# The Legendre’s Symbol Modulop and Its Some Elementary Results

Authors: Jianghua Li; Yuan Zhang
Journal: Journal of Mathematics
 (2021)

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## Abstract

The main purpose of this article is using the elementary methods and the properties of the character sums to study the calculating problem of the number of the solutions for one kind congruence equation modulop (an odd prime) and give some interesting identities and asymptotic formulas for it.


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## Body

## 1. Introduction

Letp≥3 be a prime. For any integer n, the Legendre’s symbol n/p modulo p is defined as follows:(1)np=1,ifn,p=1andnis a quadratic residue modulop,−1,ifn,p=1andnis a quadratic nonresidue modulop,0,ifp|n.This arithmetical function occupies a very important position in the elementary number theory and analytic number theory, and many classical number theory problems are closely related to it, for example, the least quadratic nonresidue problem, the class number formula of the quadratic field, and the prime number structure. In particular, ifp is a prime with p≡1mod4, then we have the identity (see Theorems 4–11 in [1])(2)p=∑a=1p−1/2a+a¯p2+∑a=1p−1/2a+r⋅a¯p2≡α2p+β2p,where r is any quadratic nonresidue modulo p, and a⋅a¯≡1modp.In addition to its wide range of applications, the Legendre’s symbol has a number of interesting and important properties of its own. The quadratic reciprocity law is one of them. That is, for any two odd primesp and q with p≠q, we have the identity (see Theorem 9.8 in [2])(3)qp⋅pq=−1p−1q−1/4.Some other papers related to character sums can be referred to [3–9]. There is no need to give detailed examples of their contents here, the interested readers may refer to them.In this paper, we will consider the following elementary number theory problem related tok-th residue modulo p, in particular, the quadratic residue modulo p.For any integern with n,p=1, let i,j=±1, and Nn,i,j;p denote the number of the solutions of the congruence equation n≡a+bmodp (1≤a,b≤p−1) such that a/p=i and b/p=j.Theorem 1.
Letp be an odd prime. Then, for any integer n with n,p=1, we have the identity(4)Nn,i,j;p=p−24−i+j4⋅np−i⋅j4⋅−1p.
For any fixed positive integerk≥3 and any prime p with p≡1modk, if we let Mn,k;p denote the number of the solutions of the congruence equation n≡a+bmodp with 1≤a,b≤p−1, both a and b are the k-th residues modulo p. Then, we also have the following.Theorem 2.
Letp be a prime with p≡1mod3. Then, for any integer n with n,p=1, we have the identity(5)Mn,3;p=p−49−2λnτ2λ+λ¯nτ2λ¯−λ¯nτ3λ−λnτ3λ¯9p,if n is a cubic residue modulo p, then we have(6)Mn,3;p=p+d−49−2τ2λ+τ2λ¯9p,where λ is any third-order character modulo p, τλ=∑a=1p−1λaea/p denotes the classical Gauss sums, 4p=d2+27⋅b2, and d is uniquely determined by d≡1mod3.Theorem 3.
Letp be a prime with p≡1mod4. If integer n is a quadratic residue modulo p, then have the identity(7)Mn,4;p=1/16⋅p−2αp−7,ifp≡1mod8,1/16⋅p+6αp−11,ifp≡5mod8,where αp is defined as the same as in (2).It is clear that from Theorem1, we can deduce the following several corollaries.Corollary 1.
Letp be an odd prime. Then, for any integer n with n,p=1, we have the identity(8)Nn,1,1;p=p−5/4,ifp≡1mod4andnp=1,p−1/4,ifp≡1mod4andnp=−1,p−3/4,ifp≡3mod4andnp=1,p+1/4,ifp≡3mod4andnp=−1.Corollary 2.
Letp≡3mod4 be an odd prime. Then, for any integer n with n,p=1, we have the identity(9)Nn,−1,−1;p=p−14,ifp≡1mod4andnp=1,p−54,ifp≡1mod4andnp=−1,p+14,ifp≡3mod4andnp=1,p−34,ifp≡3mod4andnp=−1.Corollary 3.
Letp≡3mod4 be an odd prime. Then, for any integer n with n,p=1, we have the identity(10)Nn,1,−1;p=Nn,−1,1;p=p−14,ifp≡1mod4,p−34,ifp≡3mod4.
Note that the estimateτ2λ+τ2λ¯≤2p and αp≤p, from Theorem 2 and Theorem 3; we can also deduce the following corollaries.Corollary 4.
Letp be an odd prime with p≡1mod3. Then, for any integer n with n,p=1, we have the asymptotic formula(11)Mn,3;p=19⋅p+Op.Corollary 5.
Letp be a prime with p≡1mod3. Then, for any cubic residue n modulo p, we have the identity(12)Mn,3;p=p+d9orp+d9−1.Corollary 6.
Letp be a prime with p≡1mod4. Then, for any quadratic residue nmodp, we have the asymptotic formula(13)Mn,4;p=116⋅p+Op.
Some notes: it is very interesting that the right hand side of the identity in Corollary3 does not depend on constant n. Of course, the research content and methods in this paper can be further promoted.


## 2. Several Lemmas

In this section, we will give several necessary lemmas. Of course, the proofs of the theorems and these lemmas need the knowledge of elementary and analytic number theory, in particular, the properties of the quadratic residues modulop and the classical Gauss sums. All these can be found in references [1, 2, 10], and we do not repeat them. First, we have the following.Lemma 1.
Letp be an odd prime with p≡1mod3. Then, for any three-order character λmodp, we have the identity(14)τ3λ+τ3λ¯=dp,where 4p=d2+27⋅b2, and d is uniquely determined by d≡1mod3,Proof.
See the study by Zhang and Hu [4] or Berndt and Evans [5].Lemma 2.
Letp be an odd prime with p≡1mod3. Then, we have(15)∑m=1p−1e−nmp−1+λ¯mτλ+λmτλ¯2=−2p+1−2λnτ2λ+λ¯nτ2λ¯+λ¯nτ3λ+λnτ3λ¯.
Ifn is a cubic residue modulo p, then we have(16)∑m=1p−1e−nmp−1+λ¯mτλ+λmτλ¯2=dp−2p−1−2τ2λ+τ2λ¯,where d is defined as the same as in Lemma 1.Proof.
For any integer1≤a≤p−1, note that λ¯=λ2 and τλ⋅τλ¯=p; from the properties of the three-order character λmodp and the classical Gauss sums τλ modulo p, we have(17)∑m=1p−1e−nmp−1+λ¯mτλ+λmτλ¯2=−1−2∑m=1p−1e−nmpλ¯mτλ+λmτλ¯+∑m=1p−1e−nmp2p+λmτ2λ+λ¯mτ2λ¯=−2p+1−2λnτ2λ+λ¯nτ2λ¯+λ¯nτ3λ+λnτ3λ¯.
Ifn is a cubic residue modulo p, then we have λ−1=1 and λn=λ¯n=1; from (17) and Lemma 1, we have(18)∑m=1p−1e−nmp−1+λ¯mτλ+λmτλ¯2=−2p+1−2τ2λ+τ2λ¯+τ3λ+τ3λ¯=dp−2p−1−2τ2λ+τ2λ¯.
Now, Lemma2 follows from (18).Lemma 3.
Letp be an odd prime with p≡1mod4. Then, for any four-order character ψ modulo p, we have the identity(19)τ2ψ+τ2ψ¯=2p⋅αp,where αp is defined as in (2).Proof.
See Lemma 2.2 in the study by Chen and Zhang [7].


## 3. Proofs of the Theorems

Applying three simple lemmas in Section2, we can easily complete the proofs of our theorems. First, we prove Theorem 1. Note that the trigonometric identity(20)∑m=0p−1enmp=p,ifn,p=p,0,ifn,p=1,τ2χ2=−1/p⋅p, so from the definition of Nn,i,j;p and the properties of the Legendre’s symbol, we have(21)Nn,i,j;p=1p∑a=1p−1∑b=1p−1141+iap1+jbp∑m=0p−1ema+b−np=14p∑m=1p−1e−nmp∑a=1p−1∑b=1p−11+iap1+jbpema+bp+14p∑a=1p−1∑b=1p−11+iap1+jbp=14p∑m=1p−1e−nmp−1+i∑a=1p−1apemap⋅−1+j∑b=1p−1bpembp+14pp−1+i∑a=1p−1ap⋅p−1+j∑b=1p−1bp=14p∑m=1p−1e−nmp−1+impτχ2−1+jmpτχ2+p−124p=14p−1−i−npτ2χ2−j−npτ2χ2−i⋅j⋅τ2χ2+p−124p=p−24−i+j4⋅np−i⋅j4⋅−1p,where χ2=∗/p is the Legendre’s symbol modulo p. This proves Theorem 1.Now, we prove Theorem2. For any integer 1≤a≤p−1, from the properties of the third-order character λ modulo p, we have(22)1+λa+λ¯a=3,ifais a cubic residue modulop,0,otherwise.Applying formulas (20) and (22) and Lemma 2, the definition of Mn,3;p, and the properties of the classical Gauss sums, we have(23)Mn,3;p=19p∑a=1p−1∑b=1p−11+λa+λ¯a1+λb+λ¯b∑m=0p−1ema+b−np=19p∑m=1p−1e−nmp∑a=1p−11+λa+λ¯aemap2+19p∑a=1p−1∑b=1p−11+λa+λ¯a1+λb+λ¯b=p−129p+19p∑m=1p−1e−nmp−1+λ¯mτλ+λmτλ¯2=p−49−2λnτ2λ+λ¯nτ2λ¯−λ¯nτ3λ−λnτ3λ¯9p.Ifn is a cubic residue modulo p, then we have λn=λ¯n=1. From (23) and Lemma 1, we may immediately deduce(24)Mn,3;p=p+d−49−29p⋅τ2λ+τ2λ¯.Now, Theorem2 follows from (23) and (24).Similarly, using the methods of proving Theorem2, we can also deduce Theorem 3. In fact, let ψ be a four-order character modulo p, then ψ2=χ2=∗/p. From the properties of the four-order character, we have(25)1+ψa+ψ2a+ψ¯a=4,ifais a quartic residue modulop,0,otherwise.Note thatχ2−1=1; from (25) and the methods of proving Theorem 2, we have(26)Mn,4;p=116p∑m=1p−1e−nmp∑a=1p−11+ψa+χ2a+ψ¯aemap2+116p∑a=1p−1∑b=1p−11+ψa+χ2a+ψ¯a1+ψb+χ2b+ψ¯b=116p∑m=1p−1e−nmp−1+ψ¯mτψ+χ2mp+ψmτψ¯2+p−1216p=p−316−18np−18p1+ψ−n+ψ¯−nτψτψ¯+18pψ¯−nτ2ψ+ψ−nτ2ψ¯+χ2n16pτ2ψ+τ2ψ¯.Ifp=8k+5, n is a quadratic residue modulo p, then we have ψ−1=ψn=1 and τψτψ¯=p. Thus, from (26) and Lemma 3, we have(27)Mn,4;p=p−316−18−38+2⋅αp8+2αp16=116⋅p+6αp−11.Ifp=8k+1, n is a quadratic residue modulo p, then we have ψ−1=−1, ψn=1, and τψτψ¯=−p. Thus, from (26) and Lemma 3, we have(28)Mn,4;p=p−316−18−18−2⋅αp8+2αp16=116⋅p−2αp−7.Now, Theorem3 follows from (27) and (28).This completes the proofs of all our results.



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