Binary quadratic forms and sums of triangular numbers

Transcription

1 Binary quadratic forms and sums of triangular numbers Zhi-Hong Sun( ) Huaiyin Normal University Notation: Z the set of integers, N the set of positive integers, [x] the greatest integer not exceeding x, ( m a ) the Legendre-Jacobi- Kronecker symbol, ord p m the nonnegative integer α such that p α m but p α+ m, (a, b) the greatest common divisor of a and b, (a, b, c) the form ax + bxy + cy, [a, b, c] the equivalence class containing the form (a, b, c), H(d) the form class group of discriminant d, h(d) the class number of discriminant d, R(K, n) the number of representations of n by the class K. My papers on the topic: [SW] Z.H. Sun, K.S. Williams, On the number of representations of n by ax + bxy + cy, Acta Arith. (006), 0-7.

2 [SW] Z.H. Sun and K.S. Williams, Ramanujan identities and Euler products for a type of Dirichlet series, Acta Arith. (006), [S] Z.H. Sun, The expansion of k= ( q ak )( q bk ), Acta Arith. 34(008), -9. [S] Z.H. Sun, On the number of representations of n by ax(x )/ + by(y )/, J. Number Theory 9(009), [S3] Z.H. Sun, Binary quadratic forms and sums of triangular numbers, Acta Arith. 46(0), [S4] Z.H. Sun, On the number of representations of n by ax +by(y )/, ax +by(3y )/ and ax(x )/ + by(3y )/, Acta Arith. 47(0), [S5] Z.H. Sun, Constructing x for primes p = ax + by, Advan. in Appl. Math. 48(0), 06-0.

3 . Basic concepts of binary quadratic forms A nonsquare integer d with d 0, (mod 4) is called a discriminant. For a, b, c Z, ax + bxy + cy = (a, b, c) is called a binary quadratic form with discriminant d = b 4ac. If gcd(a, b, c) =, we say that the form (a, b, c) = ax +bxy+ cy is primitive. Suppose a, b, c Z, a, c > 0 and d = b 4ac < 0. If a < b a < c or 0 b a = c, we say that (a, b, c) is a reduced form. The number of primitive reduced forms with discriminant d < 0 is called the class number of discriminant d. We denote it by h(d). Let d < 0 be a discriminant. It is well known that h(d) = d = 3, 4, 7, 8,,, 6, 9, 7, 8, 43, 67, 63 (3 numbers), 3

4 and h(d) = d = 5, 0, 4, 3, 35, 36, 40, 48, 5, 5, 60, 64, 7, 75, 88, 9, 99, 00,, 5, 3, 47, 48, 87, 3, 35, 67, 403, 47. (9 numbers) Let p be an odd prime. By the theory of binary quadratic forms, there are unique positive integers x and y such that p = 4k + = x + 4y, p = 8k +, 3 = x + y, p = 3k + = x + 3y = (L + 7M )/4, p = 0k +, 9 = x + 5y, p = 4k +, 7 = x + 6y, p = 7k +,, 4 = x + 7y, p = 40k +, 9,, 9 = x + 0y, p = 30k +, 9 = x + 5y. Let ax +bxy +cy be a primitive reduced form with h(b 4ac) = or. By genus theory, those primes p represented by ax + bxy + cy can be characterized by congruence conditions. 4

5 Two forms (a, b, c) and (a, b, c ) are equivalent ((a, b, c) (a, b, c )) if and only if there exist integers α, β, γ and δ with αδ βγ = such that the substitution x = αx + βy, y = γx + δy transforms (a, b, c) to (a, b, c ). The substitutions x = Y, y = X and x = X + ky, y = Y imply (a, b, c) (c, b, a) (a, ak + b, ak + bk + c). We denote the equivalence class of (a, b, c) by [a, b, c]. If (a, b, c) (a, b, c ), it is easy to see that gcd(a, b, c) = gcd(a, b, c ). Thus, (a, b, c) is primitive if and only if (a, b, c ) is primitive. It is also known that equivalent forms have the same discriminant. The equivalence classes of primitive, integral, binary quadratic forms of discriminant d form a finite abelian group under Gaussian composition, called the form class group. We denote this group by H(d) and its order is h(d). Lagrange: Every (primitive) form of discriminant d < 0 is equivalent to exactly one (primitive) reduced form of discriminant d. 5

6 Lagrange: If (a, b, c) (a, b, c ), then n is represented by (a, b, c) if and only if n is represented by (a, b, c ). Gauss: If (a, b, c) (a, b, c ), then {(x, y) Z : n = ax + bxy + cy } = {(x, y) Z : n = a x + b xy + c y }. For a, b, c Z with a, c > 0 and b 4ac < 0 we define R([a, b, c], n) = {(x, y) Z : n = ax + bxy + cy }, R ([a, b, c], n) = {(x, y) Z : n = ax + bxy + cy, (x, y) = }. Since [a, b, c] = [a, b, c], for K H(d) we have R(K, n) = R(K, n). If R(K, n) > 0, we say that n is represented by K. Theorem.. Let d be a discriminant, n N and a, b, c Z with b 4ac = d. Then R([a, b, c], n) = R ( n ) [a, b, c], m m N,m n 6

7 and R ([a, b, c], n) = m N,m n µ(m)r ( [a, b, c], where µ(n) is the Möbius function. n ), m. Formulas for N(n, d) For a discriminant d the conductor of d is the largest positive integer f = f(d) such that d/f 0, (mod 4). If f(d) =, we say that d is a fundamental discriminant. For discriminant d < 0 let w(d) = if d < 4, 4 if d = 4, 6 if d = 3. For n N and discriminant d < 0 define N(n, d) = K H(d) R(K, n). 7

8 Let d < 0 be a discriminant with conductor f. Let d 0 = d/f and n N. When (n, d) =, Dirichlet proved the following formula for N(n, d): (.) N(n, d) = w(d) k n ( d0 In 997 Kaplan and Williams showed that this is also true under the weaker condition (n, f) =. Taking n = in (.) we find N(, d) = w(d). k ). Sun and Williams obtained the complete formula for N(n, d), which improved the Huard- Kaplan-Williams formula. 8

9 Theorem. ([SW, 006]). Let d be a discriminant with conductor f. Let d 0 = d/f and n N. If (n, f ) is not a square, then N(n, d) = 0. If (n, f ) = m for m N, then N(n, d) w(d) = m p m = ( d 0 p )= m p m ( p ( d/m p )) + ( ) ord pn ( p ( d 0 p )= ( d/m )) p ( n + ord p m k n m ( ) d0 where in the products p runs over all distinct primes. ), k 9

10 When h(d) =, N(n, d) = R([, 0, d ], n) if d, 4 R([,, d ], n) if d. 4 Example.: d = 8, f =, d 0 = 7, h(d) =, R([, 0, 7], n) = N(n, 8) 0 if 4 n, ( 7 ) if n, = k k n ( 7 ) if 4 n. k k n 4 If for any n, n N with (n, n ) =, we have f(n n ) = f(n )f(n ), we say that f(n) is a multiplicative function. Theorem. ([SW, 006]). Let d be a discriminant. Then N(n, d)/w(d) is a multiplicative function of n N. 0

11 Theorem.3 ([SW]). Let d be a discriminant with conductor f and d 0 = d/f. Let s be a complex number with Re(s) >. Then the Dirichlet series N(n,d)/w(d) n= n s converges absolutely and n= = p f N(n, d)/w(d) p f n s ( p α p ( s) p s + pαp( s) ( p ( d 0 p )) ( p s )( ( d 0 p )p s ) ( p s )( ( d 0 p )p s ), where p runs over all primes and α p = ord p f. )

12 Example.: n= R([, 0, 7], n) n s = s + s ( s ) p,9, (mod 4) p 3,5,3 (mod 4) 7 s ( p s ) p s, where p runs over all primes satisfying the condition under the product. 3. Formulas for R(K, p t ) and R (K, p t ) Theorem 3. ([SW]. Let d be a discriminant with conductor f, and let p be a prime such that p f. Let t be a nonnegative integer and K H(d). Let I be the identity in H(d).

13 (i) If ( d p ) =, then R(K, p t ) = { w(d) if t and K = I, 0 otherwise.. (ii) If p d, then R(K, p t ) w(d) if t and K = I, = or if t and p is represented by K, 0 otherwise. (iii) Suppose ( d p ) = so that p is only represented by some class A and the inverse A in H(d). Let m be the order of A in H(d). If K is not a power of A, then R(K, p t ) = 0. If k, t 0 {0,,..., m } with t 0 t (mod m), then 3

14 R(A k, p t ) 0 if m and k t, w(d) ([ t ] ) + if t = 0 S k,m, m/(, m) w(d) [ t ] otherwise, m/(, m) where S k,m = {r k r < m, k r} {r m k r < m, k r} if m, {r min{k, m k} r < m, k r} {r max{k, m k} r < m, k r} if m. Theorem 3. ([SW]). Let d be a discriminant with conductor f. Let K H(d) and t N. Let p be a prime such that p f. (i) If ( d p ) =, then R (K, p t ) = 0. 4

15 (ii) If ( d p ) = 0, then p is represented by unique A H(d) and we have R (K, p t ) = { w(d) if t = and K = A, 0 otherwise. (iii) If ( d p ) =, then p is represented by some A H(d) and we have R (K, p t ) 0 if K A t, A t, = w(d) if K {A t, A t } and A t A t, w(d) if K = A t = A t. Theorem 3.3 ([SW]) Let d be a discriminant with conductor f. Let p be a prime such that ( d p ) = 0, and p f. Then p is represented by some class A H(d). For t N and K H(d) we have R(K, p t+ )+R(K, p t ) = R(AK, p t )+R(A K, p t ). 5

16 4. Multiplicative functions involving R(K, n) Theorem 4. ([SW]). Let d be a discriminant. If n, n,..., n r (r ) are pairwise prime positive integers and K H(d), then R(K, n n n r ) = w(d) r and K K K r =K R(K, n ) R(K r, n r ) R (K, n n n r ) = w(d) r K K K r =K R (K, n ) R (K r, n r ), where the summations are taken over all K,..., K r H(d) such that K K K r = K. Example 4. If (n, n ) = and K H(d), then R(K, n n ) = R(A, n )R(A K, n ). w(d) A H(d) 6

17 Definition 4.. Let d be a discriminant and n N. Let H(d) = {A k Ak r r 0 k < h,..., 0 k r < h r } with h h r = h(d). For K = A k Ak r r H(d) and M = A m Am r r H(d) we define and F (M, n) = w(d) [K, M] = k m h K H(d) + + k rm r h r cos π[k, M] R(K, n). Theorem 4.. n N. Let d be a discriminant and (i) If M H(d), then F (M, n) is a multiplicative function of n. (ii) If K H(d), then R(K, n) = w(d) h(d) M H(d) cos π[k, M] F (M, n). 7

18 Theorem 4.3 ([SW]). Let d be a discriminant such that H(d) is cyclic and h(d) 6 (h(d) {, 3, 5, 6} implies H(d) is cyclic). Let I be the principal class in H(d). Let A be a generator of H(d) and n N. Recall that w(d) = or according as d > 0 or d < 0. (i) If h(d) =, 3, then F (A, n) = (R(I, n) R(A, n))/w(d) is a multiplicative function of n. (ii) If h(d) = 4, then F (A, n) = (R(I, n) R(A, n))/w(d), F (A, n) = (R(I, n) + R(A, n) R(A, n))/w(d) are multiplicative functions of n. 8

19 (iii) If h(d) = 5, then F (A, n) = (R(I, n) + R(A, n) R(A, n)) / w(d), F (A, n) = (R(I, n) R(A, n) + R(A, n)) / w(d) are multiplicative functions of n. (iv) If h(d) = 6, then F (A, n) = (R(I, n) + R(A, n) R(A, n) R(A 3, n))/w(d), F (A, n) = (R(I, n) R(A, n) R(A, n) + R(A 3, n))/w(d), F (A 3, n) = (R(I, n) R(A, n) + R(A, n) R(A 3, n))/w(d) are multiplicative functions of n. 9

20 5. Formulas for F (A, p t ) when h(d) =, 3, 4 For K H(d) define R(K) = {n N : R(K, n) > 0}. Thus, n R(K) if and only if R(K, n) > 0. Theorem 5. Let d be a discriminant with conductor f and d 0 = d/f. Suppose h(d) = and H(d) = {I, A} with A = I. For n N let F (A, n) = (R(I, n) R(A, n)). Let p be a w(d) prime and let t be a nonnegative integer. If p f, then ( + ( )t ) if ( d 0 p ) =, if p d 0 and p R(I), F (A, p t ) = ( ) t if p d 0 and p R(A), t + ( ) t (t + ) if p d 0 and p R(I), if p d 0 and p R(A). Theorem 5. Let d be a discriminant with conductor f and d 0 = d/f. Suppose h(d) = 3 and H(d) = {I, A, A } with A 3 = I. For n N let F (A, n) = (R(I, n) R(A, n)). Let p be w(d) a prime and let t be a nonnegative integer. 0

21 If p f, then F (A, p t ) = if p d 0, ( + ( )t ) if ( d 0 p ) =, t + if p d 0 and p R(I), if p R(A) and 3 t, 0 if p R(A) and 3 t, if p R(A) and 3 t. Theorem 5.3 Let d be a discriminant with conductor f and d 0 = d/f. Suppose h(d) = 4 and H(d) = {I, A, A, A 3 } with A 4 = I. Let F (A, n) = w(d) (R(I, n) R(A, n)) and F (A, n) = w(d) (R(I, n)+r(a, n) R(A, n)) for n N. Let p be a prime such that p f and let t be a nonnegative integer. Then

22 F (A, p t ) = and + ( ) t if ( d 0 p ) =, if p d 0 and p R(I), t + if p d 0 and p R(I), ( ) t if p d 0 and p R(A ), ( ) t (t + ) if p d 0 and p R(A ), ( ) t/ if p R(A) and t, 0 if p R(A) and t F (A, p t ) + ( ) t if ( d 0 p ) =, = if p d 0, t + if p d 0 and p R(I) R(A ), ( ) t (t + ) if p R(A).

23 6. The method of determining R(K, n) If n = p t p t r r is the standard factorization of n, then for K H(d), F (K, n) = F (K, p t ) F (K, p t r r ). Hence, by Theorems we can determine R(K, n) for K H(d), where H(d) is cyclic and h(d) =, 3, 4. Let d be a discriminant. If H(d) = {I, A}, then { R(I, n) + R(A, n) = N(n, d), and so R(I, n) R(A, n) = w(d)f (A, n) R(I, n) = (N(n, d) + w(d)f (A, n)), R(A, n) = (N(n, d) w(d)f (A, n)). If H(d) = {I, A, A }, then R(A, n) = R(A, n) = R(A, n). We have { R(I, n) + R(A, n) = N(n, d), R(I, n) R(A, n) = w(d)f (A, n) 3

24 and so R(I, n) = (N(n, d) + w(d)f (A, n)), 3 R(A, n) = (N(n, d) w(d)f (A, n)). 3 Similarly, if H(d) = {I, A, A, A 3 }, then R(I, n) = (F (I, n) + F (A, n) + F (A, n))w(d)/4, R(A, n) = R(A 3, n) = (F (I, n) F (A, n))w(d)/4, R(A, n) = (F (I, n) F (A, n) + F (A, n))w(d)/4, where F (I, n) = N(n, d)/w(d). Let d be a discriminant and H (d) = {K : K H(d)}. Then H (d) = H(d) = C C, where C is a cyclic group of order. 4

25 Theorem 6. ([S3]). Let d < 0 be a discriminant with conductor f, d 0 = d/f and H (d) =. Let K H(d) and n N with R(K, n) > 0. Then (n, f ) = m for some m N and R(K, n) = w ( d ) n m ( + ord p ( d m ). 0 p )= related reference: N. A. Hall, The number of representations function for binary quadratic forms, Amer. J. Math. 6 (940), Euler called a positive integer n a convenient number if it satisfies the following criterion: Let m be an odd number such that (m, n) = and m = x + ny with (x, y) =. If the equation m = x + ny has only one solution with x, y 0, then m is a prime. 5

26 Euler listed 65 convenient numbers as follows:,, 3, 4, 7 h( 4n) =, 5, 6, 8, 9, 0,, 3, 5, 6, 8,, 5, 8, 37, 58 h( 4n) =,, 4, 30, 33, 40, 4, 45, 48, 57, 60, 70, 7, 78, 85, 88, 93, 0,, 30, 33, 77, 90, 3, 53 h( 4n) = 4, 05, 0, 65, 68, 0, 40, 73, 80, 3, 330, 345, 357, 385, 408, 46, 50, 760 h( 4n) = 8, 840, 30, 365, 848 h( 4n) = 6. He was interested in convenient numbers because they helped him find large primes. Gauss observed that a positive integer n is a convenient number if and only if H ( 4n) =. In 973 it was known that Euler s list is complete except for possibly one more n. 6

27 7. Formulas for t n (a, b) triangular numbers: x(x ) = ( ) x (x Z) For a, b, n N let t n (a, b) = { x, y : n = ax(x )/ + by(y )/, x, y N}. For our convenience we also define t 0 (a, b) = and t n (a, b) = 0 for n N. Let ψ(q) = Then clearly (7.) k= ψ(q a )ψ(q b ) = + q k(k )/ ( q < ). n= t n (a, b)q n ( q < ). Ramanujan conjectured and Berndt proved that (7.) qψ(q)ψ(q 7 ( ) 8 q n ) = ( q < ), n qn n= where ( m k ) is the Legendre-Jacobi-Kronecker symbol. 7

28 According to Berndt, (7.) is of extreme interest, and it would appear to be very difficult to prove it without the addition theorem for elliptic integrals. By (7.), (7.) is equivalent to (7.3) t n (, 7) = k n+, k ( k K.S. Williams and Z.H. Sun proved (7.3) and so (7.) by using the theory of binary quadratic forms. 7 ). Theorem 7. ([S, 009]) Let a, b, n N. Then 4t n (a, b) = R([a, a, a + b a + b ], n ) R([a, 0, b], n + a + b ) if 4 a + b, 4 R([a, a, a + b a + b ], 4n + ) if 4 a + b, R([4a, 4a, a + b], 8n + a + b) if a + b. 8

29 Theorem 7. ([S3,0]. Let a, b, n N. (i) If 8 a, 8 b and 4 (a + b), then t n (a, b) = R([a, 0, b], 8n + a + b). 4 (ii) If a, 8 b 4 and 4 (a + b/4), then t n (a, b) = R([a, 0, b/4], 8n + a + b). 4 Theorem 7.3 ([SW, 006]). Let n N. If n + = 3 α n 0 (3 n 0 ), then t n (3, 5) = + ( )α ( n 0 3 ) If n + = 3 α n 0 (3 n 0 ), then t n (, 5) = ( )α ( n 0 3 ) k n+, k k n+, k ( k 5 ( k 5 ). ). 9

30 Theorem 7.4 ([S]) For n N and b {5, 3, 37}, t n (, b) = k 4n+ b+ ( b k ). Theorem 7.5 ([S3]). {6, 0,,, 8, 58}. Let n N and b (i) If b {6, 0,, 58}, then t n (, b) = t n (, b/) = k 8n+b+ ( b k 8n++ b ), k ( b k ). (ii) If b {, 8}, then t n (, b) = t n (4, b/4) = k 8n+b+ ( k k 8n+4+ b 4 b/4 ), ( k b/4 ). 30

31 Theorem 7.6. Let n N. If 4n + 3 = 5 α n (5 n ), then t n (, 45) = k n 9 ( ) k ( k 5 ) if 9 n and n ± (mod 5), ( ) k ( k 5 ) k n if 3 n and n ± (mod 5), 0 otherwise. Theorem 7.7. Let n N, a {, 3}, b {7,, 9, 3, 59} and 4n+(a+3b/a)/ = 3 β n 0 (3 n 0 ). Then t n (a, 3b/a) > 0 if and only if ord q n 0 for every prime q with ( 3b q ) = and n 0 (mod 3) if b {, 59} and β a+ (mod ), (mod 3) otherwise. Moreover, if t n (a, 3b/a) > 0, then t n (a, 3b/a) = ( 3b p )=( + ord pn 0 ). 3

32 In [S,S3], t n (a, b) was determined for 38 values of (a, b). related references: A. Berkovich and H. Yesilyurt, Ramanujan s identities and representation of integers by certain binary and quaternary quadratic forms, Ramanujan J. 0 (009), M.D. Hirschhorn, The number of representations of a number by various forms, Discrete Math. 98(005),

33 8. The expansion of k= ( q ak )( q bk ) k= = + Theorem 8. ([S, 008]). Let a, b N and q R with q <. Then ( q ak )( q bk ) n= ( R(a + b, (a b), 36(a + b); 4n + a + b) R(4(a + b), (a b), 9(a + b); 4n + a + b) ) q n. In the case a + b = 4, the result was given in [SW]. 9. Constructing x and x for primes p = ax + by 33

34 . (Gauss, 85; Cauchy proved) If p (mod 4) is a prime and p = x + y with x (mod 4), then x ( (p )/) (mod p). (p )/4 (Chowla, Dwork and Evans, 986) ( (p )/) p ( + x p ) (p )/4 x (mod p ). (Cosgrave, Dilcher, 00) ( (p )/ (p )/4 ( ) ) x p x p ( 8x 3 + pq p() 8 p (q p () E p 3 ) ) (mod p 3 ), where q p () = ( p )/p and E n is the Euler number. 34

35 . (Jacobi, 87) If p (mod 3) is a prime and 4p = L + 7M with L (mod 3), then L ( (p )/3) ( (p )/) (mod p). (p )/3 (p )/6 Gauss: L ( p 3 )! 3 (mod p). (Evans, 985, unpublished; Yeung, 989, JNT) ( (p )/3 (p )/3 ) L + p L (mod p ). (Cosgrave, Dilcher, 00) ( (p )/3 (p )/3 ( L + ) pl p + )( L p B p ( 3 ) ) (mod p 3 ). Z.H.Sun: Let {U n } be given by U 0 = and U n = ( ) n n k=0 k Uk (n ). Then B p ( 3 ) 6U p 3 (mod p). 35

36 3. (Stern, 846) If p (mod 8) is a prime and so p = x + y with x (mod 4), then x ( ) p 8 ( p ) p (mod p) (Eisenstein, 848) If p 3 (mod 8) is a prime and so p = x + y with x (mod 4), then x ( ) p 3 8 ( p ) p 3 (mod p) (Cauchy) If p (mod 0) is a prime and so p = x + 5y, then 4x (p )( p ) p 3(p ) (mod p)

37 6. (Jacobi, 87) If p (mod 7) is a prime and so p = x + 7y with x (mod 7), then x (3(p ) ) 7 p (mod p) (Eisenstein, 848) If p (mod 7) is an odd prime and so p = x + 7y with x 3 (mod 7), then x (3(p ) ) 7 p (mod p) (Eisenstein, 848) If p 4 (mod 7) is a prime and so p = x + 7y with x 5 (mod 7), then x (3(p 4) ) 7 p 4 (mod p). 7 37

38 9. Let p, 3, 4, 5, 9 (mod ) be a prime and so 4p = x + y. (i) (Jacobi, 87) If p = n + and x (mod ), then x ( 3n n )( 6n )/( 4n 3n n ) (mod p). (ii) (Lee, 00) If p = n+3 and x (mod ), then x ( 3n + )( 6n + )/( 4n + ) (mod p). n 3n n (iii) (Lee, 00) If p = n+4 and x 7 (mod ), then x ( 3n + )( 6n + )/( 4n + ) (mod p). n 3n n (iv) (Lee, 00) If p = n+5 and x 8 (mod ), then x ( )( )/( ) 3n+ 6n+ 4n+ n 3n+ n (mod p). (v) (Lee, 00) If p = n+6 and x 5 (mod ), then x ( )( )/( ) 3n+ 6n+4 4n+3 n 3n+ n+ (mod p). 38

39 0. (Z. H. Sun, 0, Adv. in Appl. Math.) For a, b, n N let λ(a, b; n) Z be given by q k= ( q ak ) 3 ( q bk ) 3 = n= λ(a, b; n)q n ( q < ). () Suppose ab and p is an odd prime such that p a, b, p ab + and p = ax + by with x, y Z. Let n = ((ab + )p a b)/8. Then ( ) a+b x+b+ (4ax p) = λ(a, b; n + ). () Let a, b N with (a, b) =. Let p be an odd prime such that p ab, ab+ and p = x +aby with x, y Z. Let n = (a + b)(p )/8. If a, b, 8 b and 8 p, then ( ) y (4x p) = λ(a, b; n + ). In 89 Klein and Fricke showed that λ(, ; (p + 3)/4) = 4x p for primes p = x + y (mod 4) with x. 39

40 In his lost notebook Ramanujan conjectured that λ(, 7; p) = 4x p for primes p = x + 7y,, 4 (mod 7). In 985 Stienstra and Beukers proved that λ(, 3; (p + )/) = λ(, 6; p) = 4x p for primes p = x + 3y (mod 3). Example 9. Let p > 5 be a prime. Then λ(, ; (3p + 5)/8) = ( ) (4x y p) for p = x + y (mod 8), λ(, 5; (3p + )/4) = ( ) x (4x p) for p = x + 5y, 9 (mod 0), { 0 if p, 9 (mod 30), λ(3, 5; p) = 4x p if p = x + 5y, { 0 if p 7, 3 (mod 30), λ(3, 5; p) = p x if p = 3x + 5y. 40

41 0. Ramanujan s conjectures on Euler products for a type of Dirichlet series For k =,,..., let q m= = n= ( q km )( q (4 k)m ) φ k (n)q n ( q < ). Ramanujan conjectured that the Dirichlet series φ k (n) n= n s (k =,, 3, 4, 6, 8, ) have Euler products and gave the explicit Euler products in the cases k =,, 3. Unfortunately his formulae for k =, 3 are wrong. In 98 Rangachari outlined the proofs of the formulae for k =,, 3 using class field theory and modular forms. But Rangachari s formulae for k =, 3 are also wrong and his proofs are neither clear nor elementary. So it remains to correct the results and to give elementary proofs of them. 4

42 Let d < 0 be a discriminant with h(d) = 3. Suppose that I is the principal class and A is a generator of H(d). We recall that F (A, n) = (R(I, n) R(A, n)) is a multiplicative function of n. Define L(A, s) = n= F (A, n) n s (Re(s) > ). Theorem 0. ([SW]). Let d < 0 be a discriminant with h(d) = 3. Let f be the conductor of d and H(d) = {I, A, A } with A 3 = I. Let s C with Re(s) >. If d 9, 4, then L(A, s) = p d p f p s p R(I) p d ( d p )= p s ( p s ) p R(A) p d where p runs over all primes. + p s + p s, 4

43 Theorem 0.. Let s C with Re(s) >. Then φ (n), φ (n), φ 6 (n) are multiplicative functions. Moreover, n= φ (n) n s = 3 s ( p 3 )= p s p=x +xy+3y + p s + p s p=x +xy+6y 3 ( p s ), n= φ (n) n s = s p,6,7,8,0 (mod ) p p=3x +xy+4y + p s + p s p=x +y ( p s ) p s 43

44 and n= = φ 6 (n) n s p 5 (mod 6) p s p=x +7y ( p s ) p=4x +xy+7y + p s + p s, where p runs over all primes. Let d < 0 be a discriminant such that H(d) = {I, A, A, A 3 } with A 4 = I. Then F (A, n) = (R(I, n) R(A, n)) is multiplicative. Define L(A, s) = n= F (A, n) n s. 44

45 Theorem 0.3. Let d < 0 be a discriminant such that H(d) = {I, A, A, A 3 } with A 4 = I. Let f be the conductor of d and s C with Re(s) >. If d 0, 5, then L(A, s) = ( d p )= p d,p f p R(A ) p d p R(A ) p s p d,p f p R(I) + p s p d p R(I) ( + p s ) where p runs over all primes. p s ( p s ) p d p R(A) + p s, 45

46 Theorem 0.4. Let s C with Re(s) >. Then φ 3 (n), φ 4 (n), φ 8 (n) and φ (n) are multiplicative functions. Moreover, n= φ 3 (n) n s = + 7 s p 3,5,6 (mod 7) p 3 p,8, (mod ) + p s p=x +xy+6y ( p s ) p=4x +xy+4y 7 p s ( + p s ), 46

47 n= = n= = φ 4 (n) n s + 5 s p 3,7(mod 0) p=4x +5y 5 φ 8 (n) n s p 5,7 (mod 8) p=x +3y p,3,7,9(mod 0) p s + p s p=x +0y ( p s ) ( + p s ), p s p 3 (mod 8) ( p s ) + p s p=4x +4xy+9y ( + p s ), 47

48 n= = φ (n) n s p 3(mod 4) p 3 p=x +36y p s ( p s ) where p runs over all primes. p 5 (mod ) p=4x +9y + p s ( + p s ),. The number of representations of n by ax + by(y )/, ax + by(3y )/ and ax(x )/ + by(3y )/ Let r(n = f(x, y)) = { x, y Z : n = f(x, y)}. In 008 Z.H. Sun ([S]) determined r(n = (3x x)/+b(3y y)/) for b =,, 5. 48

49 In 0, using some results for binary quadratic forms Z.H. Sun ([S4]) determined r(n = x + by(y )/) in the cases b =,, 3, 4, 5, 6, 8, 9, 0,, 4, 5, 6,, 9, 30, 35, 39, 5, 65, 95, r(n = x + b(3y y)/) in the cases b =,, 3, 4, 5, 7, 8, 3, 7 and r(n = (x x)/ + b(3y y)/) in the cases b =,, 3, 5, 7, 0,, 4, 5, 9, 6, 3, 34, 35, 55, 59, 9, 5, 9, 455. For example, we have r ( n = x y(y ) ) + b = for b = 3, 5,, 9, r ( n = x y(3y ) + b for b = 5, 7, 3, 7 k 8n+b ) = k 4n+b ( b k ( 6b k ) ) 49

50 and r ( n = x(x ) y(3y ) +b for b = 7,, 9, 3, 59. ) = k n+ b+3 ( 3b k ) There are huge similar results conjectured by R.S. Melham in his preprint Analogues of Jacobi s two-square theorem. R.S. Melham, Analogues of Jacobi s two-square theorem: an informal account, Integers 0(00),