Let $p>3$ be a prime, and let $(\frac{\cdot}p)$ be the Legendre symbol. Let
$b\in\mathbb Z$ and $\varepsilon\in\{\pm 1\}$. We mainly prove that
$$\left|\left\{N_p(a,b):\ 1<a<p\ \text{and}\ \left(\frac
ap\right)=\varepsilon\right\}\right|=\frac{3-(\frac{-1}p)}2,$$ where $N_p(a,b)$
is the number of positive integers $x\{ax^2+b\}_p$, and
$\{m\}_p$ with $m\in\mathbb{Z}$ is the least nonnegative residue of $m$ modulo
$p$.Comment: 6 pages. Accepted by C. R. Math. Acad, Sci. Pari

Hao Pan

Qing-Hu Hou

Zhi-Wei Sun

Comptes Rendus Mathématique

arXiv

Comptes RendusMathématiqueQing-Hu Hou, Hao Pan and Zhi-Wei SunA new theorem on quadratic residues modulo primesVolume 360 (2022), p. 1065-1069https://doi.org/10.5802/crmath.371This article is licensed under theCreative Commons Attribution 4.0 International License.http://creativecommons.org/licenses/by/4.0/Les Comptes Rendus. Mathématique sont membres duCentre Mersenne pour l’édition scientifique ouvertewww.centre-mersenne.orge-ISSN : 1778-3569Comptes RendusMathématique2022, Vol. 360, p. 1065-1069https://doi.org/10.5802/crmath.371Number theory / Théorie des nombresA new theorem on quadratic residues moduloprimesQing-HuHoua, Hao Panb and Zhi-Wei Sun∗, ca School of Mathematics, Tianjin University, Tianjin 300350, People’s Republic ofChinab School of Applied Mathematics, Nanjing University of Finance and Economics,Nanjing 210046, People’s Republic of Chinac Department of Mathematics, Nanjing University, Nanjing 210093, People’s Republicof ChinaURL: http://maths.nju.edu.cn/~zwsunE-mails: qh_hou@tju.edu.cn, haopan79@zoho.com, zwsun@nju.edu.cnAbstract. Let p > 3 be a prime, and let ( ·p ) be the Legendre symbol. Let b ∈ Z and ε ∈ {±1}. We mainly provethat ∣∣∣∣{Np (a,b) : 1 < a < p and ( ap)= ε}∣∣∣∣= 3− ( −1p )2 ,where Np (a,b) is the number of positive integers x < p/2 with {x2 +b}p > {ax2 +b}p , and {m}p with m ∈Z isthe least nonnegative residue of m modulo p.2020 Mathematics Subject Classification. 11A15, 11A07, 11R11.Funding. The first, second and third authors are supported by the National Natural Science Foundation ofChina (grants 11771330, 12071208 and 11971222, respectively).Manuscript received 8 March 2022, accepted 26 April 2022.1. IntroductionThe theory of quadratic residues modulo primes plays an important role in fundamental numbertheory.Let p be an odd prime and let a ∈Zwith p - a. By Gauss’ Lemma (cf. [4, p. 52]),(ap)= (−1)|{16k6 p−12 : {ka}p> p2 }|,where( ·p)denotes the Legendre symbol, and we write {x}p for the least nonnegative residue ofan integer x modulo p.∗Corresponding author.ISSN (electronic) : 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/1066 Qing-Hu Hou, Hao Pan and Zhi-Wei SunLet n be any positive odd integer, and let a ∈Z with gcd(a(1−a),n) = 1. In 2020, Z.-W. Sun [6]proved the following new result:(−1)|{16k6 n−12 : {ka}n>k}| =(2a(1−a)n),where( ·n)is the Jacobi symbol.Let p be an odd prime and let a,b ∈Zwith a(1−a) 6≡ 0 (mod p). By [5, Lemma 2.7], we have|{x ∈ {0, . . . , p −1} : {ax +b}p > x}| = p −12.In 2019 Z.-W. Sun [5] employed Galois theory to prove that(−1)|{16i< j6 p−12 : {i 2}p>{ j 2}p }| ={1 if p ≡ 3 (mod 8),(−1)(h(−p)+1)/2 if p ≡ 7 (mod 8),where h(−p) is the class number of the imaginary quadratic fieldQ(p−p).Motivated by the above work, for an odd prime p and integers a and b, we introduce thenotationNp (a,b) :=∣∣∣∣{16 x 6 p −12 : {x2 +b}p > {ax2 +b}p}∣∣∣∣ .Example 1. We have N7(4,0) = 2 since{12}7 < {4×12}7, {22}7 > {4×22}7 and {32}7 > {4×32}7.Let p be a prime with p ≡ 1 (mod4). Then q2 ≡−1 (modp) for some integer q , hence for a, x ∈Zwe have {(qx)2}p > {a(qx)2}p if and only if {x2}p < {ax2}p . Thus, for each a = 2, . . . , p −1 there areexactly (p −1)/4 positive integers x < p/2 such that {x2}p > {ax2}p . Therefore Np (a,0) = (p −1)/4for all a = 2, . . . , p −1.In this paper we establish the following novel theorem which was conjectured by the first andthird authors [3] in 2018.Theorem 2. Let p > 3 be a prime, and let b be any integer. SetS ={Np (a,b) : 1 < a < p and(ap)= 1}andT ={Np (a,b) : 1 < a < p and(ap)=−1}.Then |S| = |T | = 1 if p ≡ 1 (mod 4), and |S| = |T | = 2 if p ≡ 3 (mod 4). Moreover, the set S does notdepend on the value of b.Example 3. Let’s adopt the notation in Theorem 2. For p = 5, we have S = {1} for any b ∈ Z, andthe set T depends on b as illustrated by the following table:b 0 1 2 3 4T {1} {0} {1} {2} {1}.For p = 7, we have S = {1,2} for any b ∈Z, and the set T depends on b as illustrated by the followingtable:b 0 1 2 3 4 5 6T {0,1} {1,2} {2,3} {1,2} {2,3} {1,2} {0,1}.Qing-Hu Hou, Hao Pan and Zhi-Wei Sun 10672. Proof of Theorem 2Lemma 4. For any prime p ≡ 3 (mod 4), we havep−1∑z=1z(zp)=−ph(−p),where h(−p) is the class number of the imaginary quadratic fieldQ(p−p).Remark 5. This is a known result of Dirichlet (cf. [1, Corollary 5.3.13]).Lemma 6. For any prime p ≡ 3 (mod 4) with p > 3, there are x, y, z ∈ {1, . . . , p −1} such that(xp)=(x +1p)= 1, −(yp)=(y +1p)= 1, and(zp)=−(z +1p)= 1.Proof. By a known result (see, e.g., [2, pp. 64–65]), we have∣∣∣∣{x ∈ {1, . . . , p −2} : ( xp)=(x +1p)= 1}∣∣∣∣= p −34 > 0.Hence∣∣∣∣{y ∈ {1, . . . , p −2} : −( yp)=(y +1p)= 1}∣∣∣∣= ∣∣∣∣{y ∈ {1, . . . , p −2} : ( y +1p)= 1}∣∣∣∣− p −34= p −12−1− p −34= p −34> 0and ∣∣∣∣{z ∈ {1, . . . , p −2} : ( zp)=−(z +1p)= 1}∣∣∣∣= ∣∣∣∣{z ∈ {1, . . . , p −2} : ( zp)= 1}∣∣∣∣− p −34= p −12− p −34= p +14> 0.Now the desired result immediately follows. Proof of Theorem 2. Let a ∈ {2, . . . , p −1}. For any x ∈Z, it is easy to see that{ax2 +bp}+{(1−a)x2p}−{x2 +bp}={0 if {x2 +b}p > {ax2 +b}p ,1 if {x2 +b}p < {ax2 +b}p ,where {α} denotes the fractional part of a real number α. ThusNp (a,b) =(p−1)/2∑x=1(1+{x2 +bp}−{ax2 +bp}−{(1−a)x2p})= p −12+(p−1)/2∑x=1{x2 +bp}−(p−1)/2∑x=1{ax2 +bp}−(p−1)/2∑x=1{(1−a)x2p}= p −12+p−1∑x=1( xp )=1{x +bp}−p−1∑y=1( yp )=( ap ){y +bp}−p−1∑z=1( zp )=( 1−ap )zp.Suppose that ( ap ) = εwith ε ∈ {±1}. ThenNp (a,b) = p −12+p−1∑x=1( xp )=1{x +bp}−p−1∑y=1( yp )=ε{y +bp}−p−1∑z=1( zp )=δεzp,where δ= ( a(1−a)p ).1068 Qing-Hu Hou, Hao Pan and Zhi-Wei SunIf ε= 1, thenNp (a,b) = p −12− 1pp−1∑z=1( zp )=δzdoes not depend on b.If p ≡ 1 (mod 4), then (−1p )= 1 and hencep−1∑z=1( zp )=1z =p−1∑z=1( p−zp )=1(p − z) = p p −12−p−1∑z=1( zp )=1z,thusp−1∑z=1( zp )=1z = p p −14andp−1∑z=1( zp )=−1z =p−1∑z=1z −p p −14= p p −14.So, if p ≡ 1 (mod 4), then |S| = |T | = 1, and moreoverS ={p −12− p −14}={p −14}.Now assume that p ≡ 3 (mod 4). We want to show that |S| = |T | = 2. By Lemma 4,p−1∑z=1z(zp)=−ph(−p) 6= 0.Thusp−1∑z=1( zp )=1z =p−1∑z=1z1+ ( zp )2= p p −14− p2h(−p)and hencep−1∑z=1( zp )=−1z =p−1∑z=1z −p−1∑z=1( zp )=1z = p p −14+ p2h(−p).By Lemma 6, for some a ∈ {2, . . . , p − 2} we have ( a−1p ) = ( ap ) = 1 and hence ( a(1−a)p ) = −1. Fora′ = p +1−a, we have (a′p)=−1 and(a′(1−a′)p)=((1−a)ap)=−1.By Lemma 6, for some a∗,b∗ ∈ {2, . . . , p −2} we have−(a∗−1p)=(a∗p)= 1 and(b∗−1p)=−(b∗p)= 1.Note that (a∗(1−a∗)p)= 1 =(b∗(1−b∗)p).Now we clearly have |S| = |T | = 2. Moreover,S ={p −12−(p −14± h(−p)2)}={p −1±2h(−p)4}.The proof of Theorem 2 is now complete. Qing-Hu Hou, Hao Pan and Zhi-Wei Sun 1069References[1] H. Cohen, A Course in Computational Algebraic Number Theory, Graduate Texts in Mathematics, vol. 138, Springer,1993.[2] H. Davenport, The Higher Arithmetic. An Introduction to the Theory of Numbers, 8th ed., Cambridge University Press,2008.[3] Q.-H. Hou, Z.-W. Sun, “Sequence A320159 at OEIS (On-Line Encyclopedia of Integer Sequences)”, 2018, http://oeis.org/A320159.[4] K. Ireland, M. Rosen, A Classical Introduction to Modern Number Theory, 2nd ed., Graduate Texts in Mathematics,vol. 84, Springer, 1990.[5] Z.-W. Sun, “Quadratic residues and related permutations and identities”, Finite Fields Appl. 59 (2019), p. 246-283.[6] ——— , “Quadratic residues and quartic residues modulo primes”, Int. J. Number Theory 16 (2020), no. 8, p. 1833-1858.

A new theorem on quadratic residues modulo primes

Hou, Qing-Hu

Pan, Hao

Sun, Zhi-Wei

Numérisation de Documents Anciens Mathématiques

arXiv.org e-Print Archive

English

Let $p>3$ be a prime, and let $(\frac{\cdot }{p})$ be the Legendre symbol. Let $b\in \mathbb{Z}$ and $\varepsilon \in \lbrace \pm 1\rbrace $. We mainly prove that
\[ \left|\left\lbrace N_p(a,b):\ 1\lbrace ax^2+b\rbrace _p$, and $\lbrace m\rbrace _p$ with $m\in \mathbb{Z}$ is the least nonnegative residue of $m$ modulo $p$

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A new theorem on quadratic residues modulo primes

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