ACTA ARITHMETICA
122.4 (2006)
Ramanujan identities and Euler products
for a type of Dirichlet series
by
Zhi-Hong Sun (Huaian) and Kenneth S. Williams (Ottawa)
1. Introduction. In his lost notebook [R2] Ramanujan stated that theDirichlet series
ββn=1 a(n)n
βs (Re(s) > 1), where the Dirichlet coefficientsa(n) (n = 1, 2, . . .) are given by
qββ
n=1
(1β q2n)(1β q22n) =ββ
n=1
a(n)qn (|q| < 1),
has an Euler product and gave an explicit formulation for the Euler product.In this paper we develop the theory of binary quadratic forms in order todetermine the Euler product for
ββn=1 a(n)n
βs, and other similarly definedDirichlet series, in a completely elementary and natural manner.Let N, Z, R and C be the sets of natural numbers, integers, real numbers
and complex numbers respectively. A nonsquare integer d with d β‘ 0, 1(mod 4) is called a discriminant. The conductor of the discriminant d is thelargest positive integer f = f(d) such that d/f2 β‘ 0, 1 (mod 4). As usual weset w(d) = 1, 2, 4, 6 according as d > 0, d < β4, d = β4 or d = β3.For integers a, b and c, we use (a, b, c) to denote the integral, binary
quadratic form ax2 + bxy + cy2. The form (a, b, c) is said to be primitiveif gcd(a, b, c) = 1. The discriminant of the form (a, b, c) is the integer d =b2 β 4ac. If d < 0, we only consider positive definite forms, that is, forms(a, b, c) with a > 0 and c > 0. Two forms (a, b, c) and (aβ², bβ², cβ²) are equivalent((a, b, c) βΌ (aβ², bβ², cβ²)) if there exist integers Ξ±, Ξ², Ξ³ and Ξ΄ with Ξ±Ξ΄ β Ξ²Ξ³ = 1such that the substitution x = Ξ±X+Ξ²Y, y = Ξ³X+Ξ΄Y transforms (a, b, c) to(aβ², bβ², cβ²). It is known that (a, b, c) βΌ (c,βb, a) and for k β Z that (a, b, c) βΌ(a, 2ak + b, ak2 + bk + c). We denote the equivalence class of (a, b, c) by[a, b, c]. The equivalence classes of primitive, integral, binary quadratic forms
2000 Mathematics Subject Classification: 11E16, 11F66, 11E25, 11B65.The first author was supported by Natural Sciences Foundation of Jiangsu Educational
Office (02KJB110007), and the second author was supported by Natural Sciences andEngineering Research Council of Canada grant A-7233.
[349]
350 Z. H. Sun and K. S. Williams
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 orderby h(d). The identity of H(d) is the so-called principal class [1, 0,βd/4] or[1, 1, (1βd)/4] according as d β‘ 0 (mod 4) or d β‘ 1 (mod 4), and the inverseof the class K = [a, b, c] is the class Kβ1 = [a,βb, c].Let (a, b, c) be an integral, binary quadratic form of discriminant d. The
positive integer n is said to be represented by (a, b, c) if there exist integersx and y with n = ax2 + bxy + cy2, and the pair {x, y} is called a represen-tation. If d < 0, every representation {x, y} is called primary. If d > 0, therepresentation {x, y} is called primary if it satisfies
2ax+ (bββd)y > 0 and 1 β€
β£
β£
β£
β£
2ax+ (b+βd)y
2ax+ (bββd)y
β£
β£
β£
β£
< Ξ΅(d)2,
where Ξ΅(d) = (x1 + y1βd)/2 and (x1, y1) is the solution in positive integers
to the equation X2 β dY 2 = 4 for which (x1 + y1βd)/2 (or equivalently y1)
is least (see [Di], [H]). For a, b, c β Z and n β N we define
(1.1) R(a, b, c;n) = |{{x, y} | n = ax2 + bxy + cy2, {x, y} is primary}|.If (a, b, c) βΌ (aβ², bβ², cβ²), by [SW, Remark 3.1] we have
R(a, b, c;n) = R(a,βb, c;n) = R(aβ², bβ², cβ²;n).From this we define R([a, b, c], n) = R(a, b, c;n) as in [SW].
Let d be a discriminant. Suppose H(d) = {Ak11 Β· Β· Β·Akrr | 0 β€ k1 <h1, . . . , 0 β€ kr < hr} with h1 Β· Β· Β·hr = h(d). For n β N andM = Am11 Β· Β· Β·Amrrβ H(d), following [SW, Definition 7.1] we define(1.2) F (M,n)
=1
w(d)
β
0β€k1<h1Β·Β·Β·
0β€kr<hr
cos 2Ο
(
k1m1h1+ Β· Β· Β·+ krmr
hr
)
Β·R(Ak11 Β· Β· Β·Akrr , n).
In particular, if h(d) = 2, 3, 4 and H(d) is cyclic with principal class I andgenerator A, then (see [SW, Theorem 7.4])
(1.3) F (A, n) =
1
w(d)(R(I, n)βR(A, n)) if h(d) = 2, 3,
1
w(d)(R(I, n)βR(A2, n)) if h(d) = 4.
Let s β C with Re(s) > 1. In this paper we introduce
(1.4) L(M, s) =ββ
n=1
F (M,n)
nsfor M β H(d).
Ramanujan identities and Euler products 351
From [SW, Theorem 7.2] we know that F (M,n) is a multiplicative functionof n β N. Thus, if Re(s) > 1, then
(1.5) L(M, s) =β
p
(
1 +
ββ
t=1
F (M,pt)pβst)
,
where p runs over all primes.From (1.5) we see that L(M, s) has an Euler product. The main pur-
pose of this paper is to give the Euler product for L(M, s). When H(d) iscyclic, in Section 5 we completely determine the Euler product for L(M, s)(M βH(d),Re(s) > 1), see Theorem 5.3. As consequences, in Sections 6β8we give explicit Euler products for L(M, s) in the cases h(d) = 2, 3 and H(d)is cyclic of order 4.For |q| < 1 let Ο(q) and Ο(q) be the theta functions defined by
(1.6)
Ο(q) =ββ
n=0
qn(n+1)/2,
Ο(q) =
ββ
m=1
(1β qm) =ββ
n=ββ(β1)nq(3n2βn)/2.
Ramanujan (see for example [B]) established many identities involving Ο(q)and Ο(q). In Section 4 of this paper we prove some of these identities fromour point of view.For k = 1, . . . , 12 let
(1.7) qΟ(qk)Ο(q24βk) =ββ
n=1
Οk(n)qn (|q| < 1).
Ramanujan ([R1], [R2]) conjectured that the Dirichlet seriesββn=1
Οk(n)ns
(k = 1, 2, 3, 4, 6, 8, 12) have Euler products and gave the explicit Euler prod-ucts in the cases k = 1, 2, 3. Unfortunately his formulae for k = 2, 3 arewrong. In [Ra] Rangachari outlined the proofs of the formulae for k = 1, 2, 3using class field theory and modular forms. But Rangachariβs formulae fork = 2, 3 are also wrong and his proofs are neither clear nor elementary. Soit remains to correct the results and to give elementary proofs of them. Forinstance, the corrected form of Ramanujanβs conjecture in his lost notebook([R2]) isββ
n=1
Ο2(n)
ns=
1
1β 11βsβ
pβ‘2,6,7,8,10 (mod 11)p 6=2
1
1β pβ2s
Γβ
p=3x2+2xy+4y2
1
1 + pβs + pβ2s
β
p=x2+11y2 6=11
1
(1β pβs)2 ,
352 Z. H. Sun and K. S. Williams
where p runs over all primes. We note that this formula also corrects theincorrect formula in [Ra].
Let Ξ΄k = (1 β (β1)k)/2 and n β N. In Section 2 we show that fork = 1, . . . , 12,
Οk(n) =1
2(R(1, Ξ΄k, (24k β k2 + Ξ΄k)/4;n)βR(4, 4β k, k + 1;n)).
Moreover, we obtain explicit formulas for Οk(n) in the cases k = 1, 2, 3, 4, 6,8, 12 (see Theorems 4.4 and 4.5). From the above it follows that for k =1, 2, 3, 4, 6, 8, 12,
ββ
n=1
Οk(n)
ns= L(Ak, s) for Re(s) > 1,
where Ak = [2, 1, 3], [3, 2, 4], [2, 1, 8], [3, 2, 7], [4, 2, 7], [3, 2, 11], [5, 4, 8] ac-cording as k = 1, 2, 3, 4, 6, 8, 12. Thus using the results for Dirichlet se-ries L(M,n) we obtain the Euler products for
ββn=1 Οk(n)n
βs in the casesk = 1, 2, 3, 4, 6, 8, 12. In this way we prove all of Ramanujan conjectures forΟk(n), and our proofs are natural and elementary. It seems that Ramanu-janβs conjecture for Ο12(n) was first proved by Mordell ([M]).
We should mention that the Euler products forββn=1 Οk(n)n
βs in thecases k = 1, 2, 3, 4, 6, 8, 12 are connected with modular forms (see for exam-ple [Ra]).
In addition to the above notation, we also use throughout this paper thefollowing notation: ordp n denotes the nonnegative integer Ξ± such that p
Ξ± |nbut pΞ±+1 β€n, pΞ± βn means pΞ± |n but pΞ±+1 β€n,
(
am
)
is the Kronecker symbol,(a, b) is the greatest common divisor of the integers a and b (not both zero),I denotes the principal class in H(d), and R(K) denotes the set of integersrepresented by forms in the class K.
Throughout this paper p denotes a prime and products (sums) over prun through all primes p satisfying any restrictions given under the product(summation) symbol. For example the condition p = x2 + 11y2 under aproduct restricts the product to those primes p which are of the form x2 +11y2 for some integers x and y.
2. Generating functions for 12 (R(I, n) β R(K,n)) when I,K βH(d). For q β R, m β N and r β Z let
(2.1) f(r,m; q) =
ββ
n=ββ(β1)nq(mn2βrn)/2 (|q| < 1).
From Jacobiβs identity (cf. [HW, Theorem 352, p. 282 (with x = qm/2, z =βqβr/2)]) we know that
Ramanujan identities and Euler products 353
(2.2) f(r,m; q)
=ββ
n=0
{(1β qmn+(mβr)/2)(1β qmn+m)(1β qmn+(m+r)/2)} (|q| < 1).
Definition 2.1. For r β Z, m β N and k β {1, . . . , 4m} with 2 | k(mβr)we define
qr2
f(r,m; qk)f(r,m; q8mβk) =ββ
n=0
fk(r,m;n)qn.
Proposition 2.1. Let r β Z and let k,m, n β N with k β€ 4m and2 | k(mβ r). Then
fk(r,m;n)
=
2β
x,yβZ4x2β2kxy+2kmy2=nxβ‘r/2 (modm)
1ββ
x,yβZ4x2βkxy+ km2 y2=nxβ‘r/2 (modm)
1 if 2 | r,
β
x,yβZx2βkxy+2kmy2=nxβ‘r (mod 2m)
1ββ
x,yβZk(mβ1)+2
2 x2+k(2mβ1)xy+2kmy2=nxβ‘r (mod 2m)
1 if 2 β€ r.
Proof. From (2.1) and Definition 2.1 we have
fk(r,m;n) =β
x,yβZk(mx2βrx)
2 + (8mβk)(my2βry)
2 +r2=n
(β1)x+y.
It is clear that
16
(
k(mx2 β rx)2
+(8mβ k)(my2 β ry)
2+ r2)
= (kx+ (8mβ k)y β 4r)2 + k(8mβ k)(xβ y)2.Thus
fk(r,m;n) =β
x,yβZ(kx+(8mβk)yβ4r)2+k(8mβk)(xβy)2=16n
(β1)xβy
=β
y,zβZ(8my+kzβ4r)2+k(8mβk)z2=16n
(β1)z
=β
x,zβZx2+k(8mβk)z2=16nxβ‘kzβ4r (mod 8m)
(β1)z =β
y,zβZy2+2kyz+8kmz2=16nyβ‘β4r (mod 8m)
(β1)z
354 Z. H. Sun and K. S. Williams
=β
x,zβZ16x2β8kxz+8kmz2=16nβ4xβ‘β4r (mod 8m)
(β1)z =β
x,yβZ2x2βkxy+kmy2=2nxβ‘r (mod 2m)
(β1)y.
If r is even, then 2 | km. By the above we must have
fk(r,m;n) =β
x,yβZ2(2x)2βk(2x)y+kmy2=2n
2xβ‘r (mod 2m)
(β1)y =β
x,yβZ4x2βkxy+ km2 y2=nxβ‘r/2 (modm)
(β1)y
= 2β
x,yβZ4x2β2kxy+2kmy2=nxβ‘r/2 (modm)
1ββ
x,yβZ4x2βkxy+ km2 y2=nxβ‘r/2 (modm)
1.
If r is odd, then 2 | k(mβ 1). From the above we obtainfk(r,m;n)
=β
x,yβZ2x2βkxy+kmy2=2nxβ‘r (mod 2m), 2|y
1ββ
x,yβZ2x2βkxy+kmy2=2nxβ‘r (mod 2m), 2β€y
1
=β
x,yβZ2x2β2kxy+4kmy2=2nxβ‘r (mod 2m)
1ββ
x,yβZ(kmβk+2)x2+k(2mβ1)x(yβx)+km(yβx)2=2n
xβ‘r (mod 2m), 2|yβx
1
=β
x,yβZx2βkxy+2kmy2=nxβ‘r (mod 2m)
1ββ
x,yβZ(kmβk+2)x2+k(2mβ1)x(2y)+km(2y)2=2n
xβ‘r (mod 2m)
1
=β
x,yβZx2βkxy+2kmy2=nxβ‘r (mod 2m)
1ββ
x,yβZk(mβ1)+2
2 x2+k(2mβ1)xy+2kmy2=nxβ‘r (mod 2m)
1.
This finishes the proof.
Definition 2.2. Let Ο(q) =ββn=0 q
n(n+1)/2. For k β {1, 2, 3, 4} wedefine Οk(n) by
qΟ(βqk)Ο(βq8βk) =ββ
n=1
Οk(n)qn (|q| < 1).
Theorem 2.1. For k β {1, 2, 3, 4} and n β N we have
Οk(n) =12 (R(1, 0, k(8β k);n)βR(4, 4β 2k, k + 1;n)).
Ramanujan identities and Euler products 355
Proof. Since
ββ
n=0
qn(n+1)/2 = 1 +ββ
n=1
q(2nβ1)(2n)/2 +ββ
n=1
q2n(2n+1)/2
=ββ
n=ββq2n
2βn (|q| < 1)
we see that
ββ
n=1
Οk(n)qn = qΟ(βqk)Ο(βq8βk)
= q(
ββ
n=ββ(βqk)2n2βn
)(
ββ
n=ββ(βq8βk)2n2βn
)
= q(
ββ
n=ββ(β1)nqk(2n2βn)
)(
ββ
n=ββ(β1)nq(8βk)(2n2βn)
)
= qf(1, 2; q2k)f(1, 2; q16β2k) =ββ
n=0
f2k(1, 2;n)qn (|q| < 1).
Thus Οk(n) = f2k(1, 2;n). Applying Proposition 2.1 we obtain
Οk(n) =β
x,yβZx2β2kxy+8ky2=nxβ‘1 (mod 4)
1ββ
x,yβZ(k+1)x2+6kxy+8ky2=n
xβ‘1 (mod 4)
1
=1
2
(
β
x,yβZx2β2kxy+8ky2=n
1ββ
x,yβZx2β2kxy+8ky2=n
2|x
1)
β 12
(
β
x,yβZ(k+1)x2+6kxy+8ky2=n
1ββ
x,yβZ(k+1)x2+6kxy+8ky2=n
2|x
1)
= 12 (R(1,β2k, 8k;n)βR(4,β4k, 8k;n)βR(k + 1, 6k, 8k;n) +R(4k + 4, 12k, 8k;n)).
Note that (4k+4, 12k, 8k) βΌ (8k,β12k, 4k+4) βΌ (8k, 4k, 4) βΌ (4,β4k, 8k),(1,β2k, 8k) βΌ (1, 0, k(8 β k)) and (k + 1, 6k, 8k) βΌ (k + 1, 2k β 4, 4) βΌ(4, 4β 2k, k + 1). We then obtain the desired result.
356 Z. H. Sun and K. S. Williams
Definition 2.3. Let Ο(q) =ββn=ββ(β1)nq(3n
2βn)/2 (|q| < 1). Fork β {1, . . . , 12} define Οk(n) by
ββ
n=1
Οk(n)qn = qΟ(qk)Ο(q24βk) (|q| < 1).
As Ο(q) = f(1, 3; q) it is clear that Οk(n) = fk(1, 3;n).
Theorem 2.2. Let k β {1, . . . , 12} and Ξ΄k = (1β (β1)k)/2. For n β Nwe have
Οk(n) =12 (R(1, Ξ΄k, (24k β k
2 + Ξ΄k)/4;n)βR(4, 4β k, k + 1;n))
=
12 (R(1, 1, 6;n)βR(2, 1, 3;n)) if k = 1,12 (R(1, Ξ΄k, (24k β k2 + Ξ΄k)/4;n)βR(k + 1, 4β k, 4;n)) if 2 β€ k β€ 3,12 (R(1, Ξ΄k, (24k β k2 + Ξ΄k)/4;n)βR(4, k β 4, k + 1;n)) if 4 β€ k β€ 8,12 (R(1, Ξ΄k, (24k β k2 + Ξ΄k)/4;n)βR(4, 12β k, 9;n)) if 9 β€ k β€ 12.
Proof. As Οk(n) = fk(1, 3;n), by Proposition 2.1 we have
Οk(n) =β
x,yβZx2βkxy+6ky2=nxβ‘1 (mod 6)
1ββ
x,yβZ(k+1)x2+5kxy+6ky2=n
xβ‘1 (mod 6)
1.
Sinceβ
x,yβZax2+bxy+cy2=nxβ‘1 (mod 6)
1
=β
x,yβZax2+bxy+cy2=n2β€x, xβ‘1 (mod 3)
1 =β
x,yβZax2+bxy+cy2=nxβ‘1 (mod 3)
1ββ
x,yβZ4ax2+2bxy+cy2=nxβ‘1 (mod 3)
1
=1
2
(
β
x,yβZax2+bxy+cy2=n
1ββ
x,yβZ9ax2+3bxy+cy2=n
1)
β 12
(
β
x,yβZ4ax2+2bxy+cy2=n
1ββ
x,yβZ36ax2+6bxy+cy2=n
1)
=1
2(R(a, b, c;n)βR(9a, 3b, c;n)βR(4a, 2b, c;n) +R(36a, 6b, c;n)),
we see that
Ramanujan identities and Euler products 357
2Οk(n) = (R(1,βk, 6k;n)βR(9,β3k, 6k;n)βR(4,β2k, 6k;n)+R(36,β6k, 6k;n))β (R(k + 1, 5k, 6k;n)βR(9k + 9, 15k, 6k;n)βR(4k + 4, 10k, 6k;n) +R(36k + 36, 30k, 6k;n)).
Note that (9k+9, 15k, 6k) βΌ (6k,β15k, 9k+9) βΌ (6k,β3k, 9) βΌ (9, 3k, 6k),(36k+36, 30k, 6k) βΌ (6k,β30k, 36k+36) βΌ (6k, 6k, 36) βΌ (36,β6k, 6k) and(4k + 4, 10k, 6k) βΌ (4k + 4, 2k β 8, 4) βΌ (4, 8 β 2k, 4k + 4) βΌ (4,β2k, 6k).We find R(9k + 9, 15k, 6k;n) = R(9,β3k, 6k;n), R(36k + 36, 30k, 6k;n) =R(36,β6k, 6k;n) and R(4k + 4, 10k, 6k;n) = R(4,β2k, 6k;n). Thus
Οk(n) =12 (R(1,βk, 6k;n)βR(k + 1, 5k, 6k;n)).
Clearly (1,βk, 6k) βΌ (1, Ξ΄k, (24k β k2 + Ξ΄k)/4) and (k + 1, 5k, 6k) βΌ (k + 1,kβ4, 4) βΌ (4, 4βk, k+1). If 9 β€ k β€ 12, then (4, 4βk, k+1) βΌ (4, 12βk, 9).Also, (4, 3, 2) βΌ (2,β3, 4) βΌ (2, 1, 3). Now combining the above we get thedesired result.
Corollary 2.1. Let n β N, 2 |n and m β {1, 2, 3, 4, 5, 6}. ThenR(1, 0,m(12βm);n) = R(4, 4β 2m, 2m+ 1;n).
Proof. Note that Ο2m(n) = 0 by Definition 2.3. Putting k = 2m inTheorem 2.2 gives the result.
Corollary 2.1 can also be deduced from [KW2, Theorem 1]. For m =1, 2, 3, 4, 5 see Corollaries 1, 4, 5, 6, 8 in [KW2] respectively.
Let Ο be the Ramanujan tau function defined by
qββ
n=1
(1β qn)24 =ββ
n=1
Ο(n)qn (|q| < 1).
Then we have
Corollary 2.2. For any positive integer n we have
Ο(n) β‘
0 (mod 23) if there is a prime p such that(
p23
)
= β1 and2 β€ ordp n, or p = 2x
2 + xy + 3y2 and
3 | ordp nβ 2,(β1)Β΅
β
p=x2+xy+6y2 6=23(1 + ordp n) (mod 23) otherwise,
where
Β΅ =β
p=2x2+xy+3y2
ordp nβ‘1 (mod 3)
1
and p runs over all primes.
358 Z. H. Sun and K. S. Williams
Proof. Eulerβs identity states that (see for example [HW, Theorem 353,p. 284])
ββ
n=ββ(β1)nq(3n2βn)/2 =
ββ
n=1
(1β qn) (|q| < 1).
Soββn=1 Ο(n)q
n
ββn=1 Ο1(n)q
n=
qββn=1(1β qn)24
qββn=1(1β qn)
ββn=1(1β q23n)
=
ββn=1(1β qn)23ββn=1(1β q23n)
=ββ
n=1
(
1 +
( 22β
k=1
(
23
k
)
(β1)kqkn)
(1β q23n)β1)
= 1 + 23
ββ
n=1
anqn (an β Z)
and hence applying Theorem 2.2 we get
Ο(n) β‘ Ο1(n) = 12 (R(1, 1, 6;n)βR(2, 1, 3;n)) = F ([2, 1, 3], n) (mod 23).Observe that H(β23) = {[1, 1, 6], [2, 1, 3], [2,β1, 3]}. Then applying [SW,Theorem 10.2(i)] we obtain the result.
We remark that Corollary 2.2 generalizes the known result 23 | Ο(n) forthose positive integers n such that
(
n23
)
= β1 (see for example [BO]).Theorem 2.3. For k = 1, . . . , 8 and n β N we have
f2k(1, 4;n) + f2k(3, 4;n) =12 (R(1, 0, k(16β k);n)βR(4, 4β 2k, 3k + 1;n)).
Proof. By Proposition 2.1 we have
f2k(1, 4;n) + f2k(3, 4;n)
=β
x,yβZx2β2kxy+16ky2=nxβ‘1,3 (mod 8)
1ββ
x,yβZ(3k+1)x2+14kxy+16ky2=n
xβ‘1,3 (mod 8)
1
=1
2
(
β
x,yβZ, 2β€xx2β2kxy+16ky2=n
1ββ
x,yβZ, 2β€x(3k+1)x2+14kxy+16ky2=n
1)
=1
2{(R(1,β2k, 16k;n)βR(4,β4k, 16k;n))
β (R(3k + 1, 14k, 16k;n)βR(12k + 4, 28k, 16k;n))}.Note that (12k + 4, 28k, 16k) βΌ (16k,β28k, 12k + 4) βΌ (16k, 4k, 4) βΌ(4,β4k, 16k), (1,β2k, 16k) βΌ (1, 0, k(16 β k)) and (3k + 1, 14k, 16k) βΌ(3k + 1, 2k β 4, 4) βΌ (4, 4β 2k, 3k + 1). We then obtain the result.
Ramanujan identities and Euler products 359
Theorem 2.4. Let k β {1, . . . , 20}, Ξ΄k = (1β (β1)k)/2, n β N. Then
fk(1, 5;n) + fk(3, 5;n)
= 12 (R(1, Ξ΄k, (40k β k2 + Ξ΄k)/4;n)βR(4, 4β k, 2k + 1;n)).
Proof. By Proposition 2.1 we have
fk(1, 5;n) + fk(3, 5;n)
=β
x,yβZx2βkxy+10ky2=nxβ‘1,3 (mod 10)
1ββ
x,yβZ(2k+1)x2+9kxy+10ky2=n
xβ‘1,3 (mod 10)
1
=1
2
(
β
x,yβZ, 2β€xx2βkxy+10ky2=n
1ββ
x,yβZ, xβ‘5 (mod 10)x2βkxy+10ky2=n
1)
β 12
(
β
x,yβZ, 2β€x(2k+1)x2+9kxy+10ky2=n
1ββ
x,yβZ, xβ‘5 (mod 10)(2k+1)x2+9kxy+10ky2=n
1)
= 12{(R(1,βk, 10k;n)βR(4,β2k, 10k;n))β (R(25,β5k, 10k;n)βR(100,β10k, 10k;n))β (R(2k + 1, 9k, 10k;n)βR(8k + 4, 18k, 10k;n))+ (R(50k + 25, 45k, 10k;n)βR(200k + 100, 90k, 10k;n))}.
Observe that (8k + 4, 18k, 10k) βΌ (8k + 4, 2k β 8, 4) βΌ (4, 8β 2k, 8k + 4) βΌ(4,β2k, 10k), (50k + 25, 45k, 10k) βΌ (10k,β45k, 50k + 25) βΌ (10k,β5k, 25)βΌ (25, 5k, 10k) and (200k + 100, 90k, 10k) βΌ (10k,β90k, 200k + 100) βΌ(10k, 10k, 100) βΌ (100,β10k, 10k). We then obtain
fk(1, 5;n) + fk(3, 5;n) =12{R(1,βk, 10k;n)βR(2k + 1, 9k, 10k;n)}.
Since (1,βk, 10k) βΌ (1, Ξ΄k, (40kβk2+Ξ΄k)/4) and (2k+1, 9k, 10k) βΌ (2k+1,k β 4, 4) βΌ (4, 4β k, 2k + 1), we obtain the desired result.Corollary 2.3. If n is even and m β {1, . . . , 10}, then
R(1, 0,m(20βm);n) = R(4, 4β 2m, 4m+ 1;n).Proof. Note that f2m(1, 5;n) = f2m(3, 5;n) = 0 by Definition 2.1. Tak-
ing k = 2m in Theorem 2.4 yields the result.
The case m = 1 of Corollary 2.3 has been given in [KW2, Corollary 3].
Corollary 2.4. For n β N we have(i) f2(1, 4;n) + f2(3, 4;n) = Ο3(n),(ii) f4(1, 5;n) + f4(3, 5;n) = Ο12(n),(iii) f8(1, 5;n) + f8(3, 5;n) = f16(1, 4;n) + f16(3, 4;n).
360 Z. H. Sun and K. S. Williams
Proof. From Theorems 2.1β2.4 we see that
f2(1, 4;n) + f2(3, 4;n) =12 (R(1, 0, 15;n)βR(4, 2, 4;n)) = Ο3(n),
f4(1, 5;n) + f4(3, 5;n) =12 (R(1, 0, 36;n)βR(4, 0, 9;n)) = Ο12(n)
and
f8(1, 5;n) + f8(3, 5;n)
= 12 (R(1, 0, 64;n)βR(4, 4, 17;n)) = 12 (R(1, 0, 64;n)βR(4,β12, 25;n))= f16(1, 4;n) + f16(3, 4;n).
So the corollary is proved.
Theorem 2.5. Let p be an odd prime. Let r be odd with p β€ r and n β N.Then
f4p(r, p;n) =
{
12 (R(1, 0, 4p
2;n)βR(4, 0, p2;n)) if n β‘ r2 (mod 4p),0 otherwise.
Proof. From Proposition 2.1 we know that
f4p(r, p;n) =β
x,yβZx2β4pxy+8p2y2=nxβ‘r (mod 2p)
1ββ
x,yβZ(1+2p(pβ1))x2+4p(2pβ1)xy+8p2y2=n
xβ‘r (mod 2p)
1.
If n 6β‘ r2 (mod 4p), then clearly f4p(r, p;n) = 0 by Definition 2.1. If n β‘ r2(mod 4p), then x2 β‘ n (mod 4p) if and only if x β‘ Β±r (mod 2p). Thus, bythe above we obtain
f4p(r, p;n) =12R(1,β4p, 8p
2;n)β 12R(2p2 β 2p+ 1, 4p(2pβ 1), 8p2;n).
Since (1,β4p, 8p2) βΌ (1, 0, 4p2) and (2p2 β 2p+ 1, 4p(2pβ 1), 8p2) βΌ (2p2 β2p+1, 4pβ 4, 4) βΌ (4, 4β 4p, 2p2β 2p+1) βΌ (4, 0, p2), we obtain the result.
3. The Euler product forββn=1
β(n,d)ns . Let d be a discriminant with
conductor f and d0 = d/f2. In view of [SW, Lemma 3.5] we introduce
(3.1) C(d) = fβ
p|f
(
1β 1p
(
d0p
))
=
h(d)w(d0)
h(d0)w(d)if d < 0,
h(d) log Ξ΅(d)
h(d0) log Ξ΅(d0)if d > 0,
where p runs over all distinct prime divisors of f . If f is a prime, then clearlyC(d) = f β
(
d0p
)
.
Definition 3.1. Let d be a discriminant with conductor f . Let d0 =d/f2 and n β N. Then we define
Ξ΄(n, d) =β
m|n
(
d
m
)
Ramanujan identities and Euler products 361
and
β(n, d) =
Ξ΄(n, d) if f β€n,
βC(d)(
d0f
)ordf nβ1Ξ΄(n, d) if f |n.
Lemma 3.1 ([SW, Lemma 4.1]). Let d be a discriminant and n β N.Then Ξ΄(n, d) is a multiplicative function of n. Moreover ,
Ξ΄(n, d) =β
( dp )=β1
1 + (β1)ordp n2
β
( dp )=1
(1 + ordp n)
andββ
n=1
Ξ΄(n, d)
ns=β
p
1
(1β pβs)(
1β(
dp
)
pβs) (Re(s) > 1).
From Definition 3.1 and Lemma 3.1 we have
Lemma 3.2. Let d be a discriminant with conductor f . Let s β C withRe(s) > 1. Set d0 = d/f2. If f is a prime, then β(n, d) is a multiplicativefunction of n β N and
ββ
n=1
β(n, d)
ns=
(
1β C(d)fβs
1β(
d0f
)
fβs
)
β
p 6=f
1
(1β pβs)(
1β(
d0p
)
pβs) .
Lemma 3.3. Let d be a discriminant such that h(d) = 1 and the con-ductor f is a prime. Set d0 = d/f
2 and Ξ΄k = (1β (β1)k)/2 for k β Z. Forn β N we have
β(n, d)
=
1
2
(
R
(
1, Ξ΄d,βd+ Ξ΄d4;n
)
βR(
f, Ξ΄d0f,βd0 + Ξ΄d04
f ;n
))
if d < 0,
R
(
1, Ξ΄d,βd+ Ξ΄d4;n
)
β C(d)R(
f, Ξ΄d0f,βd0 + Ξ΄d04
f ;n
)
if d > 0.
Proof. Let N(n, d) =β
KβH(d)R(K,n). From [SW, Theorem 4.1] weknow that
(3.2) N(n, d) =
0 if (n, f2) is not a square,
w(d) Β·mβ
p|m
(
1β 1p
(
d/m2
p
))
β
k| nm2
(
d0k
)
if (n, f2) = m2 for m β N.
362 Z. H. Sun and K. S. Williams
Observing that h(d) = 1 and f is a prime, by (3.2) we have
(3.3) R(1, Ξ΄d, (βd+ Ξ΄d)/4;n) =
0 if f βn,w(d)β
k|n
(
d0k
)
if f β€n,
w(d)C(d)β
k| nf2
(
d0k
)
if f2 |n.
Since h(d0) |h(d) (see [SW, Remark 2.2]) we see that h(d0) = 1. If f β€n,clearly R(f, Ξ΄d0f, (βd0+Ξ΄d0)f/4;n) = 0. If f |n, by (3.2), [SW, Remark 3.1]and the fact that f(d0) = 1 we have
R(f, Ξ΄d0f, (βd0 + Ξ΄d0)f/4;n) = R(1, Ξ΄d0 , (βd0 + Ξ΄d0)/4;n/f)
= w(d0)β
k|nf
(
d0k
)
.
Thus
1
w(d)
(
R
(
1, Ξ΄d,βd+ Ξ΄d4;n
)
β C(d)w(d)
w(d0)R
(
f, Ξ΄d0f,βd0 + Ξ΄d04
f ;n
))
=
β
k|n
(
d0k
)
β 0 if f β€n,
0β C(d)β
k|nf
(
d0k
)
= βC(d)β
k|n, f β€k
(
d0k
)
if f βn,
βC(d)β
k|nf , kβ€ nf2
(
d0k
)
= βC(d)β
k|n, f β€k
(
d0kfordf nβ1
)
if f2 |n.
To complete the proof, we note that
β
k|n, f β€k
(
d0k
)
=β
k|n
(
d0f2
k
)
= Ξ΄(n, d)
and
C(d)w(d)
w(d0)=
{
1 if d < 0,
C(d) if d > 0.
If q β‘ 1 (mod 4) is a prime such that h(4q) = 1, then h(q) = 1 by [SW,Remark 2.2]. From (3.1) we find C(4q) = 2
(
1 β 12(
q2
))
= 2 β (β1)(qβ1)/4.Now applying Lemmas 3.2 and 3.3 in the cases d = β12,β16,β27,β28, 4qwe get
Ramanujan identities and Euler products 363
Theorem 3.1. Let s β C with Re(s) > 1. Then
(a)ββ
n=1
12 (R(1, 0, 3;n)βR(2, 2, 2;n))
ns
=1β 21βs1 + 2βs
Β· 1
1β 3βsβ
pβ‘5 (mod 6)
1
1β pβ2sβ
pβ‘1 (mod 6)
1
(1β pβs)2 ,
(b)ββ
n=1
12 (R(1, 0, 4;n)βR(2, 0, 2;n))
ns
= (1β 21βs)β
pβ‘3 (mod 4)
1
1β pβ2sβ
pβ‘1 (mod 4)
1
(1β pβs)2 ,
(c)
ββ
n=1
12 (R(1, 1, 7;n)βR(3, 3, 3;n))
ns
= (1β 31βs)β
pβ‘2 (mod 3)
1
1β pβ2sβ
pβ‘1 (mod 3)
1
(1β pβs)2 ,
(d)ββ
n=1
12 (R(1, 0, 7;n)βR(2, 2, 4;n))
ns
=1β 21βs1β 2βs Β·
1
1β 7βsβ
pβ‘3,5,6 (mod 7)
1
1β pβ2sβ
pβ‘1,2,4 (mod 7)p 6=2
1
(1β pβs)2 .
(e) If q β‘ 1 (mod 4) is a prime such that h(4q) = 1 (for example q =5, 13, 17, 29, 41, 53, 61, 73, 89, 97, 109, 113, . . .), then
ββ
n=1
R(1, 0,βq;n)β (2β (β1)(qβ1)/4)R(2, 2, (1β q)/2;n)ns
=1β 21βs
1β (β1)(qβ1)/42βsβ
p>2
1
(1β pβs)(
1β(
qp
)
pβs) .
Theorem 3.2. For s β C with Re(s) > 1 we have
ββ
n=1
Ο1(n)
ns=
(
2β 1
1β 2βs)
Β· 1
1β 7βsβ
pβ‘3,5,6 (mod 7)
1
1β pβ2s
Γβ
pβ‘1,2,4 (mod 7)p 6=2
1
(1β pβs)2
364 Z. H. Sun and K. S. Williams
andββ
n=1
(β1)nβ1Ο1(n)ns
=1
(1β 2βs)(1β 7βs)β
pβ‘3,5,6 (mod 7)
1
1β pβ2s
Γβ
pβ‘1,2,4 (mod 7)p 6=2
1
(1β pβs)2 .
Proof. From Theorem 2.1 we know that
Ο1(n) =12 (R(1, 0, 7;n)βR(2, 2, 4;n)).
Using Lemma 3.3 with d = β28 we see that(β1)nβ1Ο1(n) = (β1)nβ1(R(1, 0, 7;n)βR(2, 2, 4;n))/2
= (β1)nβ1β(n,β28) = Ξ΄(n,β28).Thus applying Lemma 3.1 and Theorem 3.1 we obtain the result.
4. Values of Οk(n) and Οk(n) and related identities
Theorem 4.1. For n β N we have
Ο1(n) = (β1)nβ1β
m|n
(β28m
)
= (β1)nβ1β
m|n, 2β€m
(
m
7
)
,
Ο2(n) =
(β1)(nβ1)/2β
m|n
(
m
3
)
if 2 β€n,
0 if 2 |nand
Ο4(n) =
(β1)(nβ1)/4β
m|n
(β1m
)
if n β‘ 1 (mod 4),
0 if n 6β‘ 1 (mod 4).Proof. From the proof of Theorem 3.1 we see that
(β1)nβ1Ο1(n) = Ξ΄(n,β28) =β
m|n
(β28m
)
=β
m|n, 2β€m
(β7m
)
=β
m|n, 2β€m
(
m
7
)
.
According to Theorem 2.1, [SW, Theorem 9.2] and Lemma 3.1 we have
Ο2(n) =1
2(R(1, 0, 12;n)βR(3, 0, 4;n))
=
(β1)(nβ1)/2β
m|n
(β3m
)
= (β1)(nβ1)/2β
m|n
(
m
3
)
if 2 β€n,
0 if 2 |n,
Ramanujan identities and Euler products 365
Ο4(n) =12 (R(1, 0, 16;n)βR(4, 4, 5;n))
=
(
2
n
)
β
m|n
(β1m
)
if 2 β€n,
0 if 2 |n
=
(β1)(nβ1)/4β
m|n
(β1m
)
if n β‘ 1 (mod 4),
0 otherwise.
In the last step we note that if n β‘ 3 (mod 4), thenβ
m|n
(β1m
)
=β
m|nm<βn
((β1m
)
+
( β1n/m
))
=β
m|nm<βn
(β1m
)(
1+
(β1n
))
= 0.
This completes the proof.
Theorem 4.2. For n β N let n = 3Ξ±n0 (3 β€n0). Then
Ο3(n) = (β1)nβ11 + (β1)Ξ±
(
n03
)
2
β
m|n, 2β€m
(
m
15
)
.
Proof. By Theorem 2.1 we have
Ο3(n) =12 (R(1, 0, 15;n)βR(4,β2, 4;n))
=
{
12R(1, 0, 15;n) if 2 β€n,12 (R(1, 0, 15;n)βR(2, 1, 2;n/2)) if 2 |n.
Now we consider the following three cases.
Case 1: 2 β€n. By the above, [SW, Theorem 9.3] and Lemma 3.1,
Ο3(n) =1
2R(1, 0, 15;n) =
1 + (β1)Ξ±(
n03
)
2
β
m|n
(β15m
)
=1 + (β1)Ξ±
(
n03
)
2
β
m|n, 2β€m
(
m
15
)
.
Case 2: 2βn. By [SW, Theorem 9.3] and Lemma 3.1, R(1, 0, 15;n) = 0and
R(2, 1, 2;n/2) =
(
1β (β1)Ξ±(
n0/2
3
))
β
m|n2
(β15m
)
=
(
1 + (β1)Ξ±(
n03
))
β
m|n2
(
m
15
)
.
366 Z. H. Sun and K. S. Williams
Thus
2Ο3(n) = R(1, 0, 15;n)βR(2, 1, 2;n/2)
= β(
1 + (β1)Ξ±(
n03
))
β
m|n2
(
m
15
)
= β(
1 + (β1)Ξ±(
n03
))
β
m|n, 2β€m
(
m
15
)
.
Case 3: 4 |n. From [SW, Theorem 9.3] and Lemma 3.1 we know that
R(1, 0, 15;n) =
(
1 + (β1)Ξ±(
n03
))
β
m|n4
(
m
15
)
and
R(2, 1, 2;n/2) =
(
1 + (β1)Ξ±(
n03
))
β
m|n2
(
m
15
)
.
Thus
Ο3(n) =12 (R(1, 0, 15;n)βR(2, 1, 2;n/2)) = β
1 + (β1)Ξ±(
n03
)
2
β
m|n2 ,mβ€n4
(
m
15
)
.
Observing that
β
m|n2 ,mβ€n4
(
m
15
)
=β
m|n, 2β€m
(
2ord2 nβ1m
15
)
=β
m|n, 2β€m
(
m
15
)
we then get the desired result.
Summarizing the above we prove the following theorem.
Theorem 4.3. Let |q| < 1 and Ο(q) =ββn=0 qn(n+1)/2. Then
(i) qΟ(q)Ο(q7) =
ββ
n=1
(
β
m|n, 2β€m
(
m
7
))
qn.
(ii) Ο(q)Ο(q3) =ββ
n=0
(
β
m|2n+1
(
m
3
))
qn.
(iii) Ο2(q) =ββ
n=0
(
β
m|4n+1(β1)(mβ1)/2
)
qn.
(iv) Ο4(q) =
ββ
n=0
(
β
d|2n+1d)
qn.
Ramanujan identities and Euler products 367
Proof. From Definition 2.2 we obtain
qΟ(q)Ο(q7) =ββ
n=1
(β1)nβ1Ο1(n)qn.
Then, applying Theorem 4.1, we obtain (i). From Definition 2.2 and Theo-rem 4.1 we have
qΟ(βq2)Ο(βq6) =ββ
n=1
Ο2(n)qn =
ββ
n=0
(β1)n(
β
m|2n+1
(
m
3
))
q2n+1.
Thus
Ο(βq)Ο(βq3) =ββ
n=0
(β1)n(
β
m|2n+1
(
m
3
))
qn.
Replacing q by βq we then obtain (ii).We now consider (iii). It follows from Definition 2.2 and Theorem 4.1
that
qΟ2(βq4) =ββ
n=1
Ο4(n)qn =
ββ
n=0
(β1)n(
β
m|4n+1
(β1m
))
q4n+1,
from which (iii) follows.Finally we consider (iv). Let rs(n) denote the number of ways in which
n can be represented as a sum of s squares. It is well known that (cf. [IR,pp. 279, 282], [HW, pp. 242, 314])
r2(n) = 4β
d|n, 2β€d(β1)(dβ1)/2 and r4(n) =
8β
d|nd if 2 β€n,
24β
d|n, 2β€dd if 2 |n.
Set Ο4(q) =ββn=0 cnq
n. From (iii) and the formula for r2(n) we see that
16Ο4(q) = (4Ο2(q))2 =(
ββ
m=0
r2(4m+ 1)qm)2
.
Thus, using the fact that r2(4m+ 3) = 0 we derive that
16cn =β
k+m=4n+2k,mβ‘1 (mod 4)
r2(k)r2(m) =β
1β€kβ€4n+2kβ‘1 (mod 4)
r2(k)r2(4n+ 2β k)
=β
1β€kβ€4n+2kβ‘1 (mod 2)
r2(k)r2(4n+ 2β k)
=β
0β€kβ€4n+2r2(k)r2(4n+ 2β k)β
β
0β€mβ€2n+1r2(2m)r2(4n+ 2β 2m)
368 Z. H. Sun and K. S. Williams
= r4(4n+ 2)ββ
0β€mβ€2n+1r2(m)r2(2n+ 1βm)
= r4(4n+ 2)β r4(2n+ 1) = 24β
d|4n+2, 2β€ddβ 8
β
d|2n+1d
= 16β
d|2n+1d.
So (iv) is true and hence the theorem is proved.
Corollary 4.1 (Ramanujan). If |q| < 1, then
qΟ(q)Ο(q7) =q
1β q βq3
1β q3 βq5
1β q5 +q9
1β q9 +q11
1β q11 βq13
1β q13 + Β· Β· Β· ,
where the cycle of coefficients is of length 14.
Proof. Sinceβ
m|n, 2β€m
(
m
7
)
=β
m|nmβ‘1,9,11 (mod 14)
1ββ
m|nmβ‘3,5,13 (mod 14)
1,
the result follows from Theorem 4.3(i).
Remark 4.1. Corollary 4.1 was first found by S. Ramanujan. In[B, pp. 302β303], Berndt wrote: βThe first two formulas (Corollary 4.1 isthe first item in Entry 17.) are of extreme interest, since they appear toindicate that Ramanujan was acquainted with a theorem equivalent to theaddition theorem for elliptic integrals of the second kind. Although it wouldappear to be very difficult to prove (i) without this addition theorem, itis apparently not found in the notebooks.β In 1999, Williams ([W]) gavea proof of Theorem 4.3(i) and Corollary 4.1 without the use of elliptic in-tegrals. Clearly we also prove the above result of Ramanujan without theaddition theorem for elliptic integrals.
From Theorem 4.3(ii) one can easily deduce
Corollary 4.2 (Ramanujan ([B], [W, p. 378])). If |q| < 1, then
qΟ(q2)Ο(q6) =q
1β q2 βq5
1β q10 +q7
1β q14 βq11
1β q22 + Β· Β· Β· .
Remark 4.2. By equating powers of qn in Theorem 4.3 we obtainβ£
β£
β£
β£
{
(x, y)
β£
β£
β£
β£
nβ 1 = x(x+ 1)
2+ 7 Β· y(y + 1)
2, x, y β N βͺ {0}
}β£
β£
β£
β£
=β
m|n, 2β€m
(
m
7
)
,
β£
β£
β£
β£
{
(x, y)
β£
β£
β£
β£
n =x(x+ 1)
2+ 3 Β· y(y + 1)
2, x, y β N βͺ {0}
}β£
β£
β£
β£
=β
m|2n+1
(
m
3
)
,
Ramanujan identities and Euler products 369
β£
β£
β£
β£
{
(x, y)
β£
β£
β£
β£
n =x(x+ 1)
2+y(y + 1)
2, x, y β N βͺ {0}
}β£
β£
β£
β£
=β
m|4n+1(β1)(mβ1)/2,
β£
β£
β£
β£
{
(x, y, z, t)
β£
β£
β£
β£
n =x(x+ 1)
2+y(y + 1)
2+z(z + 1)
2+t(t+ 1)
2,
x, y, z, t β N βͺ {0}}β£
β£
β£
β£
=β
d|2n+1d.
The last two formulae give the number of representations of n β N as the sumof two and four triangular numbers respectively. Proofs of these formulaehave been given by Adiga [A] and Ono, Robins and Wahl [ORW]. The latterformula was known to Legendre [L]. From Definition 2.2, Theorem 4.2 and[W, (53)] we deduce that if n = 3Ξ±n0 β₯ 2 (3 β€n0), then
|{(x, y) | nβ 1 = 3x(x+ 1)/2 + 5y(y + 1)/2, x, y β N βͺ {0}}|
=1
2
(
1 + (β1)Ξ±(
n03
))
β
m|n, 2β€m
(
m
15
)
and
|{(x, y) | nβ 2 = x(x+ 1)/2 + 15y(y + 1)/2, x, y β N βͺ {0}}|
=1
2
(
1β (β1)Ξ±(
n03
))
β
m|n, 2β€m
(
m
15
)
.
Using modular equations Ramanujan proved (see [B, p. 139])
Ο2(q) =
ββ
k=0
(β1)kqk(k+1) 1 + q2k+1
1β q2k+1 , Ο4(q2) =
ββ
n=0
(2n+ 1)q2n+1
1β q4n+2 ,
qΟ8(q) =ββ
n=0
n3qn
1β q2n (|q| < 1).
Theorem 4.4. Let n β N. Then
(i)
Ο1(n) =
(β1)
β
p=2x2+xy+3y2
ordp nβ‘1 (mod 3)
1
β
p=x2+xy+6y2 6=23(1 + ordp n)
if 2 | ordp n for every prime p with(
p23
)
= β1,and if ordp n β‘ 0, 1 (mod 3) for everyprime p = 2x2 + xy + 3y2,
0 otherwise.
370 Z. H. Sun and K. S. Williams
(ii) If 2 |n, then Ο2(n) = 0. If 2 β€n, then
Ο2(n) =
(β1)
β
p=3x2+2xy+4y2
ordp nβ‘1 (mod 3)
1
β
p=x2+11y2 6=11(1 + ordp n)
if 2 | ordp n for every odd prime p β‘ 2, 6, 7, 8, 10(mod 11), and if ordp n β‘ 0, 1 (mod 3) forevery prime p = 3x2 + 2xy + 4y2,
0 otherwise.
(iii) If 2 |n or 3 |n, then Ο6(n) = 0. If 2 β€n and 3 β€n, then
Ο6(n) =
(β1)
β
p=4x2+2xy+7y2
ordp nβ‘1 (mod 3)
1
β
p=x2+27y2
(1 + ordp n)
if 2 | ordp n for every prime p β‘ 5 (mod 6),and if ordp n β‘ 0, 1 (mod 3) for everyprime p = 4x2 + 2xy + 7y2,
0 otherwise.
Proof. From Theorem 2.2 we know that
(4.1)
Ο1(n) =12 (R(1, 1, 6;n)βR(2, 1, 3;n)) = F ([2, 1, 3], n) (d = β23),
Ο2(n) =12 (R(1, 0, 11;n)βR(3, 2, 4;n)) = F ([3, 2, 4], n) (d = β44),
Ο6(n) =12 (R(1, 0, 27;n)βR(4, 2, 7;n)) = F ([4, 2, 7], n) (d = β108).
Thus applying [SW, Theorem 10.2] we obtain the result.
Theorem 4.5. Let n β N. Then
(i)
Ο3(n) =
(β1)Β΅β
pβ‘1,4,16 (mod 21)(1 + ordp n)
if 3 β€n and 2 | ordp n for everyprime p 6β‘ 1, 4, 7, 16 (mod 21),
0 otherwise,
where
Β΅ =β
pβ‘2,8,11 (mod 21)ordp nβ‘2 (mod 4)
1 +β
p=4x2+xy+4y2
ordp nβ‘1 (mod 2)
1.
Ramanujan identities and Euler products 371
(ii)
Ο4(n) =
(β1)Β΅β
pβ‘1,9 (mod 20)(1 + ordp n)
if 2 β€n and 2 | ordp nfor every prime p 6β‘ 1, 5, 9 (mod 20),
0 otherwise,
where
Β΅ =β
pβ‘3,7 (mod 20)ordp nβ‘2 (mod 4)
1 +β
p=4x2+5y2
ordp nβ‘1 (mod 2)
1.
(iii)
Ο8(n) =
(β1)Β΅β
pβ‘1 (mod 8)(1 + ordp n)
if 2 β€n and 2 | ordp nfor every prime p 6β‘ 1 (mod 8),
0 otherwise,
where
Β΅ =β
pβ‘3 (mod 8)ordp nβ‘2 (mod 4)
1 +β
p=4x2+4xy+9y2
ordp nβ‘1 (mod 2)
1.
(iv)
Ο12(n) =
(β1)Β΅β
pβ‘1 (mod 12)(1 + ordp n)
if (n, 6) = 1 and 2 | ordp nfor every prime p 6β‘ 1 (mod 12),
0 otherwise,
where
Β΅ =β
pβ‘5 (mod 12)ordp nβ‘2 (mod 4)
1 +β
p=4x2+9y2
ordp nβ‘1 (mod 2)
1.
Proof. From [SW, Proposition 11.1(i)] we know that H(d) is a cyclicgroup of order 4 when d β {β63,β80,β128,β144}. By Theorem 2.2 wehave
Ο3(n) =12 (R(1, 1, 16;n)βR(4, 1, 4;n)) = F ([2, 1, 8], n) (d = β63),
Ο4(n) =12 (R(1, 0, 20;n)βR(4, 0, 5;n)) = F ([3, 2, 7], n) (d = β80),
Ο8(n) =12 (R(1, 0, 32;n)βR(4, 4, 9;n)) = F ([3, 2, 11], n) (d = β128),
Ο12(n) =12 (R(1, 0, 36;n)βR(4, 0, 9;n)) = F ([5, 4, 8], n) (d = β144).
For any prime p it is clear that
p β R([2, 1, 8]) β p β‘ 2, 8, 11 (mod 21),p 6= 7, p β R([1, 1, 16]) βͺR([4, 1, 4]) β p β‘ 1, 4, 16 (mod 21),p β R([3, 2, 7]) β p β‘ 3, 7 (mod 20),
372 Z. H. Sun and K. S. Williams
p 6= 2, 5, p β R([1, 0, 20]) βͺR([4, 0, 5]) β p β‘ 1, 9 (mod 20),p β R([3, 2, 11]) β p β‘ 3 (mod 8),p 6= 2, p β R([1, 0, 32]) βͺR([4, 4, 9]) β p β‘ 1 (mod 8),p β R([5, 4, 8]) β p β‘ 5 (mod 12),p 6= 2, p β R([1, 0, 36]) βͺR([4, 0, 9]) β p β‘ 1 (mod 12).
Thus applying [SW, Theorem 11.1] in the cases d = β63,β80,β128,β144yields the result.
5. The Euler product for L(M, s) (M β H(d)). Let d be a discrimi-nant. Suppose
(5.1) H(d) = {Ak11 Β· Β· Β·Akrr | 0 β€ k1 < h1, . . . , 0 β€ kr < hr}with h1 Β· Β· Β·hr = h(d). For n β N and M = Am11 Β· Β· Β·Amrr β H(d), following[SW, Definition 7.1] we define
F (M,n) =1
w(d)
β
0β€k1<h1Β·Β·Β·
0β€kr<hr
cos 2Ο
(
k1m1h1+ Β· Β· Β·+ krmr
hr
)
(5.2)
ΓR(Ak11 Β· Β· Β·Akrr , n).Let N(n, d) =
β
MβH(d)R(M,n). Let s β C be such that Re(s) > 1.
From [SW, Theorem 4.1] and the same argument as in the proof of[HKW, Corollary 9.1] we know that for any Ξ΅ > 0 there exists a con-stant C(Ξ΅) such that N(n, d) β€ C(Ξ΅)nΞ΅. Letting Ξ΅ β (0,Re(s) β 1) wesee that
ββn=1N(n, d)n
βs converges absolutely. Henceββn=1R(M,n)nβs
andββn=1 F (M,n)nβs converge absolutely since R(M,n) β€ N(n, d) and
|F (M,n)|β€N(n, d)/w(d). Using the same argument, for p > 1 and Re(s)>1we see that
ββt=1 F (M,pt)pβst converges absolutely.
Definition 5.1. Let d be a discriminant and M β H(d). Let s β C besuch that Re(s) > 1. Define
Z(M, s) =
ββ
n=1
R(M,n)
nsand L(M, s) =
ββ
n=1
F (M,n)
ns.
Then Z(M, s) and L(M, s) are analytic functions of s in Re(s) > 1, andthey can be continued analytically to the whole complex plane except for asimple pole at s = 1.
Let d be a discriminant and M = [a, b, c] β H(d). For Re(s) > 1 it isclear that
Z(M, s) =ββ
n=1
R(M,n)
ns=ββ
n=1
R(a, b, c;n)
ns
Ramanujan identities and Euler products 373
=ββ
n=1
β
{x,y} is primaryn=ax2+bxy+cy2
1
ns=
β
{x,y} is primary
1
(ax2 + bxy + cy2)s.
Thus,
(5.3) Z(M, s) =β
x,yβZ(x,y) 6=(0,0)
1
(ax2 + bxy + cy2)sfor d < 0
and
(5.4) Z(M, s) =β
x,yβZ
2ax+(bββd)y>0
1β€β£
β£2ax+(b+
β
d)y
2ax+(bββ
d)y
β£
β£<Ξ΅(d)2
1
(ax2 + bxy + cy2)sfor d > 0.
For a negative discriminant d and M β H(d), by (5.3) and a classicalresult due to M. Lerch (see [D], [ZW], [SC]) we have the following functionalequation for Z(M, s):
(5.5)
(ββd2Ο
)s
Ξ (s)Z(M, s) =
(ββd2Ο
)1βsΞ (1β s)Z(M, 1β s),
where Ξ (s) is the Gamma function.For a positive discriminant d and M β H(d), we do not know if Z(M, s)
has a functional equation like (5.5).
Theorem 5.1. Let d be a discriminant. Let M β H(d) and s β C.
(i) If Re(s) > 1, then
L(M, s) =β
p
(
1 +ββ
t=1
F (M,pt)pβst)
.
(ii) If d < 0 and s 6= 0, 1, then(ββd2Ο
)s
Ξ (s)L(M, s) =
(ββd2Ο
)1βsΞ (1β s)L(M, 1β s).
Proof. From [SW, Theorem 7.2] we know that F (M,n) is a multi-plicative function of n β N. By the previous argument, L(M, s) andββt=1 F (M,pt)pβst converge absolutely if Re(s) > 1. Thus, for m β N andRe(s) > 1 we have
β
pβ€m
(
1 +ββ
t=1
F (M,pt)pβst)
=β
n
F(
M,β
pβ€mpordp n
)(
β
pβ€mpordp n
)βs
=mβ
n=1
F (M,n)
ns+Rm(M, s),
374 Z. H. Sun and K. S. Williams
where the first sum is taken over all those positive integers n whose primedivisors are less than m, and clearly
|Rm(M, s)| β€ββ
n=m+1
|F (M,n)nβs|.
Sinceββn=1 F (M,n)nβs converges absolutely, we see that if m β β, then
ββn=m+1 |F (M,n)nβs| β 0 and so Rm(M, s)β 0. Therefore
β
p
(
1 +ββ
t=1
F (M,pt)pβst)
converges and
L(M, s) =ββ
n=1
F (M,n)
ns=β
p
(
1 +ββ
t=1
F (M,pt)pβst)
for Re(s) > 1.
This proves (i).Now we consider (ii). Suppose H(d) is given by (5.1), d < 0 and M =
Am11 Β· Β· Β·Amrr . From (5.2) we see that
L(M, s) =1
w(d)
β
0β€k1<h1Β·Β·Β·
0β€kr<hr
cos 2Ο
(
k1m1h1+ Β· Β· Β·+ krmr
hr
) ββ
n=1
R(Ak11 Β· Β· Β·Akrr , n)ns
=1
w(d)
β
0β€k1<h1Β·Β·Β·
0β€kr<hr
cos 2Ο
(
k1m1h1+ Β· Β· Β·+ krmr
hr
)
Β· Z(Ak11 Β· Β· Β·Akrr , s).
Thus applying (5.5) we obtain
w(d)
(ββd2Ο
)s
Ξ (s)L(M, s)
=β
0β€k1<h1Β·Β·Β·
0β€kr<hr
cos 2Ο
(
k1m1h1+ Β· Β· Β·+ krmr
hr
)
Β·(ββd2Ο
)s
Ξ (s)Z(Ak11 Β· Β· Β·Akrr , s)
=β
0β€k1<h1Β·Β·Β·
0β€kr<hr
cos 2Ο
(
k1m1h1+ Β· Β· Β·+ krmr
hr
)
Β·(ββd2Ο
)1βs
Γ Ξ (1β s)Z(Ak11 Β· Β· Β·Akrr , 1β s)
= w(d)
(ββd2Ο
)1βsΞ (1β s)L(M, 1β s).
So (ii) is true and the proof is complete.
Ramanujan identities and Euler products 375
Let d be a discriminant with conductor f , and let H(d) be given by (5.1).ForK β H(d) we useR(K) to denote the set of integers represented by formsin K. Let p be a prime not dividing the conductor f . When p | d, from [MW,Lemma 5.3] (or [SW, Theorem 8.1(ii)]) we know that p is represented by
exactly one class A β H(d) and A = AΞ΅1h1/21 Β· Β· Β·AΞ΅rhr/2r with Ξ΅1, . . . , Ξ΅r β
{0, 1}.Let M = Am11 Β· Β· Β·Amrr β H(d) and t β N βͺ {0}. By [SW, Theorem 8.1]
we have
(5.6) F (M,pt)
=
(1 + (β1)t)/2 if
(
d
p
)
= β1,
(β1)tβrj=1 Ξ΅jmj if p | d and p β R(AΞ΅1h1/21 Β· Β· Β·AΞ΅rhr/2r ),
Ut
(
cos 2Οrβ
j=1
ajmjhj
)
if p β€ d and p β R(Aa11 Β· Β· Β·Aarr ),
where {Un(x)} is the Chebyshev polynomial of the second kind given by
(5.7) U0(x) = 1, U1(x) = 2x, Un+1(x) = 2xUn(x)βUnβ1(x) (n β₯ 1).
It is well known that (cf. [MOS, p. 259])
(5.8)ββ
n=0
Un(x)qn =
1
1β 2xq + q2 (|x| < 1, |q| < 1).
Thus, if p is a prime such that(
dp
)
= 1 and so p is represented by some class
Aa11 Β· Β· Β·Aarr β H(d), then for s β C with Re(s) > 1 we have
(5.9) 1 +ββ
t=1
F (Am11 Β· Β· Β·Amrr , pt)pβst
=ββ
t=0
Ut
(
cos 2Ο
(
a1m1h1+ Β· Β· Β·+ armr
hr
))
Β· pβst
=1
1β 2 cos 2Ο(
a1m1h1+ Β· Β· Β·+ armrhr
)
Β· pβs + pβ2s .
Now we are in a position to give
Theorem 5.2. Let d be a discriminant with conductor f . Let H(d) begiven by (5.1). Let M = Am11 Β· Β· Β·Amrr β H(d) and s β C with Re(s) > 1.Then
376 Z. H. Sun and K. S. Williams
ββ
n=1(n,f)=1
F (M,n)
ns
=β
( dp )=β1
1
1β pβ2sβ
pβR(AΞ΅1h1/21 Β·Β·Β·AΞ΅rhr/2r )p|d, pβ€f
1
1β (β1)Ξ΅1m1+Β·Β·Β·+Ξ΅rmrpβs
Γβ
pβR(Aa11 Β·Β·Β·Aarr )pβ€d
1
1β 2 cos 2Ο(
a1m1h1+ Β· Β· Β·+ armrhr
)
Β· pβs + pβ2s ,
where Ξ΅1, . . . , Ξ΅r β {0, 1} are chosen such that p β R(AΞ΅1h1/21 Β· Β· Β·AΞ΅rhr/2r )
when p | d and p β€ f , and a1, . . . , ar β Z are chosen so that p β R(Aa11 Β· Β· Β·Β· Β· Β·Aarr ) when
(
dp
)
= 1.
Proof. Since F (M,n) is a multiplicative function of n (see [SW, Theorem7.2]), by the argument similar to the proof of Theorem 5.1(i) we see that
(5.10)β
pβ€f
(
1 +
ββ
t=1
F (M,pt)pβst)
=
ββ
n=1(n,f)=1
F (M,n)
ns.
Let p be a prime not dividing f . If(
dp
)
= β1, it follows from (5.6) that
1 +ββ
t=1
F (M,pt)pβst =ββ
t=0
1 + (β1)t2
pβst =1
1β pβ2s .
If p | d, then p is represented by exactly one class A in H(d), and A =AΞ΅1h1/21 Β· Β· Β·AΞ΅rhr/2r with Ξ΅1, . . . , Ξ΅r β {0, 1}. By (5.6) we obtain
1 +
ββ
t=1
F (M,pt)pβst =ββ
t=0
(β1)tβrj=1 Ξ΅jmjpβst =
1
1β (β1)βrj=1 Ξ΅jmjpβs
.
If(
dp
)
= 1 so that p is represented by some class Aa11 Β· Β· Β·Aarr β H(d), by(5.9) we have
1 +ββ
t=1
F (M,pt)pβst =1
1β 2 cos 2Ο(
a1m1h1+ Β· Β· Β·+ armrhr
)
Β· pβs + pβ2s .
Now putting all the above together we deduce the desired result.
From Theorem 5.2 we have
Corollary 5.1. Let d be a discriminant with conductor f and 2 β€h(d).Let H(d) be given by (5.1). For M = Am11 Β· Β· Β·Amrr β H(d) and s β C withRe(s) > 1 we have
Ramanujan identities and Euler products 377
ββ
n=1(n,f)=1
F (M,n)
ns
=β
( dp )=β1
1
1β pβ2sβ
p|d, pβ€f
1
1β pβs
Γβ
pβR(Aa11 Β·Β·Β·Aarr )pβ€d
1
1β 2 cos 2Ο(
a1m1h1+ Β· Β· Β·+ armrhr
)
Β· pβs + pβ2s ,
where a1, . . . , ar β Z are determined by p β R(Aa11 Β· Β· Β·Aarr ) when(
dp
)
= 1.
Proof. If p is a prime such that p | d and p β€ f , then we must have p β R(I)since 2 β€h(d). Now the result follows from Theorem 5.2.
Let d be a discriminant with conductor f . Suppose m β N and m | f .In [SW, Lemma 2.1] we showed that any class in H(d) can be written as[a, bm, cm2], where a, b, c β Z, (a,m) = 1 and gcd(a, b, c) = 1. Following[SW] and [KW1] we define Ο1,m([a, bm, cm
2]) = [a, b, c]. From [SW, Theorem2.1] or [KW1, p. 355] we know that Ο1,m is a surjective homomorphism fromH(d) to H(d/m2). Thus, if H(d) is cyclic with generator A, then Ο1,m(A)is a generator of H(d/m2).
Theorem 5.3. Let d be a discriminant with conductor f and d0 = d/f2.
Suppose that H(d) is cyclic with generator A and order h. Let k β Z ands β C with Re(s) > 1. For a prime p let Ξ±p = ordp f , hp = h(d/p2Ξ±p),and let Ξ²k,p denote the maximum j β {0, 1, . . . , Ξ±p} such that h/h(d/p2j) | k.Then
L(Ak, s)
=β
pβ€f
1
1β(
1 +(
d0p
))
cos2Οkaph pβs +
(
d0p
)
pβ2s
β
p|fhβ€khp
1β p(1β2s)(1+Ξ²k,p)1β p1β2s
Γβ
p|fh|khp
(
1β pΞ±p(1β2s)1β p1β2s +
pΞ±p(1β2s)β1(
pβ(
d0p
))
1β(
1 +(
d0p
))
cos2Οkaph pβs +
(
d0p
)
pβ2s
)
,
where ap β Z is uniquely determined by 0 β€ ap β€ hp/2, p β R(Aapp ) andAp = Ο1,pΞ±p (A).
Proof. Let p be a prime such that p | f (that is Ξ±p β N). Since
h(d/p2Ξ²k,p) |h(d/p2j) for j β€ Ξ²k,p,
378 Z. H. Sun and K. S. Williams
we see that
h/h(d/p2j) | k for j = 0, 1, . . . , Ξ²k,p.
Thus, applying [SW, Theorem 8.4] we obtain
1 +
2Ξ±pβ1β
t=1
F (Ak, pt)pβst
= 1 +
Ξ±pβ1β
j=1
F (Ak, p2j)pβ2js =
min{Ξ²k,p,Ξ±pβ1}β
j=0
pj Β· pβ2js
=
min{Ξ²k,p,Ξ±pβ1}β
j=0
pj(1β2s) =1β pmin{Ξ²k,p+1,Ξ±p}(1β2s)
1β p1β2s .
Now suppose t β N and t β₯ 2Ξ±p. From [SW, Theorem 8.3(ii)] (with n = pt,m = pΞ±p and s = k) we know that
F (Ak, pt) =
pΞ±p(
1β 1p
(
d0p
))
F (Akpp , ptβ2Ξ±p) if Ξ²k,p = Ξ±p,
0 if Ξ²k,p < Ξ±p,
where
Ap = Ο1,pΞ±p (A) β H(d/p2Ξ±p) and kp = khp/h β Z.
Thus, if Ξ²k,p = Ξ±p, then
ββ
t=2Ξ±p
F (Ak, pt)pβst = pΞ±p(
1β 1p
(
d0p
)) ββ
t=2Ξ±p
F (Akpp , ptβ2Ξ±p)pβst
= pΞ±p(1β2s)(
1β 1p
(
d0p
)) ββ
j=0
F (Akpp , pj)pβjs.
Since Ο1,pΞ±p is a surjective homomorphism from H(d) to H(d/p2Ξ±p), we seethat Ap is a generator of H(d/p
2Ξ±p). Note that
d/p2Ξ±p = d0(f/pΞ±p)2, p β€ f/pΞ±p and k/h = kp/hp.
By the proof of Theorem 5.2 the sum
1 +ββ
j=1
F (Akpp , pj)pβjs
Ramanujan identities and Euler products 379
is equal to
1
1β pβ2s if
(
d0p
)
= β1,1
1β (β1)Ξ΅pkp pβs if p | d0 and p β R(AΞ΅php/2p ),
1
1β 2 cos 2Οapkphppβs + pβ2s
if
(
d0p
)
= 1 and p β R(Aapp ),
=1
1β(
1 +(
d0p
))
cos2Οkaph pβs +
(
d0p
)
pβ2s,
where Ξ΅p β {0, 1} is given by p β R(AΞ΅php/2p ) when p | d0, and ap β Z is
determined by 0 β€ ap β€ hp/2 and p β R(Aapp ) when(
d0p
)
= 0, 1. Thus
1 +ββ
t=1
F (Ak, pt)pβst = 1 +
2Ξ±pβ1β
t=1
F (Ak, pt)pβst +ββ
t=2Ξ±p
F (Ak, pt)pβst
=
1β p(1β2s)(1+Ξ²k,p)1β p1β2s if Ξ²k,p < Ξ±p,
1β pΞ±p(1β2s)1β p1β2s +
pΞ±p(1β2s)β1(
pβ(
d0p
))
1β(
1 +(
d0p
))
cos2Οkaph pβs +
(
d0p
)
pβ2sif Ξ²k,p = Ξ±p.
By (5.10) and Theorem 5.2 we have
β
pβ€f
(
1 +ββ
t=1
F (Ak, pt)pβst)
=β
( dp )=β1
1
1β pβ2sβ
pβR(AΞ΅h/2)p|d, pβ€f
1
1β (β1)kΞ΅pβs
Γβ
pβR(Aa)( dp )=1
1
1β 2 cos 2Οkah pβs + pβ2s
=β
pβ€f
1
1β(
1 +(
d0p
))
cos2Οkaph pβs +
(
d0p
)
pβ2s,
where Ξ΅ β {0, 1} is given by p β R(AΞ΅h/2) when p | d, and a β Z is determinedby 0 β€ a β€ h/2 and p β R(Aa) when
(
dp
)
= 0, 1.
Note that Ap = A when p β€ f . Putting all the above together with The-orem 5.1(i) gives the result.
Corollary 5.2. Let d be a discriminant with conductor f and d0 =d/f2. Suppose that H(d) is cyclic with generator A and h = h(d) β‘ 1(mod 2). Let k β Z. For a prime p let Ξ±p = ordp f , hp = h(d/p
2Ξ±p), and letΞ²k,p denote the maximum number j β {0, 1, . . . , Ξ±p} such that h/h(d/p2j) | k.
380 Z. H. Sun and K. S. Williams
For s β C with Re(s) > 1 we have
L(Ak, s)
=β
( dp )=β1
1
1β pβ2sβ
p|d, pβ€f
1
1β pβsβ
pβR(Aa)( dp )=1
1
1β 2 cos 2Οkah pβs + pβ2s
Γβ
p|fhβ€khp
1β p(1β2s)(1+Ξ²k,p)1β p1β2s
β
p|f, h|khp(d0p )=β1
(
1β pΞ±p(1β2s)1β p1β2s +
pΞ±p(1β2s)β1(p+ 1)
1β pβ2s)
Γβ
p|f, h|khpp|d0
(
1β pΞ±p(1β2s)1β p1β2s +
pΞ±p(1β2s)
1β pβs)
Γβ
p|f, h|khp(d0p )=1, pβR(A
app )
(
1β pΞ±p(1β2s)1β p1β2s +
pΞ±p(1β2s)β1(pβ 1)1β 2 cos 2Οkaph pβs + pβ2s
)
,
where a β Z is determined by 0 β€ a β€ (h β 1)/2 and p β R(Aa) when(
dp
)
= 1, Ap = Ο1,pΞ±p (A), and ap β Z is determined by 0 β€ ap β€ (hp β 1)/2and p β R(Aapp ) when p | f and
(
d0p
)
= 1.
Proof. If p is a prime such that p | d and p β€ f , from [MW, Lemma 5.3]or [SW, Lemma 5.2] we know that p is represented by unique class M inH(d) and M =Mβ1. Thus we must have p β R(I) since H(d) is cyclic and2 β€h(d). Now the result follows from Theorem 5.3.
Theorem 5.4. Let d be a discriminant such that h(d) = 5. Let f be theconductor of d, and let A be a generator of H(d). Let
F (A, n) =1
w(d)
(
R(I, n) +
β5β 12
R(A, n)ββ5 + 1
2R(A2, n)
)
,
F (A2, n) =1
w(d)
(
R(I, n)ββ5 + 1
2R(A, n) +
β5β 12
R(A2, n)
)
.
For s β C with Re(s) > 1 we have
ββ
n=1(n,f)=1
F (A, n)
ns=β
( dp )=β1
1
1β pβ2sβ
p|dpβ€f
1
1β pβsβ
pβR(I)pβ€d
1
(1β pβs)2
Γβ
pβR(A)
1
1ββ5β12 pβs + pβ2s
β
pβR(A2)
1
1 +β5+12 pβs + pβ2s
,
Ramanujan identities and Euler products 381
ββ
n=1(n,f)=1
F (A2, n)
ns=β
( dp )=β1
1
1β pβ2sβ
p|dpβ€f
1
1β pβsβ
pβR(I)pβ€d
1
(1β pβs)2
Γβ
pβR(A2)
1
1ββ5β12 pβs + pβ2s
β
pβR(A)
1
1 +β5+12 pβs + pβ2s
.
Proof. The result follows from Corollary 5.1, [SW, Theorem 7.4] and thefacts
cos2Ο
5= sin
Ο
10=
β5β 14
, cos4Ο
5= β cos Ο
5= ββ5 + 1
4.
From Theorem 5.2 and [SW, Theorem 7.4] we can easily deduce
Theorem 5.5. Let d be a discriminant such that h(d) = 6. Let f be theconductor of d, and let A be a generator of H(d). Let
F (A, n) =1
w(d)(R(I, n) +R(A, n)βR(A2, n)βR(A3, n)),
F (A2, n) =1
w(d)(R(I, n)βR(A, n)βR(A2, n) +R(A3, n)),
F (A3, n) =1
w(d)(R(I, n)β 2R(A, n) + 2R(A2, n)βR(A3, n)).
For j = 1, 2, 3 and s β C with Re(s) > 1 we have
ββ
n=1(n,f)=1
F (Aj , n)
ns
=β
( dp )=β1
1
1β pβ2sβ
pβR(I)p|d, pβ€f
1
1β pβsβ
pβR(A3)p|d, pβ€f
1
1β (β1)jpβs
Γβ
pβR(I)pβ€d
1
(1β pβs)2β
pβR(A)
1
1β cjpβs + pβ2s
Γβ
pβR(A2)
1
1β (β1)jcjpβs + pβ2sβ
pβR(A3)pβ€d
1
(1β (β1)jpβs)2 ,
where
cj = 2 cosjΟ
3=
1 if j = 1,
β1 if j = 2,β2 if j = 3.
382 Z. H. Sun and K. S. Williams
6. The Euler product for L(A, s) when H(d) = {I, A}. Let d be adiscriminant with h(d) = 2. Suppose I is the principal class and A is thegenerator of H(d). We recall that
F (A, n) =
{
R(I, n)βR(A, n) if d > 0,12 (R(I, n)βR(A, n)) if d < 0
and
L(A, s) =ββ
n=1
F (A, n)
ns(Re(s) > 1).
Putting h = 2 and k = 1 in Theorem 5.3 we deduce
Theorem 6.1. Let d be a discriminant with conductor f . Suppose h(d)= 2 and H(d) = {I, A}. For a prime p let Ξ±p be the nonnegative integersuch that pΞ±p β f , hp = h(d/p2Ξ±p), and let Ξ²p denote the maximum j β{0, 1, . . . , Ξ±p} such that h(d/p2j) = 2. Then for s β C with Re(s) > 1 wehave
L(A, s) =β
p|d, pβ€fpβR(I)
1
1β pβsβ
p|d, pβ€fpβR(A)
1
1 + pβs
β
( dp )=β1
1
1β pβ2s
Γβ
pβR(I)pβ€d
1
(1β pβs)2β
pβR(A)pβ€d
1
(1 + pβs)2
β
p|fhp=1
1β p(1β2s)(1+Ξ²p)1β p1β2s
Γβ
p|fhp=2
(
1β pΞ±p(1β2s)1β p1β2s +
pΞ±p(1β2s)β1(
pβ(
d0p
))
1β (β1)ap(
1 +(
d0p
))
pβs +(
d0p
)
pβ2s
)
,
where ap = 0 or 1 according as p is represented by the principal class inH(d/p2Ξ±p) or not.
Theorem 6.2. Let d < 0 be a discriminant with h(d) = 2, which isgiven in [SW,Table 9.1]. Let f be the conductor of d and H(d) = {I, A}with A2 = I. Let s β C with Re(s) > 1.
(i) If d 6= β60, then
L(A, s) =β
( dp )=β1
1
1β pβ2sβ
p|d, pβ€fpβR(I)
1
1β pβsβ
p|d, pβ€fpβR(A)
1
1 + pβs
Γβ
pβR(I)pβ€d
1
(1β pβs)2β
pβR(A)pβ€d
1
(1 + pβs)2,
where I, A and the conditions for p β R(I) and p β R(A) are givenin [SW, Table 9.1].
Ramanujan identities and Euler products 383
(ii) If d = β60, then I = [1, 0, 15], A = [3, 0, 5] and
L(A, s)
=1 + 21βs + 21β2s
(1 + 2βs)2Β· 1
1 + 3βsΒ· 1
1 + 5βs
β
pβ‘7,11,13,14 (mod 15)
1
1β pβ2s
Γβ
pβ‘1,4 (mod 15)
1
(1β pβs)2β
pβ‘2,8 (mod 15)p 6=2
1
(1 + pβs)2.
Proof. If d 6= β60 and p is a prime such that p | f , by [SW, Tables 9.1,9.2 and (9.3)] we see that h(d/p2) = 1. Hence hp = h(d/p2Ξ±p) = 1 forΞ±p = ordp f by [SW, Theorem 8.3(ii)]. Now applying Theorem 6.1 and [SW,Table 9.1] we obtain (i).
Now suppose d = β60. Then f = 2 and d0 = d/f2 = β15. Let p bean odd prime. By [SW, Table 9.1] we have H(β60) = {[1, 0, 15], [3, 0, 5]},H(β15) = {[1, 1, 4], [2, 1, 2]}, p β R([1, 0, 15]) β p β‘ 1, 4 (mod 15) β p βR([1, 1, 4]), and p β R([3, 0, 5]) β p β R([2, 1, 2]) β p = 3, 5 or p β‘ 2, 8(mod 15). If p is a prime such that p | f , we must have p = 2, Ξ±p = 1,hp = 2 and
(
d0p
)
=(β152
)
= 1. Thus applying Theorem 6.1 and the above
we obtain (ii). The proof is now complete.
Theorem 6.3. Let s β C with Re(s) > 1. Then
ββ
n=1
Ο2(n)
ns=
1
1 + 3βs
β
pβ‘5 (mod 6)
1
1β pβ2sβ
pβ‘1 (mod 12)
1
(1β pβs)2
Γβ
pβ‘7 (mod 12)
1
(1 + pβs)2
and
ββ
n=1
Ο4(n)
ns
=β
pβ‘3 (mod 4)
1
1β pβ2sβ
pβ‘1 (mod 8)
1
(1β pβs)2β
pβ‘5 (mod 8)
1
(1 + pβs)2.
Proof. From [SW, Table 9.1] we see that H(β48) = {[1, 0, 12], [3, 0, 4]}and H(β64) = {[1, 0, 16], [4, 4, 5]}. Thus Ο2(n) = F ([3, 0, 4], n) and Ο4(n) =F ([4, 4, 5], n) by Theorem 2.1. Now applying Theorem 6.2(i) and [SW, Table9.1] we obtain the result.
384 Z. H. Sun and K. S. Williams
Theorem 6.4. For n β N we have
f4(1, 4;n) + f4(3, 4;n) =
(β1)(nβ1)/2β
m|n
(
m
7
)
if 2 β€n,
0 if 2 |n.For s β C with Re(s) > 1 we have
ββ
n=1
f4(1, 4;n) + f4(3, 4;n)
ns
=1
1 + 7βs
β
pβ‘3,5,6 (mod 7)
1
1β pβ2sβ
pβ‘1,9,25 (mod 28)
1
(1β pβs)2
Γβ
pβ‘11,15,23 (mod 28)
1
(1 + pβs)2.
Proof. From Theorem 2.3 and [SW, Table 9.1] we see that
f4(1, 4;n) + f4(3, 4;n) =12 (R(1, 0, 28;n)βR(4, 0, 7;n)) = F ([4, 0, 7], n).
But according to [SW, Theorem 9.2] and Lemma 3.1,
F ([4, 0, 7], n) =
(β1)(nβ1)/2β
m|n
(β7m
)
= (β1)(nβ1)/2β
m|n
(
m
7
)
if 2 β€n,
0 if 2 |n.Thus the result follows from Theorem 6.2 and [SW, Table 9.1].
Remark 6.1. Comparing Theorems 4.1 and 6.4 we conclude thatf4(1, 4;n) + f4(3, 4;n) = (β1)(nβ1)/2Ο1(n) for odd n.
7. The Euler product for L(A, s) when H(d) = {I, A,A2}. Let dbe a discriminant with h(d) = 3. Suppose I is the principal class and A is agenerator of H(d). We recall that
F (A, n) =1
w(d)(R(I, n)βR(A, n)) =
{
R(I, n)βR(A, n) if d > 0,12 (R(I, n)βR(A, n)) if d < 0
and
L(A, s) =
ββ
n=1
F (A, n)
ns(Re(s) > 1).
Theorem 7.1. Let d < 0 be a discriminant with h(d) = 3. (The valuesof such d are given for example in [WH, Proposition] or [SW, Lemma 10.1].)Let f be the conductor of d and H(d) = {I, A,A2} with A3 = I. Let s β Cwith Re(s) > 1.
Ramanujan identities and Euler products 385
(i) If d 6= β92,β124, then
L(A, s) =β
p|dpβ€f
1
1β pβsβ
( dp )=β1
1
1β pβ2s
Γβ
pβR(I)pβ€d
1
(1β pβs)2β
pβR(A)
1
1 + pβs + pβ2s.
(ii) If d = β92, then I = [1, 0, 23], A = [3, 2, 8] and
L(A, s) =1 + 2βs + 21β2s
1 + 2βs + 2β2sΒ· 1
1β 23βsβ
( p23 )=β1
1
1β pβ2s
Γβ
p=x2+23y2 6=23
1
(1β pβs)2β
p=3x2+2xy+8y2
1
1 + pβs + pβ2s.
(iii) If d = β124, then I = [1, 0, 31], A = [5, 4, 7] and
L(A, s) =1 + 2βs + 21β2s
1 + 2βs + 2β2sΒ· 1
1β 31βsβ
( p31 )=β1
1
1β pβ2s
Γβ
p=x2+31y2 6=31
1
(1β pβs)2β
p=5x2+4xy+7y2
1
1 + pβs + pβ2s.
Proof. We first suppose d 6= β92,β124. Let p be a prime dividing theconductor f and pΞ±p β f . From [SW, (9.3)] and [SW, Lemma 10.1] we knowthat h(d/p2) = 1 and therefore h(d/p2j) = 1 for j = 1, . . . , Ξ±p by [SW,Remark 2.2]. Thus putting h = 3 and k = 1 in Corollary 5.2 yields theresult in this case.
If d = β92,β124 and p is a prime such that p | f , then p = f = 2,(d/4p
)
= 1 and h(d/p2) = 3. We note that H(β92) = {[1, 0, 23], [3, 2, 8],[3,β2, 8]}, H(β124) = {[1, 0, 31], [5, 4, 7], [5,β4, 7]}, H(β23) = {[1, 1, 6],[2, 1, 3], [2,β1, 3]}, H(β31) = {[1, 1, 8], [2, 1, 4], [2,β1, 4]}. Thus puttingh = 3, k = 1 and d = β92,β124 in Corollary 5.2 we obtain (ii) and (iii).The proof is now complete.
Theorem 7.2. Let s β C with Re(s) > 1. Then
ββ
n=1
Ο1(n)
ns=
1
1β 23βsβ
( p23 )=β1
1
1β pβ2sβ
p=2x2+xy+3y2
1
1 + pβs + pβ2s
Γβ
p=x2+xy+6y2 6=23
1
(1β pβs)2 ,
386 Z. H. Sun and K. S. Williams
ββ
n=1
Ο2(n)
ns=
1
1β 11βsβ
pβ‘2,6,7,8,10 (mod 11)p 6=2
1
1β pβ2s
Γβ
p=3x2+2xy+4y2
1
1 + pβs + pβ2s
Γβ
p=x2+11y2 6=11
1
(1β pβs)2
and
ββ
n=1
Ο6(n)
ns=
β
pβ‘5 (mod 6)
1
1β pβ2sβ
p=x2+27y2
1
(1β pβs)2
Γβ
p=4x2+2xy+7y2
1
1 + pβs + pβ2s.
Proof. The result follows from the fact that h(β23) = h(β44) = h(β108)= 3, (4.1) and Theorem 7.1.
Remark 7.1. Note that an odd prime p is represented by x2+xy+6y2 ifand only if p is represented by x2+23y2. The formula for Ο1(n) in Theorem7.2 was essentially conjectured by Ramanujan ([R1]). In [Ra], Rangacharioutlined a proof of this result using class field theory and modular forms.The formula for Ο2(n) in Theorem 7.2 corrects the incorrect formula ofRamanujan and Rangachari (see [Ra]).
Theorem 7.3. For s β C with Re(s) > 1 we have
ββ
n=1
f2(1, 5;n) + f2(3, 5;n)
ns
=1
1β 19βsβ
( p19 )=β1p 6=2
1
1β pβ2sβ
p=x2+19y2 6=19
1
(1β pβs)2
Γβ
p=4x2+2xy+5y2
1
1 + pβs + pβ2s.
Proof. From Theorem 2.4 we see that
f2(1, 5;n) + f2(3, 5;n) =12 (R(1, 0, 19;n)βR(4, 2, 5;n))
= F ([4, 2, 5], n).
Since h(β76) = 3, applying Theorem 7.1(i) we deduce the result.
Ramanujan identities and Euler products 387
8. Euler products for L(A, s) and L(A2, s) when H(d) = {I, A,A2,A3}. Let d be a discriminant such that H(d) = {I, A,A2, A3} with A4 = I.From [SW, Theorem 7.4] we know that for n β N,
F (A, n) = (R(I, n)βR(A2, n))/w(d)and
F (A2, n) = (R(I, n)β 2R(A, n) +R(A2, n))/w(d).Thus for s β C with Re(s) > 1 we have
L(A, s) =
ββ
n=1
F (A, n)
ns=
ββ
n=1
(R(I, n)βR(A2, n))/w(d)ns
and
L(A2, s) =ββ
n=1
F (A2, n)
ns=ββ
n=1
(R(I, n)β 2R(A, n) +R(A2, n))/w(d)ns
.
Let d be a discriminant with conductor f . If p is a prime such that p | dand p β€ f , then p is represented by a unique classM in H(d) andM =Mβ1.Thus, if H(d) = {I, A,A2, A3} with A4 = I, then either p β R(I) or p βR(A2).
Theorem 8.1. Let d < 0 be a discriminant such that H(d) = {I, A,A2,A3} with A4 = I, which is given in [SW, Proposition 11.1(i)]. Let f be theconductor of d and s β C with Re(s) > 1.
(i) If d 6= β220,β252, then
L(A, s) =β
( dp )=β1
1
1β pβ2sβ
p|d, pβ€fpβR(I)
1
1β pβsβ
p|d, pβ€fpβR(A2)
1
1 + pβs
Γβ
pβ€dpβR(I)
1
(1β pβs)2β
pβ€dpβR(A2)
1
(1 + pβs)2
β
pβR(A)
1
1 + pβ2s.
(ii) If d = β220 and so F (A, n) = 12 (R(1, 0, 55;n)βR(5, 0, 11;n)), then
L(A, s) =1 + 21β2s
1 + 2β2sΒ· 1
(1 + 5βs)(1 + 11βs)
β
( p11 )=β(p5 )
1
1β pβ2s
Γβ
( p11 )=(p5 )=β1
1
1 + pβ2s
Γβ
p=x2+55y2
1
(1β pβs)2β
p=5x2+11y2
p 6=5,11
1
(1 + pβs)2.
388 Z. H. Sun and K. S. Williams
(iii) If d = β252 and so F (A, n) = 12 (R(1, 0, 63;n)βR(7, 0, 9;n)), then
L(A, s) =1 + 21β2s
1 + 2β2sΒ· 1
1 + 7βs
β
pβ‘3,5,6 (mod 7)p 6=3
1
1β pβ2s
Γβ
pβ‘2,8,11 (mod 21)p 6=2
1
1 + pβ2s
Γβ
p=x2+63y2
1
(1β pβs)2β
p=7x2+9y2
p 6=7
1
(1 + pβs)2.
Proof. Let p be a prime such that p | f and pΞ±p β f . If d 6= β220,β252,by [SW, Proposition 11.1(i) and Remark 2.2] we have h(d/p2) = 1, 2 andso h(d/p2Ξ±p) = 1, 2. Thus putting h = 4, k = 1 and Ξ²k,p = 0 (if p | f) inTheorem 5.3 we obtain (i).
If d = β220, then f = 2, d0 = d/f2 = β55,(
d02
)
= 1 and h(d/22) =h(d0) = h(β55) = 4. Note that
H(β220) = {[1, 0, 55], [7, 2, 8], [5, 0, 11], [7,β2, 8]},
H(β55) = {[1, 1, 14], [2, 1, 7], [4, 3, 4], [2,β1, 7]},
and for any prime p, p β R(A) β p β R([7, 2, 8]) β(
p11
)
=(
p5
)
= β1.Now putting d = β220, h = 4, k = 1, a2 = 1 and a5 = a11 = 2 in Theorem5.3 and then applying the above we see that (ii) holds.
If d = β252, then f = 6, d0 = d/f2 = β7, I = [1, 0, 63], A = [8, 6, 9],A2 = [7, 0, 9], and for any prime p, p β R(A) β p > 2 and p β‘ 2, 8, 11(mod 21). Let p be a prime dividing f , then p = 2 or 3. Note that H(β63) ={[1, 1, 16], [2, 1, 8], [4, 1, 4], [2,β1, 8]},
(
d02
)
=(β72
)
= 1 and h(d/32) =h(β28) = 2. Putting d = β252, h = 4, k = 1, a2 = 1, a7 = 2 and Ξ²1,3 = 0in Theorem 5.3 and then applying the above we obtain the result.
By the above, the theorem is proved.
Theorem 8.2. Let s β C with Re(s) > 1. Then
(i)
ββ
n=1
Ο3(n)
ns=
1
1 + 7βs
β
pβ‘3,5,6 (mod 7)p 6=3
1
1β pβ2sβ
pβ‘2,8,11 (mod 21)
1
1 + pβ2s
Γβ
p=x2+xy+16y2
1
(1β pβs)2β
p=4x2+xy+4y2 6=7
1
(1 + pβs)2,
Ramanujan identities and Euler products 389
(ii)ββ
n=1
Ο4(n)
ns=
1
1 + 5βs
β
pβ‘11,13,17,19 (mod 20)
1
1β pβ2s
Γβ
pβ‘3,7 (mod 20)
1
1 + pβ2s
β
p=x2+20y2
1
(1β pβs)2
Γβ
p=4x2+5y2 6=5
1
(1 + pβs)2,
(iii)ββ
n=1
Ο8(n)
ns=
β
pβ‘5,7 (mod 8)
1
1β pβ2sβ
pβ‘3 (mod 8)
1
1 + pβ2s
Γβ
p=x2+32y2
1
(1β pβs)2β
p=4x2+4xy+9y2
1
(1 + pβs)2,
(iv)ββ
n=1
Ο12(n)
ns=
β
pβ‘3 (mod 4)p 6=3
1
1β pβ2sβ
pβ‘5 (mod 12)
1
1 + pβ2s
Γβ
p=x2+36y2
1
(1β pβs)2β
p=4x2+9y2
1
(1 + pβs)2.
Proof. By Theorem 8.1(i) and the proof of Theorem 4.5 we obtain theresult.
Remark 8.1. In his lost notebook (see [R2]), Ramanujan conjecturedthe Euler product for
ββn=1 Ο3(n)n
βs. But his formula is erroneous. SoRangachariβs proof of this result is also somewhat wrong. In [R1] Ramanujanconjectured Theorem 8.2(iv). We have established these results in a unifiedand natural way.
Theorem 8.3. Let s β C with Re(s) > 1. Then
(i)ββ
n=1
f1(1, 5;n) + f1(3, 5;n)
ns
=1
1 + 3βsΒ· 1
1 + 13βs
β
( p13 )=β(p3 )
1
1β pβ2sβ
( p13 )=(p3 )=β1
1
1 + pβ2s
Γβ
p=x2+xy+10y2
1
(1β pβs)2β
p=3x2+3xy+4y2 6=3,13
1
(1 + pβs)2,
390 Z. H. Sun and K. S. Williams
(ii)ββ
n=1
f8(1, 5;n) + f8(3, 5;n)
ns
=β
pβ‘3 (mod 4)
1
1β pβ2sβ
pβ‘5 (mod 8)
1
1 + pβ2s
β
p=x2+64y2
1
(1β pβs)2
Γβ
p=4x2+4xy+17y2
1
(1 + pβs)2,
(iii)
ββ
n=1
f20(1, 5;n) + f20(3, 5;n)
ns
=β
pβ‘3 (mod 4)
1
1β pβ2sβ
pβ‘13,17 (mod 20)
1
1 + pβ2s
β
p=x2+100y2
1
(1β pβs)2
Γβ
p=4x2+25y2
1
(1 + pβs)2.
Proof. From [SW, Proposition 11.1(i)] we know that H(d) is a cyclicgroup of order 4 for d β {β39,β256,β400}. Actually,
H(β39) = {[1, 1, 10], [2, 1, 5], [3, 3, 4], [2,β1, 5]},H(β256) = {[1, 0, 64], [5, 2, 13], [4, 4, 17], [5,β2, 13]},H(β400) = {[1, 0, 100], [8, 4, 13], [4, 0, 25], [8,β4, 13]}.
Thus, by Theorem 2.4 we have
f1(1, 5;n) + f1(3, 5;n) = (R(1, 1, 10;n)βR(4, 3, 3;n))/2 = F ([2, 1, 5], n),f8(1, 5;n) + f8(3, 5;n) = (R(1, 0, 64;n)βR(4,β4, 17;n))/2
= F ([5, 2, 13], n)
and
f20(1, 5;n) + f20(3, 5;n) = (R(1, 0, 100;n)βR(4,β16, 41;n))/2= F ([8, 4, 13], n).
For a prime p it is clear that
p β R([2, 1, 5]) β(
p
13
)
=
(
p
3
)
= β1,
p β R([5, 2, 13]) β p β‘ 5 (mod 8),p β R([8, 4, 13]) β p β‘ 13, 17 (mod 20).
Now combining all the above with Theorem 8.1(i) in the cases d =β39,β256,β400 yields the result.
Ramanujan identities and Euler products 391
Remark 8.2. By [SW, Theorems 10.2 and 11.1] and the proofs of Theo-rems 7.3 and 8.3, we may obtain explicit formulas for fk(1, 5;n)+fk(3, 5;n)in the cases k = 1, 2, 8, 20.
Theorem 8.4. Let d<0 be a discriminant such thatH(d)={I, A,A2, A3}with A4 = I. (The values of such d are given in [SW, Proposition 11.1(i)].)Let f be the conductor of d and s β C with Re(s) > 1. Then
L(A2, s) = c(d, s)β
p|d, pβ€f
1
1β pβsβ
( dp )=β1
1
1β pβ2s
Γβ
pβR(I)βͺR(A2)pβ€d
1
(1β pβs)2β
pβR(A)
1
(1 + pβs)2,
where
c(d, s) =
1 + 21β2s if d = β128,β256,1 + 21βs + 21β2s
(1 + 2βs)2if d = β220,β252,
1 + 2βs + 21β2s
1 + 2βsif d = β80,β144,β208,β400,β592,
1 otherwise.
Proof. By Theorem 5.3, the result is true when f = 1. Now supposef > 1 and d0 = d/f
2. Let p be a prime such that p | f . Suppose pΞ±p β f , hp =h(d/p2Ξ±p) and Ξ²2,p is the maximum j β {0, 1, . . . , Ξ±p} such that 2 |h(d/p2j).From [SW, Proposition 11.1(i)] we know that all possible (d, f) with f > 1are given below:
(d, f) = (β63, 3), (β80, 2), (β128, 22), (β144, 2 Β· 3), (β171, 3),(β196, 7), (β208, 2), (β220, 2), (β252, 2 Β· 3), (β256, 23),(β275, 5), (β363, 11), (β387, 3), (β400, 2 Β· 5), (β475, 5),(β507, 13), (β592, 2), (β603, 3), (β1467, 3).
From [SW, Proposition 11.1, Lemma 9.2 and (9.3)] we also have
hp = h
(
d
p2Ξ±
)
=
1 if d 6β {β80,β144,β208,β220,β252,β400,β592}or p > 2,
2 if d β {β80,β144,β208,β400,β592} and p = 2,4 if d β {β220,β252} and p = 2.
If p > 2, then hp = h(d/p2) = 1. Thus 4 β€ 2hp, Ξ²2,p = 0 and hence(1 β p(1β2s)(1+Ξ²2,p))/(1 β p1β2s) = 1. Thus applying Theorem 5.3 we seethat the result is true in the cases d = β63,β171,β196,β275,β363,β387,β475,β507,β603,β1467.
392 Z. H. Sun and K. S. Williams
If d = β128,β256, then p = 2, p | d0, hp = 1, 4 β€ 2hp and Ξ²2,p = 1, thusputting d = β128,β256 and k = 2 in Theorem 5.3 we obtain the result.If d = β220, then p = f = 2, hp = h(β55) = 4, 4 | 2hp,
(
d0p
)
=(β552
)
= 1,thus putting d = β220, k = 2 and a2 = 1 in Theorem 5.3 yields the result.If d = β252, then f = 6, d0 = β63 and so p = 2, 3. Observe that
h2 = h(β63) = 4, 4 | 2h2 and(
d02
)
=(β632
)
= 1. Putting d = β252 andk = 2 in Theorem 5.3 gives the result.If d = β80,β144,β208,β400,β592, then 2 | f . Let p = 2. Then Ξ±p = 1,
hp = 2, 4 | 2hp, p | d0 and p is represented by the generator of H(d/p2) by[SW, Table 9.1]. Thus applying Theorem 5.3 and the above we obtain theresult.Combining the above we prove the theorem.We remark that the conditions for p β R(I) βͺ R(A2) or p β R(A) in
Theorem 8.4 can be described by certain congruence conditions.
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Department of MathematicsHuaiyin Teachers CollegeHuaian, Jiangsu 223001, P.R. ChinaE-mail: [email protected]://www.hytc.cn/xsjl/szh
Centre for Research in Algebra and Number TheorySchool of Mathematics and Statistics
Carleton UniversityOttawa, Ontario K1S 5B6, CanadaE-mail: [email protected]
http://mathstat.carleton.ca/Λwilliams
Received on 24.2.2005
and in revised form on 10.6.2005 (4943)