The Quadratic Residues and Some of Their New Distribution Properties

Abstract

In this paper, we give some interesting identities and asymptotic formulas for one kind of counting function, by studying the computational problems involving the symmetry sums of one kind quadratic residues and quadratic non-residues $\mathrm{mod} p$. The main methods we used are the properties of the Legendre’s symbol $\mathrm{mod} p$, and the estimate for character sums. As application, we solve two open problems proposed by Zhiwei Sun.

1. Introduction

Let p be an odd prime. For any integer a satisfying $(a,p)=1$ (i.e., a co-prime to p), if there exists an integer x such that the congruence $x^2\equiv a\mathrm{mod} p$, then a is called a quadratic residue $\mathrm{mod} p$. Otherwise, a is called a quadratic non-residue $\mathrm{mod} p$. In order to facilitate the study of the properties of quadratic residues $\mathrm{mod} p$, Legendre first introduced the characteristic function of the quadratic residues $\mathrm{mod} p$. The defination of Legendre’s symbol $\left(\frac{\ast }{p}\right)$ is that, for any integer a,

$$\left(\frac{a}{p}\right)=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}a\mathrm{is}a\mathrm{quadratic}\mathrm{residues} \mathrm{mod} p;\hfill \\ -1,\hfill & \mathrm{if}a\mathrm{is}a\mathrm{quadratic}\mathrm{non-residues} \mathrm{mod} p;\hfill \\ 0\hfill & \mathrm{if}p\mid a.\hfill \end{array}\right.$$

We can find various interesting properties of the quadratic residues and Legendre’s symbol in number theory books such as [1,2].

In fact, the properties of quadratic residues and Legendre’s symbol $\mathrm{mod} p$ are very meaningful in number theory, which attract attention of many experts and scholars. Valuable research results about them have been obtained. For example, Burgess [3,4] proved that the least quadratic non-residue $\mathrm{mod} p$ is less than $p^{\frac{1}{4}+\epsilon }$, where $\epsilon$ denotes any fixed positive number. The constant $\frac{1}{4}$ has been improved by some authors, see references [5,6,7].

Let $p\ge 3$ be a prime satisfying $p=4k+1$. For any quadratic residue $r\mathrm{mod}p$ and quadratic non-residue $s\mathrm{mod}p$, it holds that (see [2]: Theorems 4–11)

$${\left(\frac{1}{2}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a+r\overline{a}}{p}\right)\right)}^2+{\left(\frac{1}{2}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a+s\overline{a}}{p}\right)\right)}^2=p,$$

(1)

where $\overline{a}$ is satisfied with the equation $\overline{a}·a\equiv 1\mathrm{mod}p$.

Some papers as regards quadratic residues and non-residues $\mathrm{mod} p$ can be found in references [8,9,10,11,12,13,14,15,16].

Recently, Zhiwei Sun proposed two conjectures during his communication with us:

A. For any prime $p\ge 101$, there is at least one integer a, such that a, $a+\overline{a}$ and $a-\overline{a}$ are all quadratic residues $\mathrm{mod} p$?

B. For any prime $p\ge 18$, there is at least one quadratic non-residue $a\mathrm{mod}p$, such that $a+\overline{a}$ and $a-\overline{a}$ are quadratic residues $\mathrm{mod} p$?

Zhiwei Sun carried out some numerical tests to support his conjectures but cannot prove them. In this paper, we concern about problems involving quadratic residues and non-residues $\mathrm{mod} p$.

We think Zhiwei Sun’s conjectures not only show the symmetry of quadratic residues and non-residues but also reveal their profound distribution properties of the quadratic residues and non-residues $\mathrm{mod} p$.

As application of our results, we solve them thoroughly. We actually proved two stronger conclusions. For narrative purposes, let p denote any odd prime $S(p,1)$ denote the number of all integers $1\le a\le p-1$ such that a, $a+\overline{a}$, and $a-\overline{a}$ are all quadratic residues $\mathrm{mod} p$, $S(p,-1)$ denotes the number of all integers $1\le a\le p-1$ such that a is a quadratic non-residue $\mathrm{mod} p$, $a+\overline{a}$ and $a-\overline{a}$ are quadratic residues $\mathrm{mod} p$.

2. Results

The notations are as above. We prove our main results using elementary methods, the estimate for character sums, and the properties of Legendre’s symbol $\mathrm{mod} p$.

Theorem 1.

For any prime p with$p\equiv 3 \mathrm{mod} 4$, we have the identities

$$\begin{array}{c}\hfill S(p,1)=\left\{\begin{array}{cc}\frac{1}{8}\left(p-3\right),\hfill & \mathrm{if} p\equiv 3 \mathrm{mod} 8;\hfill \\ \frac{1}{8}\left(p-7\right),\hfill & \mathrm{if} p\equiv 7 \mathrm{mod} 8\hfill \end{array}\right.\end{array}$$

and

$$\begin{array}{c}\hfill S(p,-1)=\left\{\begin{array}{cc}\frac{1}{8}\left(p-3\right),\hfill & \mathrm{if} p\equiv 3 \mathrm{mod} 8;\hfill \\ \frac{1}{8}\left(p+1\right),\hfill & \mathrm{if} p\equiv 7 \mathrm{mod} 8.\hfill \end{array}\right.\end{array}$$

Theorem 2.

For any prime p with$p\equiv 1 \mathrm{mod} 4$, we have the asymptotic formulas

$$\begin{array}{c}\hfill S(p,1)=\left\{\begin{array}{cc}\frac{1}{8}\left(p-3\right)+K(p,1),\hfill & \mathrm{if} p\equiv 5 \mathrm{mod} 8;\hfill \\ \frac{1}{8}\left(p-17\right)+K_1(p,1),\hfill & \mathrm{if} p\equiv 1 \mathrm{mod} 8\hfill \end{array}\right.\end{array}$$

and

$$\begin{array}{c}\hfill S(p,-1)=\left\{\begin{array}{cc}\frac{1}{8}\left(p+3\right)+K(p,-1),\hfill & \mathrm{if} p\equiv 5 \mathrm{mod} 8;\hfill \\ \frac{1}{8}\left(p+3\right)+K_1(p,-1),\hfill & \mathrm{if} p\equiv 1 \mathrm{mod} 8,\hfill \end{array}\right.\end{array}$$

where we have the estimates$\left|K(p,1)\right|\le \frac{3}{4}·\sqrt{p}$,$|K_1(p,1)|\le \frac{5}{4}·\sqrt{p}$,$\left|K(p,-1)\right|\le \frac{3}{4}·\sqrt{p}$and$|K_1(p,-1)|\le \frac{5}{4}·\sqrt{p}$.

From Theorems 1 and 2 and some simple calculations, we can deduce the following two corollaries:

Corollary 1.

For any prime$p\ge 101$, there is at least one integer a, such that a,$a+\overline{a}$and$a-\overline{a}$are all quadratic residues$\mathrm{mod} p$.

Corollary 2.

For any prime$p\ge 18$, there is at least one quadratic non-residue$a \mathrm{mod} p$, such that$a+\overline{a}$and$a-\overline{a}$are quadratic residues$\mathrm{mod} p$.

Thus, we solved two problems proposed by professor Zhiwei Sun.

3. Several Lemmas

To complete the proofs of our main results, we need the following two basic lemmas.

Lemma 1.

Let p be a prime with$p\equiv 1 \mathrm{mod} 4$; then, for any integer k with$(k,p)=1$, we have the identity

$${\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^2+k}{p}\right)=-1-\left(\frac{k}{p}\right).$$

Proof of Lemma 1.

Note that, for any integer a with $(a,p)=1$, there is $\left(\frac{a}{p}\right)=\left(\frac{\overline{a}}{p}\right)$. Using the properties of the Legendre’s symbol and reduced residue system $\mathrm{mod} p$, we have

$$\begin{array}{cc}& {\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^2+k}{p}\right)={\displaystyle \sum _{a=1}^{p-1}}\left(1+\left(\frac{a}{p}\right)\right)\left(\frac{a+k}{p}\right)={\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a+k}{p}\right)+{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^2+ak}{p}\right)\hfill \\ =& {\displaystyle \sum _{a=0}^{p-1}}\left(\frac{a+k}{p}\right)+{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{1+\overline{a}k}{p}\right)-\left(\frac{k}{p}\right)={\displaystyle \sum _{a=1}^{p-1}}\left(\frac{1+a}{p}\right)-\left(\frac{k}{p}\right)\hfill \\ =& {\displaystyle \sum _{a=0}^{p-1}}\left(\frac{1+a}{p}\right)-\left(\frac{k}{p}\right)-1=-1-\left(\frac{k}{p}\right).\hfill \end{array}$$

This proves Lemma 1. □

Lemma 2.

Let p be a prime with$p\equiv 1 \mathrm{mod} 4$; then, we have the estimate

$$\left|{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a\pm \overline{a}}{p}\right)\right|\le 2·\sqrt{p} and \left|{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^4-1}{p}\right)\left(\frac{a}{p}\right)\right|\le 4·\sqrt{p}.$$

Proof of Lemma 2.

Let $f\left(x\right)=x^3\pm x$; then, $f\left(x\right)$ is not a complete square of an integral coefficient polynomial $g\left(x\right)$. Thus, from Weil’s important work [17], we have

$$\left|{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a\pm \overline{a}}{p}\right)\right|=\left|{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a\pm \overline{a}}{p}\right)\left(\frac{a^2}{p}\right)\right|=\left|{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^3\pm a}{p}\right)\right|\le 2·\sqrt{p}.$$

Of course, for any prime $p=4k+1$, note that 1 and $-1$ are two quadratic residues $\mathrm{mod} p$, so, from identity (1), we can also deduce the above estimate.

This proves the first estimate in Lemma 2.

Now, let ${\chi }_2=\left(\frac{\ast }{p}\right)$, $e\left(y\right)=e^{2\pi iy}$, $\tau \left(\chi \right)={\displaystyle \sum _{a=1}^{p-1}}\chi \left(a\right)e\left(\frac{a}{p}\right)$ denote the classical Gauss sums. Then, note that $\tau \left({\chi }_2\right)=\sqrt{p}$; from the properties of $\tau \left(\chi \right)$, we have

$$\begin{array}{cc}& {\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)\left(\frac{a^4-1}{p}\right)=\frac{1}{\tau \left({\overline{\chi }}_2\right)}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right){\displaystyle \sum _{b=1}^{p-1}}\left(\frac{b}{p}\right)e\left(\frac{b\left(a^4-1\right)}{p}\right)\hfill \\ =& \frac{1}{\sqrt{p}}{\displaystyle \sum _{b=1}^{p-1}}\left(\frac{b}{p}\right)e\left(\frac{-b}{p}\right){\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)e\left(\frac{b{a}^4}{p}\right).\hfill \end{array}$$

(2)

Let g denote a primitive root $\mathrm{mod} p$, $r=g^{\frac{p-1}{4}}$. It is clear that, if $p=8k+5$, then r is a quadratic non-residue $\mathrm{mod} p$ and $r^4\equiv 1 \mathrm{mod} p$. Thus, in this case, for any integer $1\le b\le p-1$, we have

$$\begin{array}{c}\hfill {\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)e\left(\frac{b{a}^4}{p}\right)={\displaystyle \sum _{a=1}^{p-1}}\left(\frac{ar}{p}\right)e\left(\frac{b{\left(ra\right)}^4}{p}\right)=-{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)e\left(\frac{b{a}^4}{p}\right)\end{array}$$

or

$${\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)e\left(\frac{b{a}^4}{p}\right)=0.$$

(3)

If $p=8k+1$, then r is a quadratic residue $\mathrm{mod} p$. Thus, in this case, there must be a character ${\chi }_1 \mathrm{mod} p$ such that ${\chi }_1^4=\left(\frac{\ast }{p}\right)$. Let $\lambda$ be a fourth-order character $\mathrm{mod} p$ (That is, ${\lambda }^4={\chi }_0$ is the principal character $\mathrm{mod} p$); then, from the properties of the Gauss sums and character sums $\mathrm{mod} p$, we have

$$\begin{array}{cc}& {\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)e\left(\frac{b{a}^4}{p}\right)={\displaystyle \sum _{a=1}^{p-1}}{\chi }_1^4\left(a\right)e\left(\frac{b{a}^4}{p}\right)={\displaystyle \sum _{a=1}^{p-1}}{\chi }_1\left(a^4\right)e\left(\frac{b{a}^4}{p}\right)\hfill \\ =& {\displaystyle \sum _{a=1}^{p-1}}{\chi }_1\left(a\right)\left(1+\lambda \left(a\right)+{\lambda }^2\left(a\right)+{\lambda }^3\left(a\right)\right)e\left(\frac{ba}{p}\right)={\overline{\chi }}_1\left(b\right)\tau \left({\chi }_1\right)\hfill \\ & +{\overline{\chi }}_1\left(b\right)\overline{\lambda }\left(b\right)\tau \left({\chi }_1\lambda \right)+{\overline{\chi }}_1\left(b\right){\overline{\lambda }}^2\left(b\right)\tau \left({\chi }_1{\lambda }^2\right)+{\overline{\chi }}_1\left(b\right){\overline{\lambda }}^3\left(b\right)\tau \left({\chi }_1{\lambda }^3\right).\hfill \end{array}$$

(4)

For any character $\chi \mathrm{mod} p$, note that the estimate $\left|\tau \left(\chi \right)\right|\le \sqrt{p}$; from (2) and (4), we have the estimate

$$\begin{array}{cc}& \left|{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)\left(\frac{a^4-1}{p}\right)\right|=\left|\frac{1}{\sqrt{p}}{\displaystyle \sum _{b=1}^{p-1}}\left(\frac{b}{p}\right)e\left(\frac{-b}{p}\right){\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)e\left(\frac{b{a}^4}{p}\right)\right|\hfill \\ \le & \frac{\left|\tau \left({\chi }_1\right)\tau \left({\overline{\chi }}_1{\chi }_2\right)\right|}{\sqrt{p}}+\frac{\left|\tau \left({\chi }_1\lambda \right)\tau \left({\overline{\chi }}_1\overline{\lambda }{\chi }_2\right)\right|}{\sqrt{p}}+\frac{\left|\tau \left({\chi }_1{\lambda }^2\right)\tau \left({\overline{\chi }}_1{\overline{\lambda }}^2{\chi }_2\right)\right|}{\sqrt{p}}\hfill \\ & +\frac{\left|\tau \left({\chi }_1{\lambda }^3\right)\tau \left({\overline{\chi }}_1{\overline{\lambda }}^3{\chi }_2\right)\right|}{\sqrt{p}}\le 4·\sqrt{p}.\hfill \end{array}$$

(5)

Now, the second estimate in Lemma 2 follows from (2), (3) and (5). □

4. Proofs of the Theorems

In this section, we shall complete the proofs of our main results.

Proof of Theorem 1.

For any prime p with $p\equiv 3 \mathrm{mod} 4$, note that $\left(\frac{-1}{p}\right)=-1$, so, for all integers $1\le a\le p-1$, we have $\left(a+\overline{a},p\right)=1$ or $\left(a^2+1,p\right)=1$. If $a=1$ or $p-1$, then $a-\overline{a}=0$, and $\left(a^2-1,p\right)=1$ for all integers $2\le a\le p-2$. Note that the identities

$${\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^4-1}{p}\right)={\displaystyle \sum _{a=1}^{p-1}}\left(\frac{1-{\overline{a}}^4}{p}\right)={\displaystyle \sum _{a=1}^{p-1}}\left(\frac{1-a^4}{p}\right)=-{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^4-1}{p}\right)=0,$$

$${\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)\left(\frac{a^2-{\overline{a}}^2}{p}\right)={\displaystyle \sum _{a=1}^{p-1}}\left(\frac{-a}{p}\right)\left(\frac{a^2-{\overline{a}}^2}{p}\right)=-{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)\left(\frac{a^2-{\overline{a}}^2}{p}\right)=0,$$

$${\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a+\overline{a}}{p}\right)={\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a-\overline{a}}{p}\right)={\displaystyle \sum _{a=1}^{p-1}}\left(\frac{-a-\overline{a}}{p}\right)={\displaystyle \sum _{a=1}^{p-1}}\left(\frac{-a+\overline{a}}{p}\right)=0,$$

from Lemma 1, the definitions of S(p,1) and quadratic residues $\mathrm{mod} p$ we have

$$\begin{array}{cc}& S(p,1)=\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{a+\overline{a}}{p}\right)\right)\left(1+\left(\frac{a-\overline{a}}{p}\right)\right)\hfill \\ =& \frac{p-3}{8}+\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(\frac{a}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(\frac{a+\overline{a}}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(\frac{a-\overline{a}}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(\frac{a^2-{\overline{a}}^2}{p}\right)\hfill \\ & +\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(\frac{a^2+1}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(\frac{a^2-1}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(\frac{a}{p}\right)\left(\frac{a^2-{\overline{a}}^2}{p}\right)\hfill \\ =& \frac{p-3}{8}+\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a+\overline{a}}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a-\overline{a}}{p}\right)-\frac{1}{4}\left(\frac{2}{p}\right)\hfill \\ & +\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^2+1}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^2-1}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)\left(\frac{a^2-{\overline{a}}^2}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^4-1}{p}\right)\hfill \\ =& \frac{p-3}{8}-\frac{1}{4}\left(\frac{2}{p}\right)-\frac{1}{8}+\frac{1}{8}{\displaystyle \sum _{a=0}^{p-1}}\left(\frac{a^2+1}{p}\right)\hfill \\ =& \frac{1}{8}\left(p-5-2\left(\frac{2}{p}\right)\right)=\left\{\begin{array}{cc}\frac{1}{8}\left(p-3\right),\hfill & \mathrm{if} p\equiv 3 \mathrm{mod} 8;\hfill \\ \frac{1}{8}\left(p-7\right),\hfill & \mathrm{if} p\equiv 7 \mathrm{mod} 8.\hfill \end{array}\right.\hfill \end{array}$$

(6)

Similarly, from the methods of proving (6), we have the computational formula

$$\begin{array}{cc}& S(-1,p)=\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(1-\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{a+\overline{a}}{p}\right)\right)\left(1+\left(\frac{a-\overline{a}}{p}\right)\right)\hfill \\ =& \frac{1}{4}{\displaystyle \sum _{a=2}^{p-2}}\left(1+\left(\frac{a+\overline{a}}{p}\right)\right)\left(1+\left(\frac{a-\overline{a}}{p}\right)\right)-S(p,1)\hfill \\ =& \frac{p-3}{4}+\frac{1}{4}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a+\overline{a}}{p}\right)+\frac{1}{4}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a-\overline{a}}{p}\right)+\frac{1}{4}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^2-{\overline{a}}^2}{p}\right)-S(p,1)\hfill \\ =& \frac{1}{8}\left(p-1+2\left(\frac{2}{p}\right)\right)=\left\{\begin{array}{cc}\frac{1}{8}\left(p-3\right),\hfill & \mathrm{if} p\equiv 3 \mathrm{mod} 8;\hfill \\ \frac{1}{8}\left(p+1\right),\hfill & \mathrm{if} p\equiv 7 \mathrm{mod} 8.\hfill \end{array}\right.\hfill \end{array}$$

(7)

It is clear that Theorem 1 follows from Formulas (6) and (7). □

Proof of Theorem 2.

If $p\equiv 1 \mathrm{mod} 4$, then $a+\overline{a}\equiv 0 \mathrm{mod} p$ has two solutions $\lambda$ and $-\lambda$ with $\lambda \ne \pm 1$. Thus, note that $\left(\frac{-1}{p}\right)=1$; from the definition of $S(p,1)$ and the properties of the Legendre’s symbol $\mathrm{mod} p$, we have

$$\begin{array}{cc}& S(p,1)=\frac{1}{8}{\displaystyle \underset{\left(a+\overline{a},p\right)=1}{{\displaystyle \sum _{a=2}^{p-1}}}}\left(1+\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{a+\overline{a}}{p}\right)\right)\left(1+\left(\frac{a-\overline{a}}{p}\right)\right)\hfill \\ =& \frac{p-5}{8}+\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(\frac{a}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(\frac{a+\overline{a}}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(\frac{a-\overline{a}}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(\frac{a^2-{\overline{a}}^2}{p}\right)\hfill \\ & +\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(\frac{a^2+1}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(\frac{a^2-1}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=2}^{p-2}}\left(\frac{a}{p}\right)\left(\frac{a^2-{\overline{a}}^2}{p}\right)\hfill \\ & -\frac{1}{4}\left(1+\left(\frac{\lambda }{p}\right)\right)\left(1+\left(\frac{2\lambda }{p}\right)\right)\hfill \\ =& \frac{p-7}{8}+\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a+\overline{a}}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a-\overline{a}}{p}\right)-\frac{1}{2}\left(\frac{2}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^2+1}{p}\right)\hfill \\ & +\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^2-1}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)\left(\frac{a^2-{\overline{a}}^2}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^4-1}{p}\right)\hfill \\ & -\frac{1}{4}\left(1+\left(\frac{\lambda }{p}\right)+\left(\frac{2\lambda }{p}\right)+\left(\frac{2}{p}\right)\right)\hfill \\ =& \frac{p-13}{8}+\frac{1}{8}\left(1+\left(\frac{\lambda }{p}\right)\right){\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a+\overline{a}}{p}\right)-\frac{3}{4}\left(\frac{2}{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)\left(\frac{a^4-1}{p}\right)\hfill \\ & -\frac{1}{4}\left(1+\left(\frac{2}{p}\right)\right)\left(\frac{\lambda }{p}\right)+\frac{1}{8}{\displaystyle \sum _{a=1}^{p-1}}\left(1+\left(\frac{a}{p}\right)\right)\left(\frac{a^2-1}{p}\right).\hfill \end{array}$$

(8)

When $p=8k+5$, we have $\left(\frac{2}{p}\right)=\left(\frac{\lambda }{p}\right)=-1$; from Lemma 2, we have

$$\begin{array}{c}\left|{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)\left(\frac{a^4-1}{p}\right)\right|+\left|{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)\left(\frac{a^2-1}{p}\right)\right|\hfill \\ \le 4·\sqrt{p}+\left|{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a-\overline{a}}{p}\right)\right|\le 6·\sqrt{p}.\hfill \end{array}$$

(9)

Combining (8) and (9) and Lemma 1, we have the asymptotic formula

$$\begin{array}{c}\hfill S(p,1)=\frac{1}{8}(p-9)+K(p,1),\end{array}$$

(10)

where $\left|K(p,1)\right|\le \frac{3}{4}·\sqrt{p}$.

If $p=8k+1$, then $\left(\frac{2}{p}\right)=\left(\frac{\lambda }{p}\right)=1$; from (8) and (9), Lemmas 1 and 2, we have the asymptotic formula

$$S(p,1)=\frac{1}{8}(p-25)+K_1(p,1),$$

(11)

where $|K_1(p,1)|\le \frac{5}{4}·\sqrt{p}$.

On the other hand, note that the identity

$$\begin{array}{cc}& S(p,-1)=\frac{1}{8}{\displaystyle \underset{\left(a+\overline{a},p\right)=1}{{\displaystyle \sum _{a=2}^{p-1}}}}\left(1-\left(\frac{a}{p}\right)\right)\left(1+\left(\frac{a+\overline{a}}{p}\right)\right)\left(1+\left(\frac{a-\overline{a}}{p}\right)\right)\hfill \\ =& \frac{1}{4}{\displaystyle \underset{\left(a+\overline{a},p\right)=1}{{\displaystyle \sum _{a=2}^{p-1}}}}\left(1+\left(\frac{a+\overline{a}}{p}\right)\right)\left(1+\left(\frac{a-\overline{a}}{p}\right)\right)-S(p,1)\hfill \\ =& \frac{p-5}{4}+\frac{1}{4}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a+\overline{a}}{p}\right)+\frac{1}{4}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a-\overline{a}}{p}\right)-\left(\frac{2}{p}\right)-\frac{1}{2}\hfill \\ & +\frac{1}{4}{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a}{p}\right)\left(\frac{a^2-1}{p}\right)-S(p,1).\hfill \end{array}$$

(12)

From (8)–(12), we have the asymptotic formulas

$$S(p,-1)=\left\{\begin{array}{cc}\frac{1}{8}\left(p+3\right)+K(p,-1),\hfill & \mathrm{if} p\equiv 5 \mathrm{mod} 8;\hfill \\ \frac{1}{8}\left(p+3\right)+K_1(p,-1),\hfill & \mathrm{if} p\equiv 1 \mathrm{mod} 8,\hfill \end{array}\right.$$

(13)

where $\left|K(p,-1)\right|\le \frac{3}{4}·\sqrt{p}$ and $|K_1(p,-1)|\le \frac{5}{4}·\sqrt{p}$.

It is clear that Theorem 2 follows from asymptotic Formulas (10), (11) and (13). □

To prove Corollary 1, we only require $S(p,1)>0$. That is,

$$\begin{array}{c}\hfill \frac{1}{8}(p-17)>\frac{5}{4}·\sqrt{p}.\end{array}$$

This inequality implies that $S(p,1)>0$ for all primes $p>144$.

It is easy to verify $S(p,1)>0$ for all $101\le p\le 143$ by simple calculation.

Thus, Zhiwei Sun’s problem A is correct for all primes $p\ge 101$.

Similarly, we can also prove that the problem B is also correct for all primes $p\ge 18$. This completes the proofs of our all results.

5. Conclusions

The main results of this paper are two theorems. Theorem 1 establishes two exact formulas for $S(p,1)$ and $S(p,-1)$ with $p=4k+3$. Theorem 2 establishes two asymptotic formulas for $S(p,1)$ and $S(p,-1)$ with $p=4k+1$. At the same time, we give two sharp upper bound estimates for the error terms. As application, we obtain two conclusions as follows:

i

For any prime $p\ge 101$, there is at least one integer a, such that a, $a+\overline{a}$ and $a-\overline{a}$ are all quadratic residues $\mathrm{mod} p$.

ii

For any prime $p\ge 18$, there is at least one quadratic non-residue $a \mathrm{mod} p$, such that $a+\overline{a}$ and $a-\overline{a}$ are quadratic residues $\mathrm{mod} p$.

Therefore, we solved two problems proposed by Zhiwei Sun.