Quadratic residues of the norm group in sectorial domains

In the article the distribution of quadratic residues in the ring Gpn, in the norm subgroup En of multiplicative group G∗pn, is investigated. The asymptotic formula for the number R(x,ϕ) of quadratic residues in the sectorial domain of a special form has been constructed.

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Hauptverfasser:	Balyas, L., Varbanets, P.
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spelling	irk-123456789-1557362019-06-18T01:25:39Z Quadratic residues of the norm group in sectorial domains Balyas, L. Varbanets, P. In the article the distribution of quadratic residues in the ring Gpn, in the norm subgroup En of multiplicative group G∗pn, is investigated. The asymptotic formula for the number R(x,ϕ) of quadratic residues in the sectorial domain of a special form has been constructed. 2016 Article Quadratic residues of the norm group in sectorial domains / L. Balyas, P. Varbanets // Algebra and Discrete Mathematics. — 2016. — Vol. 22, № 2. — С. 153-170. — Бібліогр.: 11 назв. — англ. 1726-3255 2010 MSC:11L05, 11L07, 11N25, 11S40. http://dspace.nbuv.gov.ua/handle/123456789/155736 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України
institution	Digital Library of Periodicals of National Academy of Sciences of Ukraine
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language	English
description	In the article the distribution of quadratic residues in the ring Gpn, in the norm subgroup En of multiplicative group G∗pn, is investigated. The asymptotic formula for the number R(x,ϕ) of quadratic residues in the sectorial domain of a special form has been constructed.
format	Article
author	Balyas, L. Varbanets, P.
spellingShingle	Balyas, L. Varbanets, P. Quadratic residues of the norm group in sectorial domains Algebra and Discrete Mathematics
author_facet	Balyas, L. Varbanets, P.
author_sort	Balyas, L.
title	Quadratic residues of the norm group in sectorial domains
title_short	Quadratic residues of the norm group in sectorial domains
title_full	Quadratic residues of the norm group in sectorial domains
title_fullStr	Quadratic residues of the norm group in sectorial domains
title_full_unstemmed	Quadratic residues of the norm group in sectorial domains
title_sort	quadratic residues of the norm group in sectorial domains
publisher	Інститут прикладної математики і механіки НАН України
publishDate	2016
url	http://dspace.nbuv.gov.ua/handle/123456789/155736
citation_txt	Quadratic residues of the norm group in sectorial domains / L. Balyas, P. Varbanets // Algebra and Discrete Mathematics. — 2016. — Vol. 22, № 2. — С. 153-170. — Бібліогр.: 11 назв. — англ.
series	Algebra and Discrete Mathematics
work_keys_str_mv	AT balyasl quadraticresiduesofthenormgroupinsectorialdomains AT varbanetsp quadraticresiduesofthenormgroupinsectorialdomains
first_indexed	2025-07-14T07:58:39Z
last_indexed	2025-07-14T07:58:39Z
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fulltext	Algebra and Discrete Mathematics RESEARCH ARTICLE Volume 22 (2016). Number 2, pp. 153–170 © Journal “Algebra and Discrete Mathematics” Quadratic residues of the norm group in sectorial domains Lyubov Balyas and Pavel Varbanets Communicated by D. Simson Abstract. In the article the distribution of quadratic residues in the ring Gpn , in the norm subgroup En of multiplicative group G∗ pn , is investigated. The asymptotic formula for the number R(x, φ) of quadratic residues in the sectorial domain of a special form has been constructed. 1. Introduction In 1918 I.M. Vinogradov and G. Polya built the asymptotic formula for the number of quadratic residues modulo prime number on the seg- ment 1 6 n 6 x < p, which was nontrivial for every x > √ p log p. It was the first result about incomplete residue system in analytic number theory. Henceforth Vinogradov-Polya theorem was firstly sharpened by D. Burgess [1]. After this on the assumption under extended Weil hypoth- esis H. Montogomery and R. Vaughan [3] got the unimprovable result for the theorem. The research of analogous issue over the ring of Gaussian integers is, evidently, a difficult problem by the virtue of the fact, that geometry of points of a plane is richer than geometry of points on a line. In this article the distribution of quadratic residues in the norm subgroup En 2010 MSC: 11L05, 11L07, 11N25, 11S40. Key words and phrases: quadratic residue, the norm group, Hecke Z-function,the gamma function, Dirichlet series,functional equation, Gaussian integers. 154 Quadratic residues of multiplicative group G∗ pn , is investigated. Here p is a prime rational number of the type p = 3 + 4k and En can be written in the form: En := {α ∈ Gpn \|N(α) ≡ ±1 (mod pn)} . This subgroup is cyclic, its order is equal to 2(p+1)pn−1. The numbers p = 2 and p ≡ 1 (mod 4) are not prime in G. Thus, for p ≡ 1 (mod 4) we have p = π · π̄;π, π̄ ∈ Z[i], and the residue class rings in Z[i] modulo pn (respectively, πn) are isomorphic. So, this case was investigated in the works mentioned above. Similarly we have for p = 2. That is why we don’t consider these p. If (u0+iv0) is a generating element of the group En, then N(u0+iv0) ≡ −1 (mod pn). It follows that only the elements of the type (u0 + iv0)2a, where a = 0, 1, . . . , (p+ 1)pn−1, are quadratic residues modulo pn in En. Our aim is to prove Theorems 1 and 2 stated in Section 3, and to obtain an asymptotic formula for the number R(x, φ) of quadratic residues in the sectorial domain S(x, φ) = { φ1 6 argw < φ2, 0 < N(w) 6 x, φ2 − φ1 = φ < π 2 } . (1) The formula for R(x, ϕ) is contained in Theorem 2 and has the fol- lowing form R(x;φ) = φ2 − φ1 2 · p+ 1 p · x pn +O ( 3nx 1−s pn log x ) . The most interesting case is the case, when φ2 − φ1 → 0 with x → ∞, because the case φ2 − φ1 > C, C > 0 is a fixed constant, follows from the work [5] about the distribution of values of the function r(n) (the number of representations of n by the sum of two squares) in the arithmetic progression. Notations. We will use the following notations: • G := { a+ bi\|a, b ∈ Z, i2 = −1 } is the ring of Gaussian integers; • Gγ is the ring of residues of Gaussian integers modulo γ; • G∗ γ = {α ∈ Gγ , (α, γ) = 1}; • for α ∈ G we denote N(α) = \|α\|2 ,Sp(α) = 2ℜ(α); • En ⊂ Gpn is the norm group; • χ stands for a character of the group En; • for a ∈ Z (or α ∈ G) νp(a) (or νp(α)) stands that pνp(a)\|a, pνp(a)+1 does not divide a; L. Balyas, P. Varbanets 155 • s ∈ C, s = σ + it, σ = ℜs, t = ℑs; • Γ(z) is the Euler gamma-function; • by f ≪ g(f = O(g)) for x ∈ X, where X is an arbitrary set, on which f and g are defined, we mean that there exists a constant C > 0 such that \|f(x)\| 6 C · g(x) for all x ∈ X; • exp (x) = ex for x ∈ C (sometimes, instead of ex we will use exp(x)). Let us denote E+ n := { α ∈ G∗ pn \|N(α) ≡ 1 (mod pn) } = { α ∈ En\|α = (u0 + iv0)2a, a = 0, 1, . . . , (p+ 1)pn−1 } . Then R(x, φ) = ∑ α∈E+ n ∑ w∈G w≡α (mod pn) w∈S(x,φ) 1. (2) We consider Dirichlet series Fm(s) = ∑ α∈E+ n ∑ w≡α (mod pn) e4mi arg w N(w)s , ℜs > 1. We have Fm(s) = ∑ α∈E+ n 1 N(pn)s ζm ( s; α pn , 0 ) , ℜs > 1, (3) where ζm ( s; α pn , 0 ) is a special case of Hecke zeta-function ζm (s; δ0, δ) with a shift. In the domain ℜs > 1 the last is defined by absolutely convergent Dirichlet series ζm (s; δ0, δ) = ∑ w∈G e4mi arg (w+δ0) N(w + δ0)s eπiSp(δw), where δ0, δ are Gaussian numbers from the field Q(i); Sp(β) is a trace of an element β from Q(i) to Q. 2. Auxiliary results In the following lemmas we bring necessary information about Hecke zeta-function for the next steps. 156 Quadratic residues Lemma 1. The Hecke zeta-function ζm (s; δ0, δ) satisfies the functional equation π−sΓ(2\|m\| + s)ζm (s; δ0, δ) = π−(1−s)Γ (2 \|m\| + 1 − s) · ζ−m (1 − s; δ0,−δ) e−πi Sp(δδ̄0). Moreover, ζm (s; δ0, δ) is an entire function if m 6= 0 or m = 0 and δ is not a Gaussian integer. For m = 0 and δ ∈ G it is holomorphic except for the point s = 1, where it has a simple pole with the residue π. Proof. For δ0 = δ = 0 and m = 4m1, we get the well-known Hecke zeta-function Zm(s) with the Hecke character of the first kind with the exponent m (see, [2]). In [8] this lemma has been stated without a proof. But for the completeness of treatment we restore a proof of this statement. In the general case, for the proof of statement of the lemma we start from the relation Γ(s) \|wδ0\|−2s = ∞ ∫ 0 exp (−x\|w + δ0\|2)xs−1dx. It is evident that for ℜs > 1 and m ∈ Z we can write Γ(2\|m\| + s)Zm(s; δ0; δ) = δ ∫ 0 ∑ w∈G w 6=−δ0 e−x\|w+δ0\|2xs−1dx. Let us denote δ0 = δ01 + δ02. Then a groundtruthing shows that the functions f(u1, u2) = exp (−x ( u2 1 + u2 2 ) + 2πi(δ01u1 + δ02u2)), f̂(v1, v2) = π x exp ( −π2 x [ (δ01 + v1)2 + (δ02 + v2)2 ] ) satisfy the conditions of Poisson summation formula (see, e.g. [6], Ch. VII, Corollary 2.6). Hence, denoting Θm(x, δ0, δ) = ∑ w∈G exp ( −x\|w + δ0\|2 ) (w + δ0)4m exp ( πiSp(δw) ) and using Poisson summation formula, we find Θ0(x, δ0, δ) = π x Θ0 ( π2 x , δ,−δ0 ) exp ( −πiSp(δ0δ) ) . L. Balyas, P. Varbanets 157 Consider the operator d dδ0 := ∂ ∂δ01 + i ∂ ∂δ02 , δ0 = δ01 + δ02. Then the following equalities for m > 0 (−2x)4mΘm(x, δ0, δ) = dm dδm Θ0(x, δ0, δ) and π x (−2πi)4mΘm ( π2 x , δ0,−δ ) exp ( −πiSp ( δ0δ )) = dm dδm 0 ( π x Θ0 ( π2 x , δ,−δ0 ) exp ( −πiSp ( δ0δ )) ) hold. So, for any m ∈ Z the following functional equation Θm(x, δ0, δ) = ( π x )4m+1 Θm ( π2 x , δ, δ0 ) exp ( −πiSp ( δ0δ )) (4) is true. Now, applying reasoning used for the proof of the functional equa- tion for Riemann zeta-function by the functional equation for a theta- function Θm we easily infer Γ(2\|m\| + s)ζm(s, δ0, δ) = π−(1−2s) exp ( −πiSp ( δ0δ )) Im(δ0, δ), where Im(δ0, δ) = ∫ ∞ 0 ∑ w w 6=−δ0 exp (−x\|w + δ0\|2)(w + δ0)4m exp (πiSp(δw))xs+2m−1dx = π ∫ 0 + ∞ ∫ π := Im,1 + Im,2. 158 Quadratic residues In the integral Im,1 we apply the functional equation (4) for Θm(x, δ0,δ) and make the substitution x = π2y−1. This gives the equality Γ(2\|m\| + s)ζm(s, δ0, δ) = π2s−1 exp (−πiSp(δ0δ))× × ∫ ∞ π ∑ w∈G w 6=−δ exp (−x\|w + δ\|2)(w + δ)4m exp (−πiSp(δ0w))x−s+2mdx + ∫ ∞ π ∑ w∈G w 6=−δ0 exp (−x\|w + δ0\|2)(w + δ0)4m exp (−πiSp(δw))xs+2m−1dx + ε(m, δ) πs s− 1 − ε(m, δ0) exp (−πiSp(δ0, δ)) πs s , (5) where ε(m, a) = { 1 if m = 0 and a ∈ G 0 otherwise. The equality (5) was obtained for ℜs > 1. However, the right part of this equality is an analytic function in all complex s-planes except maybe the points s = 0 and s = 1, which can be the poles. Now, multiplying the equality (5) by exp(πiSp(δ0δ))π −2s+1 and mak- ing the substitution s → 1 − s, δ0 → δ, δ → δ0, we obtain that the right part doesn’t vary, and hence, we have proved the following functional equation π−sΓ(2\|m\| + s)ζm(s; δ0, δ) = π−(1−s)Γ(2\|m\| + 1 − s)ζm(1 − s; −δ, δ0) exp(−πiSp(δ0δ)). If m = −m′, m′ > 0, we put δ0 = −δ′ 0, δ = −δ′, and then we have ζm(s, δ, δ0) = ζm′(s,−δ,−δ0) ⇒ ζm′(1 − s, δ0,−δ) = ζm(1 − s,−δ0, δ). So, for any m ∈ Z, π−sΓ(2\|m\| + s)ζm(s; δ, δ0) = π−(1−s)Γ(2\|m\| + 1 − s)ζ−m(1 − s,−δ0, δ) = π−(1−s)Γ(2\|m\| + 1 − s)ζ−m(1 − s; δ0,−δ). This completes the proof of Lemma 1. Corollary 1. If δ is not a Gaussian integer, then ζ0(0; δ0, δ) = 0. L. Balyas, P. Varbanets 159 Lemma 2. In the strip ε 6 ℜs 6 1 + ε, ε > 0, the following estimate (s− 1) · ζm(s; δ0, δ) ≪ (\|t\| + 1)(t2 +m2) (1−2σ)(1+ε−σ) 1+2ε \|N(δ)\|− σ+ε 1+2ε holds. This lemma follows from Phragmen-Lindelof principle and the esti- mates for ζm (s; δ0, δ) on the boundaries of the strip ε 6 ℜs 6 1+ε, which can be received with the usage of the functional equation for ζm (s; δ0, δ) and Stirling formula for Γ(z). Lemma 3. Let y > k ∈ {0, 1, 2}. Let a be a real number, −1 < a 6 5 4 , η(a) = minj=0,1,...,k \|a − j\| 6= 0. Then for any real numbers u, v the following estimate a+iv ∫ a+iu ysψ(s,m) s(s+ 1) . . . (s+ k) ds ≪ N(γ) 1 2M ( ( y N(γ) · 1 M )a (η−1(a)+logM) + ( y N(γ)M ) 1 2 − 2k+1 4 ) , holds, where ψ(s,m) = ( 1 π N(γ) 1 2 )1−2s Γ(2\|m\|+1−s) Γ(2\|m\|+s) , M = \|m\| + 10. Proof. Apply [3, Lemma 8]. Lemma 4 ([7], Theorem 1). Upon the condition D 1 2 6 x < D2 the asymptotic formula ∑ n≡1 (mod D) n6x r(n) = πx D γ0 ∏ p\|D ( 1 − χ4(p) p ) +O  D 1 2 exp  c (logD) 1 2 log logD    +O ( x 1 2 D 1 2 τ(D) ) , γ0 = { 1 if D 6≡ 0 (mod 4), 2 if D ≡ 0 (mod 4) is true. Lemma 5. Let p ≡ 3 (mod 4). Then for n = 1, 2, 3, . . . the estimate ∑ α∈E+ n e πi Sp α2 pn ≪ p n 2 holds. 160 Quadratic residues Proof. In the articles [5] and [9] the following description of elements α ∈ E+ n , n > 2 was given: α = (u0 + iv0)2(p+1)t+k ≡ n−1 ∑ j=0 (Aj(k) + iBj(k)) tj (mod pn). Here (u0 + iv0) is a generator of the group E+ n , t = 0, 1, . . . , pn−1, k = 0, 1, . . . , 2p+ 1. Moreover, A0(k) = u(k), B0(k) = v(k); A1(k) ≡ −py0v(k) (mod p3), B1(k) ≡ py0u(k) (mod p3), A2(k) ≡ −1 2 p2y2 0u(k) (mod p3), B2(k) ≡ −1 2 p2y2 0v(k) (mod p3), (u0 + iv0)k ≡ u(k) + iv(k) (mod pn), (y0, p) = 1. Furthermore, u(k) ≡ 0 (mod p), when k = p+ 1 2 , k = 3(p+ 1) 2 ; v(k) ≡ 0 (mod p), when k = 0, k = p+ 1; Aj(k) ≡ Bj(k) ≡ 0 (mod p3), j = 3, 4, . . . ,m− 1, k = 0, 1, . . . , 2p+ 1. Hence we easily conclude ℜ(α2) ≡ (A2 0(k) −B2 0(k)) + 2(A0(k)A1(k) −B0(k)B1(k))t + (A2 1(k) −B2 1(k)) +A0(k)A2(k) −B0(k)B2(k)t2 (mod p3). Then ℜ(α2) ≡ C0 + C1t+ C2t 2 (mod p3) with the coefficients C1 ≡ −2py0u(k)v(k) (mod p3), C2 ≡ 1 2 p2y2 0(u2(k) − v2(k)) (mod p3) or C2 ≡ 1 2 p2y2 0(1 − 2v2(k)) (mod p3). Let us note that u(k)2 + v(k)2 ≡ (−1)k (mod p). Therefore, it follows that u(k) and v(k) can not divide p simultaneously. It is obvious that νp(C2) > 2 (the strict inequality holds for the cases k=0, p+1 2 , 3(p+1) 2 , p+1). That is why, when νp(C1) < νp(C2), S = 0. So, from the well-known relation, for (b, p) = 1, f(x) ∈ Z[x], ∣ ∣ ∣ ∣ ∣ ∣ ∑ x∈Zpn e 2πi ax+pbx2+p2f(x) pn ∣ ∣ ∣ ∣ ∣ ∣ = { 0 if (a, p) = 1, 2p n−1 2 if a ≡ 0 (mod p) L. Balyas, P. Varbanets 161 we get ∣ ∣ ∣ ∣ ∣ ∣ ∑ α∈E+ n e πi Sp α2 pn ∣ ∣ ∣ ∣ ∣ ∣ 6 4p n 2 . In case n = 1 we take into account that E1 = { ±1,±i, a− i a+ i , i a− i a+ i ∣ ∣ a = 1, 2, . . . , p− 1 } . Thus, we conclude that Sp(α2) can be represented as the ratio of the polynomials of degree 2. Then, following Weil [11], we have ∣ ∣ ∣ ∣ ∣ ∣ ∑ α∈E1 e πi Sp α2 p ∣ ∣ ∣ ∣ ∣ ∣ 6 2 √ p. Hence, the assertion of lemma follows. Lemma 6 ([7], Lemma 5). Let p be a prime number, let u1, u2 be integers and (u1, u2, p n) = pm. Then ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∑ l21+l22≡1 (mod pn) e 2πi u1l1+u2l2 pn ∣ ∣ ∣ ∣ ∣ ∣ ∣ 6 2p n+m 2 . Corollary 2. For m 6= 0 the following estimate ∑ α∈E+ n ζm ( 0; α pn , 0 ) ≪ p 3 2 nM logM, M = \|m\| + 10, holds. This statement follows immediately from the functional equation for ζm(s; δ0, δ) for m 6= 0 and Lemma 6. The following Lemma was proved in [10] (see Lemma 11, pp. 259–260). Lemma 7 (Vinogradov’s ‘glasses’). Let r ∈ N, Ω > 0, 0 < ∆ < 1 2Ω and let φ1, φ2 be real numbers, ∆ 6 φ2 − φ1 6 Ω − 2∆. Then there exists a periodic function f(φ) with the period Ω such that: (i) f(φ) = 1, in the segment φ ∈ [φ1, φ2]; 0 6 f(φ) 6 1 in the segments [φ1 − ∆, φ1] and [φ2, φ2 + ∆]; f(φ) = 0, in the segment [φ2 + ∆, φ1 + Ω − ∆]; 162 Quadratic residues (ii) f(φ) has the expansion in a Fourier series f(φ) = +∞ ∑ m=−∞ ame 2πi mφ Ω , where a0 = 1 Ω(φ2 − φ1 + ∆), and for m 6= 0 and r ∈ N each of the following inequalities holds \|am\| 6        1 Ω(φ2 − φ1 + ∆), 2 π\|m\| , 2 π\|m\| ( rΩ π\|m\|∆ )r . 3. Main results Let us consider the function of a natural argument rm(k) = ∑ u,v∈Z u2+v2=k e4mi arg(u+iv). In view of (3) we can write Fm(s) = ∞ ∑ k6x k≡1 (mod pn) rm(k) ks . Theorem 1. Let m 6= 0, pn 6 x 6 p2n. Then ∑ k6x k≡1 (mod pn) rm(k) ≪ √ x p n 2 + p n 2 log x+ p n 2M logM. Proof. Our assertion is trivial for x ≪ pnM . That is why we will assume that x > C ·Mpn, C > 0. It follows from Lemma 1 that ζm (s; δ0, δ) is an entire function. In view of the fact 1 p2ns ζm ( s; α pn , 0 ) = ∑ w∈G w≡α (mod pn) e 4mi arg w N(w)s for ℜs > 1 and every α ∈ G the usage of the theorem of the residues gives 1 2πi ∫ 2+i·∞ 2−i·∞ xs+2ζm ( s; α pn , 0 ) p2nss(s+ 1)(s+ 2) ds = x2 2 δm ( s; α pn , 0 ) + 1 2πi a+i·∞ ∫ a−i·∞ xs+2ζm ( s; α pn , 0 ) p2nss(s+ 1)(s+ 2) (6) L. Balyas, P. Varbanets 163 for every −1 < a < 0. Let us denote S2(x, α) = 1 2 ∑ 0<N(w)6x w≡α (mod pn) e4mi arg w(x−N(w))2. (7) Using the relation 1 2πi 2+i·∞ ∫ 2−i·∞ ys+l s(s+ 1) . . . (s+ k) = { 1 l!(y − 1)l if y > 1 0 if 0 < y < 1 and taking into account the uniform convergence of the series for zeta- function ζm ( s; α pn , 0 ) in the semiplane ℜs > 1 + ε, ε > 0, we get 1 2πi 2+i·∞ ∫ 2−i·∞ xs+2ζm ( s; α pn , 0 ) p2nss(s+ 1)(s+ 2) ds = ∑ w w≡α (mod pn) e4mi arg w N(w)−2 · 1 2πi 2+i·∞ ∫ 2−i·∞ ( x N(w) )s+2 s(s+ 1)(s+ 2) ds = 1 2 ∑ w≡α (mod pn) N(w)6x e4mi arg w(x−N(w))2 = S2(x, α). (8) The application of the functional equation for ζm(s; δ0, δ) (see Lemma 1) and the estimate ζm(s; δ0, δ) in critical strip (see Lemma 2) give 1 2πi a+i·∞ ∫ a−i·∞ xs+2ζm ( s; α pn , 0 ) p2nss(s+ 1)(s+ 2) = ∑ w∈G\{0} e−4mi arg we πi Sp ( αw pn ) N(w)−s 1 2πi a+i·∞ ∫ a−i·∞ (xN(w))s+2 Γ(2\|m\|+1−s) Γ(2\|m\|+s) π1−2ss(s+1)(s+2)p2ns ds. (9) From (6)–(9) we deduce the formula: S2(x, α) = x2 2 ζm ( 0; α pn , 0 ) + ∑ w∈G\{0} e−4mi arg we πi Sp ( αw pn ) N(w)−sW ( xN(w) p2n ) p2(n+1), 164 Quadratic residues where W (y) = 1 2πi a+i·∞ ∫ a−i·∞ ys+2Γ(2\|m\| + 1 − s) s(s+ 1)(s+ 2)Γ(2\|m\| + s) ds. We consider the following operator ∆zF (x) = 2 ∑ j=0 (−1)jF (x+ jz) = x+z ∫ x dy1 y1+z ∫ y1 F ′′ (y2)dy2. Then ∆z ( x2 2 ζm ( 0; α pn , 0 ) ) = z2ζm ( 0; α pn , 0 ) . It is obvious that for every b, −1 < b < 0, we have W (y) = 1 2πi b+i·∞ ∫ b−i·∞ ys+2Γ(2\|m\| + 1 − s) s(s+ 1)(s+ 2)Γ(2\|m\| + s) ds. We put b = −1 + 1 log y , if y > 1. Using Lemma 3, we conclude that W (y) ≪ K(y,m), (10) where K(y,m) = p3nM3y (log y + logM) . It means that ∆zW ( xN(w) p2n ) ≪ K ( xN(w) p2n ,m ) , (11) if only z ≪ xN(w) p2n . The value ∆zW ( xN(w) p2n ) may be defined in a different way. We put Φ(y) = 1 2πi c+i·∞ ∫ c−i·∞ ys+2Γ(2\|m\| + 1 − s) s(s+ 1)(s+ 2)Γ(2\|m\| + s) ds, c > 1. Then Φ(y) = y2 2 Γ(2\|m\| + 1) Γ(2\|m\|) +W (y). L. Balyas, P. Varbanets 165 For all y > 0 the integrals 1 2πi c+i·∞ ∫ c−i·∞ ys+2Γ(2\|m\| + 1 − s) s · . . . · (s+ 2 − j)Γ(2\|m\| + s) ds, j = 0, 1, 2, converge absolutely and uniformly. Hence, for the derivatives of Φ(y) we have Φ(j)(y) = 1 2πi c+i·∞ ∫ c−i·∞ ys+2−jΓ(2\|m\| + 1 − s) s · . . . · (s+ 2 − j)Γ(2\|m\| + s) ds, j = 0, 1, 2. Thus, W ′′ (y) = −Γ(2\|m\| + 1) Γ(2\|m\|) + 1 2πi c+i·∞ ∫ c−i·∞ ys s Γ(2 \|m\| + 1 − s) Γ(2\|m\| + s) ds. (12) Now we will take into account that the subintegral function doesn’t have singularities in the semiplane ℜs > 0. Then, transfering the contour of the integration in (12) to the line ℜs = 1 log y and using Lemma 3, Stirling formula for the gamma-function, we get W ′′ (y) ≪ L(y,m), where L(y,m) = pn(M logM + y 1 4 ). But then ∆z ( W ( N(w)x p2n )) = N(w) p2n (x+z) ∫ N(w) p2n x dy1 y1+ N(w) p2n z ∫ y1 W ′′ (y2)dy2 ≪ L ( xN(α) p2n ,m ) z2N(w)2 p4n . (13) Let us denote as S2(x) the following sum S2(x) = ∑ α∈E+ n S2(x, α). 166 Quadratic residues We have S2(x) = x2 2 ∑ α∈E+ n ζm ( 0; α pn , 0 ) + ∑ α∈E+ n ∑ w≡α (mod pn) e4mi arg we πi Sp ( αw pn )W ( xN(w) p2n ) N(w)3 = x2 2 ∑ α∈E+ n ζm ( 0; α pn , 0 ) + ∑ χEn 1 \|En\| ∑ α∈En χ̄(α) × ∑ N(w)≡1 (mod pn) χ(w)e4mi arg w N(w)3 · e πi Sp ( αw pn ) W ( xN(w) pn ) . (14) Applying the operator ∆z to both parts of (14), we obtain ∆z (S2(x)) = z2 ∑ α∈E+ n ζm ( 0; α pn , 0 ) + ∑ w∈G N(w)≡1 (mod pn) e−4mi arg wN(w)−3W ( xN(w) p2n ) ∑ α∈E+ n e πi Sp α2 pn . In virtue of (10), (11) and (13), Lemma 5 and Corollary 2 we infer ∆z (S2(x)) ≪ z2p 3 2 nM logM + p n 2 z2 ∑ N(w)6x N(w)−1L ( xN(w) p2n ,m ) + p n 2 ∑ N(w)>x N(w)−3K ( xN(w) p2n ,m ) ≪ z2p n 2 pnM logM + z2p n 2 ∑ N(w)6x pn ( M logM+ N(w) 1 4x 1 4 p n 2 ) N(w)−1 + z2p n 2 ∑ N(w)>x p3nM3N(w)−2p−2n logN(w). (15) From this we get ∆z (S2(x)) ≪ ≪ p 3 2 { z2M logM + z2M logM log x+ z2p− n 2 √ x+M3 log x } . (16) L. Balyas, P. Varbanets 167 The application of the estimates (10), (14) requires that z ≪ xN(w) p2n . Thus the condition N(w) > x in the second sum of (15) allows to assume z = pnM 6 x2 p2n for M ≪ x2 p2n . Then the following inequality ∆z (S2(x)) ≪ z2p 3 2 nM logM holds. Let H2(x) stands for the sum H2(x) = ∑ α∈E+ n ∑ w∈G w≡α (mod pn) N(w)6x e4mi arg w. (17) Then from the definition of S2(x) we easily find H2(x) = d2 dx2 (S2(x)). It is clear that x+z ∫ x dy1 y1+z ∫ y1 H2(y2)dy2 = ∆z (S2(x)) . By x 6 y1 6 x+ 2z and Lemma 4 we have \|H2(y2) −H2(x)\| = \|E+ n \| · ∣ ∣ ∣ ∣ ∣ ∑ x<N(w)6y2 N(w)≡1 (mod pn) e4mi arg w ∣ ∣ ∣ ∣ ∣ 6 (p+ 1)pn−1 ∑ x<n6x+2z n≡1 (mod pn) r(n) 6 πz pn · p+1 p +O (√ x p n 2 ) +O  p n 2 exp  c (log pn) 1 2 log log pn     . Consequently, \|H2(y2)−H2(x)\| = O ( z pn ) +O (√ xp− n 2 ) +O  p n 2 exp  c (log pn) 1 2 log log pn     . It follows that H2(y2) = H2(x) +O ( z pn ) +O (√ xp− n 2 ) +O  p n 2 exp  c (log pn) 1 2 log log pn     . (18) 168 Quadratic residues Now from (17) and (18) we get z2  H2(x) +O ( z pn ) +O (√ xp− n 2 ) +O  p n 2 exp  c (log pn) 1 2 log log pn       = O(z2p n 2M logM). Thus, H2(x) = x 1 2 p− n 2 + p n 2 exp  c (log pn) 1 2 log log pn  + p n 2M logM. So, the proof of Theorem 1 is completed. Now we can investigate the distribution of quadratic residues mod- ulo pn in narrow sectors. Theorem 2. Let p 3 2 n 6 x 6 p2n, 0 6 φ1 < φ2 6 π 2 and let 0 < s 6 1 8 . Then for φ2 − φ1 > x−s the asymptotic formula R(x;φ) = φ2 − φ1 2 · p+ 1 p · x pn +O ( 3nx 1−s pn log x ) holds. Proof. It is known that the distribution of the arguments of Gaussian integers (being considered up to the association) has the period π 2 . In view of this fact the application of Lemma 7 with Ω = π 2 gives for every T > 1 ∑ α∈E+ n φ16α<φ2 N(α)6x 1 = Φ(φ1, φ2) + θ1Φ(φ1 − ∆, φ1) + θ2Φ(φ2, φ2 + ∆), \|θi\| 6 1, i = 1, 2, Φ(φ1, φ2) = 1 4 ∑ w∈E+ n N(w)6x) f(argw) and f(φ) is the function from Lemma 7, 0 < ∆ = 1 2Ω. Furthermore Φ(φ1, φ2) = ∑ w∈E+ n N(w)6x) +∞ ∑ m=−∞ ame4mi arg α = +∞ ∑ m=−∞ am ∑ k≡1 (mod pn) k6x rm(x), where am is the Fourier coefficient from Lemma 7. L. Balyas, P. Varbanets 169 We put δ = xs, 0 < s < 1 (we will find the more precise estimate for s later). Let us use the estimates for the coefficients am (see Lemma 7 with r = 2): \|am\| ≪    1 \|m\| , \|m\| 6 δ = ∆−1; 1 \|m\|3∆2 , \|m\| > δ. After simple calculations we get Φ(φ1, φ2) = φ2 − φ1 2 · p+ 1 p · x pn +O ( x1−s pn ) +O ( s √ x p n 2 log2 x ) +O ( sp n 2 log2 x ) +O ( x 1 2 +s p n 2 ) +O ( 3np n 2 xs log x ) . In view of the assumption of the theorem the following inequalities x1−s pn ≫ p n 2 xs, x1−s pn ≫ x 1 2 +s p n 2 hold. Therefore, Φ(φ1, φ2) = φ2 − φ1 2 · p+ 1 p · x pn +O ( 3nx 1−s pn log x ) . (19) It follows from (19) that Φ(φ1 − ∆, φ1),Φ(φ2, φ2 + ∆) ≪ 3nx 1−s pn log x. (20) The relations (19) and (20) show that Theorem 2 is proved for every s, 0 < s 6 1 8 . Acknowledgment The authors would like to thank the referee for useful remarks and suggestions. References [1] D. Burgess, The Distribution of quadratic residues and non-residues, Mathematika, N.4, 1957, pp.106–112. [2] E. Hecke, Eine neue Art von Zetafunktionen und ihre Beziehungen zur Verteilung der Primzahlen I-II, Math.Z., T.1, 1918, ss.357-376; T.6, 1920, pp.11–51. 170 Quadratic residues [3] J. Kubilius, On one problem of multidimensional analytic number theory, Proc. Vilnius Univ., N.4, 1955, pp.5–41 (in Lithuanian). [4] H. Montgomery, R. Vaughan, Exponential sums with multiplicative coefficients, Inventiones math., N.43, 1977, pp.69–82. [5] S. Sergeev, P. Varbanets, Exponential sums over norm group, Siauliai Math.Seminar, N.9(17), 2014, pp.83–92. [6] Elias M. Stein, Guido Weiss, Introduction to Fourier analysis on Euclidian spaces, Princeton, New Jersey, Princeton University Press, 1971, P.312. [7] P. Varbanets, Problem of circle in aritmetic progression, Mat. Zametki, N.8(6), 1970, pp.787–798 (in Russian). [8] P. Varbanec, P. Zarzycki, Divisors of the Gaussian integers in an Arithmetic Progression, Journal of Number Theory, V.33, N.2, 1989, pp.152–169. [9] S. Varbanets, General Klosterman sums over ring of Gaussian integers, Ukr. Math. J., N.59(9), 2007, pp.1179–2000. [10] I.M. Vinogradov, Selectas, Academic Press USSR, Moscow, 1952 (in Russian). [11] A. Weil, On some exponential sums, Proc. Nat. Acad. Sci. U.S.A., 34, 1948, pp. 204–207. Contact information L. Balyas, P. Varbanets Department of Computer Algebra and Dis- crete Mathematics, I. I. Mechnikov Odessa Na- tional University, Dvoryanskaya 2 65026 Odessa, Ukraine E-Mail(s): balyas@ukr.net, varb@sana.od.ua Web-page(s): onu.edu.ua Received by the editors: 29.01.2016 and in final form 29.12.2016.

Quadratic residues of the norm group in sectorial domains

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