we obtain by 1., 1.1 and 1. n 1 Ga, b, c;; n l x0 n l 1 e πiax / n n e πiax / n n i /4 n/ n 1 / n. y0 e πiahy / n l n l l A i /4 n/ Ah i n l 1 /4 n l/ A l h i n+l /4+ n l 1 /4 n+l/

11 A Double Gauss Sus 17 l h n+l 1 n l 1/4 i n l 1 /4 n+l/ A l h l+1n A n+l nn l n i l 1 /4 l/ l h l+n n 1 l 1/ l. This coletes the roof in view of We now illustrate Theore 1.1 by alying it to soe secific binary quadratic fors ax + bxy + cy. Corollary.1. Let be an odd rie and let n N. Let Z satisfy 0 od. Then n if, i G1, 1, 1; ; n n n if ; 7 n 7 if 7, ii G1, 1, ;; n 7 7 n n if 7; n 11 if 11, iii G1, 1, ;; n 1 n+1 iv G1, 1, 4; ; n 1 n n n if 11; n if, 5 n 5 if 5, 5 n n if, 5; n if, 1 n v G, 1, ;; n 1 n n 5 5 if 5, vi G,, ; ; n 5 n n if 5; 5 n 5 if 5, n n if, 5;

12 18 Ayşe Alaca, Şaban Alaca and Kenneth S. Willias if, n 1, vii G,, 5;; n 1 n n+1 if, n, n n if ; viii G,, ;; n n n ; 7 n 7 7 if 7, ix G,, 5; ; n 4 n n if 7; x G4, 4, 5;; n n n ; 1 n n if, xi G5,, 5; ; n 6 n n if. Proof. Throughout this roof denotes an odd rie and is an integer not divisible by. i Here a, b, c 1, 1, 1 and 4ac b. The binary quadratic for x + xy + y reresents 1 so we ay take A 1. We have { l 1, h 1 if, l 0, h if. Theore 1.1 with gives G1, 1, 1;; n and for G1, 1, 1; ; n n n n n n n. ii Here a, b, c 1, 1, and 4ac b 7. We ay take A 1. We have { l 1, h 1 if 7, l 0, h 7 if 7. Theore 1.1 with 7 gives G1, 1, ;;7 n 7 7 n 7

13 Double Gauss Sus 19 and for 7 G1, 1, ; ; n n 7 n n 7 n n. iii The roof is siilar to that of i and ii. iv Here a, b, c 1, 1, 4 and 4ac b 15. We ay take A 1. We have Theore 1.1 gives l 1, h 5 if, l 1, h if 5, l 0, h 15 if, 5. 5 G1, 1, 4;; n n+1 n 1 n+1 G1, 1, 4;;5 n 5 5 5n+1 n 5 1 n+1 5 n 15 G1, 1, 4;; n n n n, 5 n 5, 5 n n if, 5. v Here a, b, c, 1, and 4ac b 15. We ay take A. We have Theore 1.1 gives l 1, h 5 if, l 1, h if 5, l 0, h 15 if, 5. 5 G, 1, ;; n n+1 n 1 n G, 1, ;;5 n 5 5 5n+1 n 5 1 n 5 n 15 G, 1, ;; n n n n, 5 n 5, 5 n n if, 5. vi Here a, b, c,, and 4ac b 0. We ay take A. We have { l 1, h 4 if 5, l 0, h 0 if 5. Theore 1.1 gives G,, ;;5 n 5 5 n 5 5 n 5, 5 G,, ;; n n0 n n 0 n n 5 n n if 5.

14 140 Ayşe Alaca, Şaban Alaca and Kenneth S. Willias vii Here a, b, c,, 5 and 4ac b 6. We ay take A. We have { l, h 4 if, With Theore 1.1 gives G,, 5;; n For Theore 1.1 gives l 0, h 6 if. G,, 5;; n if n 1, n n+1 1 n n+1 if n. n n. viii Here a, b, c,, and 4ac b. We ay take A 8. We have l 0 and h. Theore 1.1 gives G,, ;; n n n n ix Here a, b, c,, 5 and 4ac b 56. We ay take { 5 if, A if. We have { l 1, h 8 if 7, l 0, h 56 if 7. Theore 1.1 gives n 56 G,, 5;; n G,, 5;;7 n 7 G,, 5;; n n n n n. 4 n n, 7 8 7n+1 7 n 7 n 56 n n 7 n 7, 4 n n if, 7. x Here a, b, c 4, 4, 5 and 4ac b 64. We can take A 4. We have l 0 and h 64. Theore 1.1 gives G4, 4, 5;; n n n. xi Here a, b, c 5,, 5 and 4ac b 96. We can take A 8. We have { l 1, h if, l 0, h 96 if.

15 Theore 1.1 gives 8 G5,, 5; ; n G5,, 5; ; n Double Gauss Sus 141 n+1 n 96 n n n 1 n 6 n n if. n,. Proofs of Theores 1. and 1. We begin by roving Theore 1., which evaluates the double Gauss su Ga, b, c;; n when and b is even. Proof of Theore 1.. Define the integers r, s, t and u by Then r 1, s b, t 0, u a. ar + brt + ct a 1 od,.1 ars + bru + st + ctu 0,. as + bsu + cu a ac b l ah, 4. ru st a 1 od,.4 where we aealed to 1.16 for.. Let Z n denote the ring of residue classes odulo n. We write k for the residue class od n containing the integer k. The aing λ : Z n Z n Z n Z n given by λ x, ȳ rx + sy, tx + uy is well-defined. It is both injective and surjective in view of.4. Hence it is a bijection. As e πix/n is a eriodic function of x Z with eriod n, we have by.1-. Ga, b, c;; n x,y0 x,y0 x0 e πiarx+sy +brx+sytx+uy+ctx+uy / n e πiax + l ahy / n n 1 e πiax / n y0 If n 1 the first su in this roduct is 0 by 1. so we have Ga, b, c; ; n 0 if n 1. e πiahy / n l. Now suose n. If n l the second su in the roduct is n and we have by 1. Ga, b, c;; n n 1 + i a n/ n a

16 14 Ayşe Alaca, Şaban Alaca and Kenneth S. Willias so Ga, b, c;; n If n l + 1 the second su is n 1 + i a n/ if n l. a so l+1 1 y0 e πiahy / l 1 e πiahy / 0 y0 Finally if n l + we have by 1. Ga, b, c; ; n Ga, b, c; ; n 0 if n l + 1. a al h This coletes the roof of Theore 1.. n 1 + i a n/ l n l 1 + i ah n l/ ah n+l 1 + i a 1 + i ah n+l/. We now aly Theore 1. to the binary quadratic fors ax + bxy + cy considered in Corollary.1 having b even. Corollary.1. Let n N. Let be an odd integer. Then 0 if n 1, i G,, ;; n 1 1 n+1 n+1 i if n ; ii G,, 5; ; n iii G,, ;; n 0 if n 1, n+1 i if n ; 0 if n 1, 4, 8 1 i if n, 16 1 i if n, n+ i if n 5; 0 if n 1,, iv G,, 5;; n n+1 if n ;

17 Double Gauss Sus 14 0 if n 1, 5, i if n, v G4, 4, 5;; n i if n, if n 4, vi G5,, 5;; n i n+ i if n 6; 0 if n 1, 4, i if n, i if n, 1 n n+ if n 5. Proof. i Here we choose a, b, c,, so ac b/ 5. Thus l 0 and h 5. By Theore 1. we have 0 if n 1, G,, ;; n 1 + i 5n 1 + i 15 n if n. Now by 1.4 we have so 1 + i 1 + i 15 G,, ; ; n i 1 5n 1 + i 1 + i 15 n 1 n+1 n+1 1 for n. ii Here we choose a, b, c 5,, so ac b/ 9. Thus l 0 and h 9. By Theore 1. we have 0 if n 1, G,, 5;; n 1 + i 9n i 45 n if n. Now by 1.4 we have 1 + i i 45 i 5 1

18 144 Ayşe Alaca, Şaban Alaca and Kenneth S. Willias so G,, 5;; n n+1 1 for n. iii Here we choose a, b, c,, so ac b/ 8. Thus l and h 1. By Theore 1. we have 0 if n 1, 4, G,, ; ; n 1 + i if n, 1 + i 9/ if n, 1 + i n+/ if n 5, 0 if n 1, 4, if n, 16 if n, n+ if n 5. iv Here we choose a, b, c,, 5 so ac b/ 14. Thus l 1 and h 7. By Theore 1. we have G,, 5;; n 0 if n 1,, n i 1 + i 1 n+1/ if n, 7 0 if n 1,, n+/ if n, 0 if n 1,, n+1 if n. v Here we take a, b, c 5, 4, 4 so ac b/ 16. Thus l 4 and h 1. By Theore 1. we have 0 if n 1, 5, 81 + i if n, G4, 4, 5; ; n i if n, i if n 4, 1 + i n+ if n 6,

19 Double Gauss Sus if n 1, 5, if n, 16 + if n, if n 4, 1 n+ 1 if n 6. vi Here we take a, b, c 5,, 5 so ac b/ 4. Thus l and h. By Theore 1. we have 0 if n 1, 4, G5,, 5;; n 81 + i 5 if n, 1 + i 5 9/ if n, 5 n i i 15 n+/ if n 5, 0 if n 1, 4, if n, 16 + if n, 1 n n+ if n 5. Next we turn to the evaluation of the double Gauss su Ga, b, c;; n when and b is odd. We begin with a lea. Lea.1. Let n, k N. Let Z be odd. Let a Z. If n k + then and Proof. We have x 0 x 0 od k x 0 x a od k x 0 x 0 od k e πix / n n 1 + i n/ e πix / n 0 for a 0 od k. e πix / n n k 1 y0 e πiy / n k

20 146 Ayşe Alaca, Şaban Alaca and Kenneth S. Willias n k 1 k y0 n k by 1. as n k. Now let a Z be such that a 0 od k. We have Now x 0 x a od k x0 e πix / n 1 n 1 k e πiy / n k k 1 + i n k/ 1 + i n n/, x0 1 k 1 k t0 e πix n + πitx k As is odd there is an integer s such that Hence x0 s 1 od n. k 1 e πix / n t0 n 1 e πiat/k x0 e πix + n k tx n. e πitx a/k e πix n + πitx k. x0 e πix + n k tx/ n x0 x0 x0 y0 e πix + n k stx/ n e πix+n k 1 st n k s t / n e πix+n k 1 st / n as n k + e πiy / n n 1 + i n/

21 Double Gauss Sus 147 by 1.. Finally x 0 x a od k as a 0 od k. e πix / n n k i n/ k We now ake use of Lea.1 to rove Theore 1.. t0 e πiat/k 0, Proof of Theore 1.. Define α N 0 and a 1 Z with a 1 1 od by a α a 1. As a 1 b 1 od there exists an odd integer w such that so that Ga, b, c;; n b a 1 w od n x,y0 Case i: n α. Here, as n N, α 1. Fro.5 we obtain Ga, b, c;; n y0 y0 x0 z0 y 0 y 0 od n e πiα a 1 x +a 1 wxy+cy / n..5 e πiya 1wx+cy/ n e πiyz/n as a 1 w 1 od n n 1 acn n. Case ii: n α + 1. Here α 0. We have fro.5 Ga, b, c;; n y0 y0 y0 x,y0 α+1 1 x0 e πiα a 1 x +a 1 wxy+cy / α+1 1 x e πiya 1wx+cy/ α+1 as a 1 1od x 0 x 0 od x0 e πiya 1wx+cy/ α+1 e πiya 1wx+cy/ α+1

22 148 Ayşe Alaca, Şaban Alaca and Kenneth S. Willias Now, and y0 y0 z 0 z 0 od y0 α 1 x0 z0 y0 y0 e πiya 1wx+cy/ α+1 e πiyz/α+1 as a 1 w 1od z 0 z cy od e πiyz/α+1 z 0 z 0 od z 0 z y od α 1 t0 e πiyz/α+1 y 0 y 0 od α+1 α+1 e πiyz/α+1 α+1 if c 0 od, e πiyz/α+1 α+1 if c 1 od. e πiyt/α { α if y 0 od α, 0 if y 0 od α, z 0 z 1 od e πiyz/α+1 Thus if c 0 od we have z0 e πiyz/α+1 z 0 z 0 od { α+1 if y 0 od α+1 } 0 if y 0 od α+1 { α if y 0 od α } 0 if y 0 od α α if y 0 od α+1, α if y α od α+1, 0 if y 0 od α. e πiyz/α+1 Ga, b, c;; n y 0 y 0 od α α + α α+1 α α+1

23 if c 1 od and α 1 we have Ga, b, c;; n + Double Gauss Sus 149 y 0 y 0 od α+1 α+1 n 1 acn n ; y 0 y 1 od y 0 y 0 od α { α if y 0 od α α + 0 α+1 α + α α+1 α+1 n 1 acn n ; and if c 1od and α 0 so that n 1 we have 0 if y 0 od α } α if y 0 od α+1 α if y α od α+1 0 if y 0 od α Ga, b, c;; n acn n. Case iii: n α +. Here α 0. Aealing to.5 we obtain Ga, b, c; ; n that is x,y0 1 α+ 1 α+ 1 α+ e πiα a 1 x +a 1 wxy+cy / n n+α+ 1 y0 n+α+ 1 y0 n+α+ 1 y0 n+1 1 x0 n+1 1 x0 n+1 1 x0 e πiα a 1 x +a 1 wxy+cy / n e πiα+ a 1 x + α+ a 1 wxy+ α+ cy / n+α+ e πia 1 α+1 x+wy + α+ c a 1 w y / n+α+ 1 n+α+ 1 n+1 1 α+ e πiα+ c a 1 w y / n+α+ y0 Ga, b, c;; n x0 e πia 1 α+1 x+wy / n+α+,

24 150 Ayşe Alaca, Şaban Alaca and Kenneth S. Willias 1 n+α+ 1 α+ y0 e πiα+ c a 1 w y / n+α+ n+α+ 1 z 0 z wy od α+1 In this case n α + so that n + α + α Hence n+α+ 1 z 0 z wy od α+1 e πia 1z / n+α+ 0 for y 0 od α+1, by Lea.1. Thus, by Lea.1 again, we obtain Ga, b, c;; n 1 n+α+ 1 α+ y 0 y 0 od α+1 a 1 n+α 1 + i a 1 n α a 1 n+α e πia 1z / n+α+..6 e πiα+ c a 1 w y / n+α+ a 1 n+α+ 1 + i a 1 n+α+/ 1 + i a 1 n+α n+1 1 e πiα+ c a 1 w z / n α z0 n α 1 1 e πiα+ c a 1 w z / n α z0 a 1 n+α 1 + i a 1 n+α 1 α+ c a 1 w 1 + i α+ c a 1 w n α/. n α Now and α+ c a 1 w n α n if α 0, a 1 4c n+α if α 1, a 1 i α+ c a 1 w 1. i a1 a 1 Thus for α 0 and n we have Ga, b, c;; n n n/ 1 n 1 1 n/ a a a 4c a n n n n 1 acn n, aa 4c 1 + 4ac

25 and for α 1 and n α + we have Ga, b, c;; n Double Gauss Sus 151 n+α n+α/ 1 a 1 a n α/ a 1 n 1 acn n. This coletes the roof of Theore 1.. a 1 n+α We conclude this section by deterining Ga, b, c;; n for those binary quadratic fors in Corollary.1 with b odd. Corollary.. Let n N. Let be an odd integer. Then i G1, 1, 1;; n 1 n n, ii G1, 1, ;; n n, iii G1, 1, ; ; n 1 n n, iv G1, 1, 4;; n n, v G, 1, ;; n n. Proof. This follows iediately fro Theore Proofs of Theores We just rove Theore 1.4 as the reaining theores can be roved siilarly. We first observe that N na, b, c, d,e, f;k 1 n n + 1 n n + 1 n x,y,z,w0 1 n 1 x,y,z,w0 r0 n + 1 n 1 n r 1 n r0 M 1 M x,y,z,w0 0 e πiax +bxy+cy +dz +ezw+fw k/ n e πiax +bxy+cy +dz +ezw+fw k/ n n r 1 M 1 M e πi r Max +bxy+cy +dz +ezw+fw k/ n e πimk/n r x,y0 e πimax +bxy+cy / n r z,w0 e πimdz +ezw+fw / n r 4.1

26 15 Ayşe Alaca, Şaban Alaca and Kenneth S. Willias n + 1 n 1 n r0 n r 1 4r M 1 M Noting that k α K with K, 4.1 becoes e πimk/n r Ga, b, c; M; n r Gd, e, f;m; n r. N na, b, c, d,e, f;k 4. n + 1 n 1 n r0 n r 1 4r M 1 M e πimk/n r α Ga, b, c; M; n r Gd, e, f;m; n r. We need the following two forulas fro [1,. 1670, 1675]. n r 1 e πimk/n r α M 1 M n r n r 1 if r n α, α if r n α 1, 0 if r n α, 4. n r 1 M 1 M M e πimk/n r α 0 if r n α, K α if r n α 1, 0 if r n α. 4.4 First we rove the theore when,. By 4., 1. and Corollary.1vii, we obtain N n1, 0, 1,,, 5; k n + 1 n 1 n r0 n + n n n 1 r r 1 r0 By substituting 4. into 4.5, we obtain n r 1 4r M 1 M M n r i n r 1 e πimk/n r α n r 1 n r 1 M 1 M n r N n1, 0, 1,,,5;k + 1 n 1 n α, n r 1 n r e πimk/n r α 4.5 which is the asserted result. We now rove the theore when and α 0. Fro 4., 1. and Corollary.1vii we obtain N n1, 0, 1,,,5;k n + n M M1 e πimk/. 4.6

27 By 4.4 and 4.6 we obtain Double Gauss Sus 15 N n1, 0, 1,,,5;k n + n 1 K which is the asserted result. Now let and α 1. Fro 4., 1. and Corollary.1vii we obtain n N n1, 0, 1,,,5;k n + n+1 By 4. and 4.7 we obtain r0 n r 1 r M 1 M N n1, 0, 1,,,5;k 4 n 1 n 1 α,, e πimk/n r α. 4.7 which is the asserted result. We now rove the theore when and α 0. Fro 4., 1. and Corollary.1ii we obtain n N n1, 0, 1,,,5;k n n+ By 4. and 4.8 we obtain r0 N n1, 0, 1,,,5;k n, n r 1 r M 1 M e πimk/n r. 4.8 which is the asserted result. Now let and α 1. Fro 4., 1. and Corollary.1ii we obtain n N n1, 0, 1,,,5;k n n+ By 4. and 4.9 we obtain which is the asserted result. r0 n r 1 r M 1 M N n1, 0, 1,,,5;k n α, e πimk/n r α. 4.9 References [1] A. Alaca and K. S. Willias, On the quaternary fors x + y + z + t, x + y + z + 6t, x + y + z + 6t and x + y + 6z + 6t, Int. J. Nuber Theory 8 01, no. 7, [] B. C. Berndt, R. J. Evans and K. S. Willias, Gauss and Jacobi Sus, Canad. Math. Soc. Ser. of Monograhs and Adv. Texts, John Wiley & Sons, Inc., New York, 1998.