c Pleiades Publishing, Ltd., 2007. ISSN 0001-4346, Mathematical Notes, 2007, Vol. 81, No. 2, pp. 172–182.  c S. A. Gritsenko, 2007, published in Matematicheskie Zametki, 2007, Vol. 81, No. 2, pp. 193–204. Original Russian Text 

Goldbach’s Ternary Problem Involving Prime Numbers Expressible by Given Quadratic Forms S. A. Gritsenko* Belgorod State University Received May 17, 2006

Abstract—In this paper, we solve Goldbach’s ternary problem involving primes expressible by given primitive positive definite binary quadratic forms whose discriminants coincide with the discriminants of imaginary quadratic fields in which quadratic forms split into linear multipliers. DOI: 10.1134/S0001434607010208 Key words: Goldbach’s ternary problem, binary quadratic form, imaginary quadratic field, arithmetic progression, simple ideal, Dirichlet series.

INTRODUCTION In 1937, Vinogradov solved Goldbach’s ternary problem by obtaining an asymptotic formula for the number of solutions of the equation p1 + p2 + p3 = N

(1)

in primes p1 , p2 , p3 with N odd. In the present paper, we obtain an asymptotic formula for the number of solutions of Eq. (1) in primes p1 , p2 , p3 such that each prime pi , i = 1, 2, 3 can be expressed by a binary quadratic form. Note that if these quadratic forms are one-class, then the problem can be reduced, essentially, to the solution of Goldbach’s ternary problem involving primes lying in certain arithmetic progressions. For the case in which the differences of these arithmetic progressions are constants or increase with N not too fast, the solution of such a problem does not differ significantly from that of the classical Goldbach ternary problem. The same applies to the problem of obtaining an asymptotic formula for the number of solutions of (1) in primes expressible by all quadratic forms of given discriminants. If quadratic forms representing primes, are multiclass, the question of expressing positive integers by such forms cannot be reduced to the question of whether these numbers belong to some classes of residues. (in this connection, see [1]). Therefore, to solve the problem under consideration, one must have some information about the distribution of primes expressible by quadratic forms in arithmetic progressions as well as estimates of some special trigonometric sums. In what follows, by quadratic forms we shall mean primitive positive definite binary quadratic forms whose discriminants coincide with the discriminants of imaginary quadratic fields in which these forms split into linear multipliers. The discriminants of quadratic forms are assumed to be constant everywhere, although it can be seen from the proofs of the main theorem and the lemmas that they may increase as the parameter N increases. In what follows, the following notation will be used: • p, p1 , p2 are primes; • P are simple ideals of an imaginary quadratic field; *

E-mail: [email protected]

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• N (A) is the norm of an integer ideal A of an imaginary quadratic field;  1 if q ≡ l (mod k), • χ(q; k, l) = 0 otherwise; • χi (l) is a character of a quadratic field in which the quadratic form Qi splits into linear multipliers, i = 1, 2, 3; • SQ (α) = form Q; •

b−1 ∗



e2πiαp , where the prime stands for summation over primes expressible by a quadratic

denotes summation over a’s coprime to b in the specified limits.

a=0

The main results of the paper are the following theorems. Theorem 1. Suppose that J(N, Q1 , Q2 , Q3 ) is the number of solutions of Eq. (1) in primes p1 , p2 , p3 expressible by quadratic forms Q1 , Q2 , Q3 with discriminants −D1 , −D2 , −D3 , respectively, h1 , h2 , h3 are the numbers of classes of equivalent forms with discriminants −D1 , −D2 , −D3 , respectively. Then, for any constant c > 0, the following formula holds: J(N, Q1 , Q2 , Q3 ) = σ(N, D1 , D2 , D3 )I(N ) + O(N 2 log−c N ), where I(N ) =

 n1 +n2 +n3 =N 3≤n1 ,n2 ,n3 ≤N

N2 1 , ∼ log n1 log n2 log n3 2 log3 N

q−1 ∞  1 1 ∗ −2πiaN/q e σ(N, D1 , D2 , D3 ) = h1 h2 h3 q=1 ϕ3 (q) a=0       q q × μ(q) + χ(q; D1 , 0)χ1 (a)μ χ1 τ χ1 D1 D1       q q χ2 τ χ2 × μ(q) + χ(q; D2 , 0)χ2 (a)μ D2 D2       q q χ3 τ χ3 × μ(q) + χ(q; D3 , 0)χ3 (a)μ D3 D3

and χ1 , χ2 , χ3 are the characters of the imaginary quadratic fields with discriminants −D1 , −D2 , −D3 , respectively. Theorem 2. Suppose that J(N, Q1 , Q2 , Q3 ) is the number of solutions of Eq. (1) in primes p1 , p2 , p3 expressible by quadratic forms Q1 , Q2 , Q3 with discriminant −D and h is the number of classes of equivalent forms with discriminant −D. Then, for any constant c > 0, the following formula holds: J(N, Q1 , Q2 , Q3 ) = σ(N, D)I(N ) + O(N 2 log−c N ), where 

I(N ) =

n1 +n2 +n3 =N 3≤n1 ,n2 ,n3 ≤N

N2 1 , ∼ log n1 log n2 log n3 2 log3 N

σ(N, D) = σ0 (N ) + σ1 (N, D) + σ2 (N, D) + σ3 (N, D), MATHEMATICAL NOTES

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σ0 (N ) =

∞ 1  μ(q)γ0 (N, q) , h3 ϕ3 (q)

σ1 (N, D) =

q=1

σ2 (N, D) =

3 h3

γ0 (N, q) =

∞  μ(Dq2 )γ2 (N, Dq2 ) , ϕ3 (q2 )

3 h3

∞  q2 =1 q2 ,D)=1

σ3 (N, D) =

q2 =1 q2 ,D)=1 q 

e−2πiaN/q ,

γ1 (N, q) =

a=1 a,q)=1

γ2 (N, q) =

q 

μ2 (Dq2 )γ1 (N, Dq2 ) , ϕ3 (q2 )

1 h3

q 

∞  γ3 (N, Dq2 ) , ϕ3 (q2 )

q2 =1 q2 ,D)=1

S(a, q)e−2πiaN/q ,

a=1 a,q)=1 2

−2πiaN/q

S (a, q)e

,

γ3 (N, q) =

a=1 a,q)=1

q 

S 3 (a, q)e−2πiaN/q ;

a=1 a,q)=1

here S(a, q) =

q 

χ1 (l)e2πial/q

l=1 l,q)=1

and χ1 is the character of the imaginary quadratic field with discriminant −D. The proofs are carried out by the circular method. Besides, an essential role is played by the functional equation of the Dirichlet series of special form from [2] on which the asymptotic formula from [3] and the estimate from [4] are based. To prove Theorems 1 and 2, we need the following lemmas. 1. LEMMAS Lemma 1. Suppose that (a, D1 q2 ) = 1, (D, q2 ) = 1, D | D1 ; moreover, each prime divisor D1 divides D. Then the following identity holds:  χ1 (a)χ1 (q2 )μ(q2 )τχ1 if D1 = D, S(a, D1 q2 ) = 0 if D1 = D,  2πil/D is the Gauss sum. where τχ1 = D l=1 χ1 (l)e Proof. Since the character χ1 is real, we have S(a, D1 q2 ) = χ1 (a)S(1, D1 q2 ). Further, S(1, D1 q2 ) =

D1 

q2 

χ1 (l1 q2 + l2 D1 )e2πil1 /D1 e2πil2 /q2

l1 =1 l2 =1 l1 ,D1 )=1 l2 ,q2 )=1

= χ1 (q2 )μ(q2 )

/D−1 D D1  l1 =1

χ1 (l + Dl2 )e2πil1 /D1 e2πil2 /(D1 /D) .

l2 =0

Since the sum over l2 is equal to 1 if D1 = D and zero otherwise, we obtain the proof of Lemma 1. MATHEMATICAL NOTES

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Lemma 2. Suppose that (D, q2 ) = 1. Then the following identities hold: γ1 (N, Dq2 ) = Dμ(q2 )χ1 (N )γ0 (N, q2 ), γ2 (N, Dq2 ) = χ1 (−1)Dμ2 (q2 )γ0 (N, D)γ0 (N, q2 ), γ3 (N, Dq2 ) = χ1 (−1)D2 μ(q2 )χ1 (N )γ0 (N, q2 ). Proof. Let us use Lemma 1. We have Dq2 

γ1 (N, Dq2 ) = χ1 (q2 )μ(q2 )τ (χ1 )

χ1 (a)e−2πiaN/(Dq2 )

a=1 a,Dq2 )=1 D 

= χ1 (q2 )μ(q2 )τ (χ1 )

q2 

χ1 (a1 q2 + a2 D)e−2πia1 N/D e−2πia2 N/q2

a1 =1 a2 =1 a,D)=1 a2 ,q2 )=1

= χ21 (q2 )μ(q2 )τ (χ1 )γ0 (N, q2 )

D 

χ1 (a)e−2πiaN/D .

a=1

Let us apply the well-known identity D 

χ1 (a)e−2πiaN/D = χ1 (−1)χ1 (N )τχ1 ;

a=1

then we obtain γ1 (N, Dq2 ) = χ21 (q2 )μ(q2 )χ1 (−1)τχ21 χ1 (N )γ0 (N, q2 ). Further, χ21 (q2 ) = 1 and χ1 (−1)τχ21 = D, because χ1 is a real character and (q2 , D) = 1. Thus, we have obtained an identity for γ1 (N, Dq2 ). Let us also derive an identity for γ2 (N, Dq2 ). By Lemma 1, we have γ2 (N, Dq2 ) = μ2 (q2 )τχ21 γ0 (N, Dq2 ). Now, the required identity follows from the fact that the function γ0 (N, q) is multiplicative with respect to q and from equality τχ21 = χ1 (−1)D. Let us derive an identity for γ3 (N, Dq2 ). We again apply Lemma 1, obtaining γ3 (N, Dq2 ) = τχ21 γ1 (N, Dq2 ) = χ1 (−1)D2 μ(q2 )χ1 (N )γ0 (N, q2 ). Lemma 2 is proved. Lemma 3. The following identities hold:     1  3Dμ2 (D)χ1 (N )  γ0 (N, p) γ0 (N, p) , σ1 (N, D) = , 1− 1− σ0 (N ) = 3 h p ϕ3 (p) h3 ϕ3 (D) ϕ3 (p) pD   3Dμ(D)χ1 (−1)γ0 (N, D)  γ0 (N, p) , 1− σ2 (N, D) = h3 ϕ3 (D) ϕ3 (p) pD   2  D χ1 (−1)χ1 (N ) γ0 (N, p) . 1− σ3 (N, D) = h3 ϕ3 (D) ϕ3 (p) pD

Proof. Lemma 3 immediately follows from Lemma 2 and the fact that the function γ0 (N, q) is multiplicative with respect to q. MATHEMATICAL NOTES

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Lemma 4. Suppose that (l, q) = 1, 1 ≤ q ≤ logc1 x, and C is the class of ideals. Then the following asymptotic formula holds: q  ∗ 1 + χ(q; D, 0)χ1 (l) γ Li x + O(N e−c0 (log N ) /2 ), π1 (x; q, l, C ) = hϕ(q) l=1

where c0 = c0 (c1 ) > 0, γ = 1/(20c1 ). Proof. The proof is given in [3]. Lemma 5. Suppose that (a, q) = 1, α = a/q + z, q ≤ logc1 N , |z| ≤ 1/(qτ ), and τ = N log−c2 N . Then, for SQ (α), the following formula holds: q  ∗ 1 + χ(q; D, 0)χ1 (l) 2πial/q γ e M (z) + O(N e−c0 (log N ) /2 ), SQ (α) = hϕ(q) l=1

where c0 and γ are the constants from the lemma and M (z) =

N  e2πizn n=3

log n

.

Proof. By assumption, the form Q can be expressed as Q(x, y) =

N (xω1 + yω2 ) , N (A)

where ω1 , ω2 is the basis of an integer ideal A. Suppose that A lies in the class of ideals A and B is an arbitrary ideal from the class A −1 . Then AB = (ξB ) is the principal ideal, ξB ∈ A. The mapping B → ξB induces a bijection between the proper ideals from the class A −1 and the classes of equivalent numbers from the ideal A (two numbers are equivalent if their ratio is ±1). Since N (ξB ) = N (B), N (A) it follows that only those primes p that are the norms of simple ideals from the class A −1 can be expressed by the form Q. Hence we have the equality  √ √ e2πiαN (P ) + O N = S1Q (α) + O N . SQ (α) = P ∈A −1 N (P )≤N

√ √ The remainder O( N ) is due to the fact that there exist O( N ) simple ideals whose norms are equal to p2 ≤ N . Now we consider the sum S1Q (α). We have    q ∗ 2πial/q a +z = e Tl (z), S1Q (α) = S1Q q l=1

where Tl (z) =



e2πizN (P ) .

P ∈A −1 N (P )≤N N (P )≡l (mod q)

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Applying to Tl (z) the partial summation formula, we obtain

N √  Tl (z) = − √ π1 (u; q, l, A −1 ) − π1 ( N ; q, l, A −1 ) de2πizu N √ + π1 (N ; q, l, A −1 )e2πizN + O N , where



π1 (u; q, l, A −1 ) =

1.

−1

P ∈A N (P )≤u N (P )≡l (mod q)

Integrating by parts and using Lemma 4, we find

1 + χ(q; D, 0)χ1 (l) N e2πizu γ du + O(N e−c0 (log N ) /2 ). Tl (z) = hϕ(q) log u 3 The assertion of the lemma now follows from the well-known equality

N 2πizu e du + O(1). M (z) = log u 3 Lemma 6. Suppose that a(n) =



X(A),

A N (A)=n

where X is an arbitrary nonprincipal character of the group of classes of ideals. Suppose that c > 0 is an arbitrary constant. Then there exist positive constants c1 and c2 such that, for α = a/q + z, |z| ≤ 1/(qτ ), (a, q) = 1, τ = N (log N )−c2 , (log N )c1 < q ≤ τ , the following estimate holds: N 

 Λ(n)a(n)e2πiαn = O N (log N )−c .

n=1

Proof. The proof is given in [4]. Corollary 1. Suppose that c > 0 is an arbitrary constant. Then there exist positive constants c1 and c2 such that, for α = a/q + z, |z| ≤ 1/(qτ ), (a, q) = 1, τ = N (log N )−c2 , (log N )c1 < q ≤ τ , the following estimate holds:  SQ (α) = O N (log N )−c . Proof. The following relations are valid:  √ 1  e2πiαN (P ) + O N = X(A ) SQ (α) = h −1 P ∈A N (P )≤N

X



X(P )e2πiαN (P ) + O

√

N ,

N (P )≤N

where X ranges over all characters of the group of classes of ideals and h is the order of the group of classes of ideals. Further,   X(A ) X(P )e2πiαN (P ) X

N (P )≤N

=



e2πiαp +

p≤N

MATHEMATICAL NOTES



χ1 (p)e2πiαp +

p≤N

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2007

X(A )

 N (P )≤N

X(P )e2πiαN (P ) ,

178

GRITSENKO

where χ1 is the character of the quadratic field and X0 is the principal character of the group of classes of ideals. The estimate    X(A ) X(P )e2πiαN (P ) = O N (log N )−c X =X0

N (P )≤N

follows from Lemma 6. The estimate



 e2πiαp = O N (log N )−c

p≤N

is well known (see, for example, [5, Chap. 10]). The estimate   χ1 (p)e2πiαp = O N (log N )−c p≤N

 also holds and can be obtained, essentially, in the same way as the estimate of the sum p≤N e2πiαp , because the modulus of the Dirichlet character χ1 is a constant (nonincreasing as N increases). 2. PROOF OF THE MAIN THEOREM Proof of Theorem 1. Suppose that τ = N log−c2 N and c2 > 1. We write J(N, Q1 , Q2 , Q3 ) in integral form:

1−1/τ SQ1 (α)SQ2 (α)SQ3 (α)e−2πiαN dα. J(N, Q1 , Q2 , Q3 ) = −1/τ

Let us define the sets E1 and E2 :

  

1 a

1 1 c1

: α − ≤ , (a, q) = 1, q ≤ log N , E1 = α ∈ − , 1 − τ τ q qτ  1 1 \ E1 . E2 = − , 1 − τ τ Choose the parameters c1 and c2 so that c + 1 < c1 , 2c1 < c2 ; moreover, c1 and c2 satisfy the assumptions of Lemma 6. Then if N is a sufficiently large number, then the sets E1 and E2 consist of nonintersecting intervals and J(N, Q1 , Q2 , Q3 ) = J1 (N, Q1 , Q2 , Q3 ) + J2 (N, Q1 , Q2 , Q3 ), where

Ji (N, Q1 , Q2 , Q3 ) =

SQ1 (α)SQ2 (α)SQ3 (α)e−2πiαN dα,

i = 1, 2.

Ei

By Corollary 1, we have

 J2 (N, Q1 , Q2 , Q3 ) = O π(N ) max |SQ1 (α)| = O(N 2 log−c N ). α∈E2

Suppose that α ∈ E1 . Applying Lemmas 1 and 5, we obtain q−1  1 1 ∗ −2πiaN/q e J1 (N, Q1 , Q2 , Q3 ) = h1 h2 h3 ϕ3 (q) a=0 q≤logc1 N       q q × μ(q) + χ(q; D1 , 0)χ1 (a)μ χ1 τ χ1 D1 D1       q q χ2 τ χ2 × μ(q) + χ(q; D2 , 0)χ2 (a)μ D2 D2

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GOLDBACH’S TERNARY PROBLEM





× μ(q) + χ(q; D3 , 0)χ3 (a)μ

×

1/(qτ )







q q χ3 τ χ3 D3 D3



179

M 3 (z)e−2πizN dz.

−1/(qτ )

Following arguments from [5, Chap. 10], we can obtain the equality

1/(qτ )

1/2  M 3 (z)e−2πizN dz = M 3 (z)e−2πizN dz + O N 2 (log N )−2c2 +2c1 −1/(qτ )

−1/2

 = I(N ) + O N 2 (log N )−2c2 +2c1 .

Finally, trivially estimating the sum over q > logc1 N , we obtain the assertion of Theorem 1. Proof of Theorem 2. Theorem 2 is proved in the same way, the only exception being that the sum over a is calculated with the help of Lemmas 2 and 3. 3. LOWER BOUNDS FOR THE SUMS OF SINGULAR SERIES With regard to the sum of the singular series from Theorem 1, we restrict ourselves to the following assertion. Proposition 1. Suppose that the numbers D1 , D2 , and D3 are sufficiently large and N is an odd number. Then the following inequality holds: σ(N, D1 , D2 , D3 ) >

1 . 2h1 h2 h3

Proof. Let us remove the brackets in the product       q q χ1 τ χ1 μ(q) + χ(q; D1 , 0)χ1 (a)μ D1 D1       q q χ2 τ χ2 × μ(q) + χ(q; D2 , 0)χ2 (a)μ D2 D2       q q χ3 τ χ3 . × μ(q) + χ(q; D3 , 0)χ3 (a)μ D3 D3 We obtain eight summands, which correspond to eight singular series. The first summand is equal to μ(q); it corresponds to the series ∞  1 μ(q)γ0 (q, N ) . σ1 (N ) = h1 h2 h3 q=1 ϕ3 (q)

It is well known that σ1 (N ) > 1/(h1 h2 h3 ) (see, for example, [5, Chap. 10]). Let us find an upper bound for the sums of the other series. Since all the estimates are obtained identically, we restrict ourselves to the series q−1 ∞   ∗ −2πiaN/q χ(q; D1 , 0)χ(q; D2 , 0)χ(q; D3 , 0) 1 e σ8 (N, D1 , D2 , D3 ) = h1 h2 h3 ϕ3 (q) q=1 a=0       q q q μ μ × χ1 (a)χ2 (a)χ3 (a)μ D1 D2 D3       q q q χ2 χ3 τ χ1 τ χ2 τ χ3 . × χ1 D1 D2 D3

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Using the root estimates for the Gauss sums, we find that there exists an absolute constant c3 > 0 such that √ c3 D1 D2 D3 , |σ8 (N, D1 , D2 , D3 )| < h1 h2 h3 [D1 , D2 , D3 ]2 where [D1 , D2 , D3 ] is the least common multiple of the numbers D1 , D2 , D3 . √ 3/2 If D1 ≤ D2 ≤ D3 , then D1 D2 D3 ≤ D3 , [D1 , D2 , D3 ]2 ≥ D32 ; therefore, the following inequality holds: 1 c3  ≤ |σ8 (N, D1 , D2 , D3 )| < 14h1 h2 h3 h1 h2 h3 max(D1 , D2 , D3 ) if min(D1 , D2 , D3 ) ≥ 196c23 . Using similar procedures with regard to the six remaining singular series, we obtain our assertion. Now consider the singular series from Theorem 2. Proposition 2. If N is an even number, then σ0 (N ) = σ1 (N, D) = σ2 (N, D) = σ3 (N, D) = 0. Proof. If D is an even number, then, as is known, D is divisible by 4; hence σ1 (N ) = σ2 (N ) = 0. Also, χ1 (N ) = 0, i.e., σ3 (N ) = 0. Finally, it is also well known that σ0 (N ) = 0 for an even N . But if D is an odd number, then, for an even N , we have   γ0 (N, p) = 0, σ0 (N ) = σ1 (N, D) = σ2 (N, D) = σ3 (N, D) = 0. 1− ϕ3 (p) pD

Proposition 3. Suppose that N is an odd number and −D = δF is the discriminant of a quadratic form which is not one-class and expands into linear multipliers in the imaginary quadratic field with discriminant δF . Suppose that D > 15. Then the inequality σ(N, D) > 2/(5h3 ) holds. Proof. Since −D is the discriminant of a quadratic field, it suffices to consider the following cases. Case 1. D = 8d, where d is an odd square-free number. Then μ(D) = 0, σ1 (N, D) = σ2 (N, D) = 0, and hence σ0 (N, D) = σ0 (N ) + σ3 (N, D). If (N, D) > 1, then χ1 (N ) = 0, σ3 (N, D) = 0; therefore, in this case, σ(N, D) = σ0 (N ). Suppose that (N, D) = 1; then         1 p2 1 γ0 (N, p) 1 ≥ σ0 (N ), 1+ − 1 − σ(N, D) ≥ 3 2 h (p − 1)3 (p − 1)3 ϕ3 (p) 2 p|d

p|d

pD

because, for any p ≥ 3, the following inequality holds: 1+

p2 1 ≥ . (p − 1)3 (p − 1)3

Case 2. D = 4d, where d is an odd square-free number, d ≥ 5 (otherwise, the quadratic form with discriminant −D is one-class). Then if (N, D) > 1, we have σ(N, D) = σ0 (N ). Suppose that (N, D) = 1; then         p2 1 γ0 (N, p) 2  . 1+ − 1− σ(N, D) ≥ 3 h (p − 1)3 (p − 1)3 ϕ3 (p) p|d

p|d

pD

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By assumption, d is an odd square-free number, d ≥ 5; therefore, there exists a prime p ≥ 5 dividing d. But, for p ≥ 5, the following inequality holds:   p2 1 1 ≥ 1+ , 2 (p − 1)3 (p − 1)3 whence σ(N, D) ≥ (1/2)σ0 (N ). Case 3. D is an odd square-free number, D ≡ 3 (mod 4), D > 15. We split this case into subcases. 3.1. χ1 (N ) = 1. Then χ1 (−1) = −1;

γ0 (N, D) = μ(D), therefore,

      1 1 γ0 (N, p) p2 1  ≥ σ0 (N ), 1+ − 1− σ(N, D) ≥ 3 3 3 3 h (p − 1) (p − 1) ϕ (p) 2 p|D

p|D

pD

because there exists a prime divisor D greater than 5. 3.2. χ1 (N ) = −1. Then

    1 −6D + D2  γ0 (N, p) > σ0 (N ), 1− σ(N, D) = σ0 (N ) + 3 h ϕ3 (D) ϕ3 (p) pD

because −6D + D2 > 0. 3.3. χ1 (N ) = 0. In this case, σ1 (N, D) = σ3 (N, D) = 0. Suppose that (N, D) = D1 > 1, D = D1 D2 , (N, D2 ) = 1. Then γ0 (N, D) = μ(D2 )ϕ(D1 ),      1 1 1  1− 1+ σ(N, D) = 3 h (p − 1)2 (p − 1)3 p|D1 p|D2       p p γ0 (N, p) . 1− −3 (p − 1)2 (p − 1)3 ϕ3 (p) p|D1

p|D2

pD

By assumption, D = D1 D2 is an odd square-free number, D > 15. Therefore, either D1 or D2 has a prime divisor greater than 5. If p | D1 , p ≥ 7, then we have the inequality   1 p 1 1− , ≥ 2 5 (p − 1) (p − 1)2 from which it follows that, in this case, σ(N, D) ≥ (2/5)σ0 (N ). But if p | D2 , p ≥ 7, then the following inequality holds:   1 p 1 1+ ≥ 3 10 (p − 1) (p − 1)3 and, therefore, σ(N, D) ≥ (7/10)σ0 (N ). The resulting estimates imply that, for all N and D satisfying the conditions, the following inequality holds: 2 2 σ(N, D) ≥ σ0 (N ) > 3 5 5h (the inequality σ0 (N ) > 1/h3 is well known; see, for example, [5, Chap. 10]). MATHEMATICAL NOTES

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In the previous proposition, we did not consider the case D = 15. As will be seen from the next proposition, the quadratic forms with the discriminant −15 which are not one-class are the unique exception from the following general rule: there exist infinitely many odd numbers N for which σ(N, 15) = 0. Proposition 4. Suppose that N is an odd number. If (N, 15) = 1, then 1 . 13 If (N, 3) = 1, (N, 5) > 1 or (N, 5) = 1, (N, 3) > 1, then σ(N, 15) >

1 . 8 If N ≡ 0 (mod 15), then σ(N, 15) = 0; moreover, the equation σ(N, 15) >

p1 + p2 + p3 = N is insoluble in primes p1 , p2 , p3 expressible by the quadratic forms with discriminant −15. Proof. The first three assertions immediately follow from the equality     45χ1 (N ) 45γ0 (N, 15) 225χ1 (N ) γ0 (N, p) 1  + − − 1− σ(N, 15) = 8 ϕ3 (p) 512 512 512 p|15   γ0 (N, p) 1− × ϕ3 (p) p15

(we have taken into account the fact that h = 2). Also, any quadratic form with discriminant −15 is equivalent to one of the forms f1 (x, y) = x2 + xy + 4y 2 ,

f2 (x, y) = 2x2 + xy + 2y 2 .

Simple calculations show that the forms f1 and f2 can represent only primes congruent to 1, 2, 4, and 8 modulo 15, while the sums of any three such numbers are not divisible by 15. ACKNOWLEDGMENTS This work was supported by the Ministry of Education and Science of the Russian Federation under the program “Development of the Scientific Potential of Higher School” (grant no. RNP.2.1.1.3263) and by Belgorod State University (grant no. VGK 007-04). REFERENCES 1. Z. I. Borevich and I. R. Shafarevich, Number Theory (Nauka, Moscow, 1985) [in Russian]. 2. S. A. Gritsenko, “On the functional equation of an arithmetic Dirichlet series,” Chebyshev Sbornik 4 (2), 55–67 (2003). 3. S. A. Gritsenko, “On the distribution of the norms of simple ideals from a given class and arithmetic progressions,” Zapiski Nauchn. Sem. POMI 322, 45–62 (2005). 4. S. A. Gritsenko, “Estimate of a linear trigonometric sum over primes expressible by a given quadratic form,” Chebyshev Sbornik 5 (4), 82–87 (2005). 5. A. A. Karatsuba, Foundations of Analytic Number Theory (Nauka, Moscow, 1983) [in Russian].

MATHEMATICAL NOTES

Vol. 81 No. 2 2007