On the Quartic Residues and Their New Distribution Properties

Abstract

In this paper, we use the analytic methods, the properties of the fourth-order characters, and the estimate for character sums to study the computational problems of one kind of special quartic residues modulo p, and give an exact calculation formula and asymptotic formula for their counting functions.

1. Introduction

Let p be an odd prime and $k\ge 2$ be an integer. For any integer a with $(a,p)=1$, if the congruence equation $x^k\equiv amodp$ has solution, then we call a is a k-th residue modulo p. Otherwise, a is called a k-th non-residue modulo p. If $k=2$, then we call a quadratic residue or quadratic non-residue modulo p. Legendre first introduced the characteristic function of the quadratic residues $\left(\frac{a}{p}\right)$, which was later called Legendre’s symbol. It is defined as follows:

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

In particular, we have $\left(\frac{-1}{p}\right)={(-1)}^{\frac{p-1}{2}}$, $\left(\frac{2}{p}\right)={(-1)}^{\frac{p^2-1}{8}}$ and the quadratic reciprocity law

$$\left(\frac{q}{p}\right)·\left(\frac{p}{q}\right)={(-1)}^{\frac{(p-1)(q-1)}{4}},$$

where p and q are two different odd primes.

The study of k-th residues modulo p is one of the important pieces of content in elementary number theory and analytic number theory, and many number theory problems are closely related to them. Because of this, many scholars have been engaged in the research work in this field, and have made rich research results. It is worth mentioning that Sun Zhihong [1,2,3,4,5] has done a lot of profound research on the quartic residues; these bring us to the study of the distribution properties of various k-th residues modulo p. Some other papers related to quadratic residues and cubic residues modulo p can be found in references [6,7,8,9,10,11,12,13,14,15,16,17,18]. For example, recently, Wang Tingting and Lv Xingxing [6] studied the distribution properties of some special quadratic residues and non-residues modulo p, who obtained an exact calculation formula and a sharp asymptotic formula for its counting function. As some applications, they solved two problems proposed by Sun Zhiwei. That is, they proved the following two interesting results:

(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 modulo p.

(B). For any prime $p\ge 18$, there is at least one quadratic non-residue $amodp$, such that $a+\overline{a}$ and $a-\overline{a}$ are quadratic residues modulo p, where $\overline{a}$ is defined as $a\overline{a}\equiv 1modp$.

As an extension of Wang Tingting and Lv Xingxing’s work [6], a natural problem is the quartic residues modulo p. It is clear that, if $p\equiv 3mod4$, then the quadratic residue is the same as the quartic residue modulo p. In this time, the problem is trivial. Thus, we just consider the non-trivial case $p\equiv 1mod4$.

Let p be a prime with $p\equiv 1mod4$, and $N(p)$ denotes the number of all integers $1<a<p-1$ such that $a+\overline{a}$ and $a-\overline{a}$ are quartic residues modulo p.

In this paper, we will use the analytic methods, the properties of the classical Gauss sums, and the estimate for character sums to study the computational problems of $N(p)$, and give an exact calculation formula and asymptotic formula for it. That is, we will prove the following two results:

Theorem 1.

Let p be an odd prime with $p\equiv 5mod8$, then we have

$$N(p)=\frac{1}{16}·\left(p-7-2\sum _{a=1}^{\frac{p-1}{2}}\left(\frac{a+\overline{a}}{p}\right)\right).$$

Theorem 2.

Let p be an odd prime with $p\equiv 1mod8$, then we have the identity

$$\begin{array}{c}\hfill N(p)=\frac{1}{16}·\left(p-17\right)+E(p),\end{array}$$

where we have the estimates $\left|E(p)\right|\le \frac{15}{4}·\sqrt{p}$.

From our theorems, we can also deduce the following two corollaries:

Corollary 1.

Let p be an odd prime with $p\equiv 5mod8$, and then we have the congruence

$$p\equiv 7+2\sum _{a=1}^{\frac{p-1}{2}}\left(\frac{a+\overline{a}}{p}\right)mod16,$$

which implies that

$$\sum _{a=1}^{\frac{p-1}{2}}\left(\frac{a+\overline{a}}{p}\right)\equiv \left\{\begin{array}{ll}-1mod8,& ifp\equiv 5mod16;\\ 3mod8,& ifp\equiv 13mod16.\end{array}\right.$$

Corollary 2.

Let $p>3700$ be a prime with $p\equiv 1mod4$, then there exists at least one integer a such that $a+\overline{a}$ and $a-\overline{a}$ are quartic residues modulo p.

Some notes: Prior work gave us great inspiration for the research of this paper, but the methods we used is completely different from the methods in [6] or [1,2,3,4,5], where they all use elementary methods, so they can only get some qualitative results or asymptotic formulas. We used some analytic methods and the properties of the classical Gauss sums. Thus, an accurate calculation formula is obtained.

In addition, Corollary 2 is a very rough estimate deduced directly from our theorems. If we use some mathematical software, then the constant 3700 in Corollary 2 can be made much smaller.

2. Several Lemmas

In this section, we need to prove several simple lemmas. For ease of understanding, we first define the symbols that appear below: $\tau \left(\chi \right)$ denotes the classical Gauss sum

$$\tau \left(\chi \right)=\sum _{a=1}^q\chi (a)e\left(\frac{a}{q}\right)$$

where $q\ge 3$ is an integer, $\chi$ is any Dirichlet character $modq$, and $e(y)=e^{2\pi iy}$.

$\lambda$ denotes any fourth-order character $modp$, which is $\lambda \ne {\chi }_0$, ${\lambda }^4={\chi }_0$.

${\chi }_2=\left(\frac{*}{p}\right)$ denotes the Legendre’s symbol $modp$.

The basic knowledge required in this section can also be found in references [19,20]. We will decompose the proofs of our theorems into the following several lemmas by means of the characteristic function of the fourth-order character $\lambda modp$. In the end, we only deal with some estimate for a certain character sums or calculations for some special Gauss sums. First, we have:

Lemma 1.

Let p be a prime with $p\equiv 1mod4$. Then, for any fourth-order character $\lambda modp$, one has the identity

$${\tau }^2\left(\lambda \right)+{\tau }^2\left(\overline{\lambda }\right)=2\sqrt{p}\sum _{a=1}^{\frac{p-1}{2}}\left(\frac{a+\overline{a}}{p}\right),$$

where $\left(\frac{*}{p}\right)$ denotes the Legendre’s symbol $modp$ and $a\overline{a}\equiv 1modp$.

Proof. 

This result is Theorem 1 in Chen and Zhang [21]. □

Lemma 2.

Let p be an odd prime with $p\equiv 5mod8$. Then, for any fourth-order character $\lambda modp$, we have the identity

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

and

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

Proof. 

Note that $p\equiv 5mod8$, ${\chi }_2={\lambda }^2={\overline{\lambda }}^2$ and $\lambda (-1)=-1$, so, from the properties of the Legendre’s symbol and complete residue system $modp$, we have

$$\begin{array}{ll}& {\displaystyle \sum _{a=1}^{p-1}}\lambda \left(a^2-{\overline{a}}^2\right)={\displaystyle \sum _{a=1}^{p-1}}\lambda \left(a-\overline{a}\right)+{\displaystyle \sum _{a=1}^{p-1}}{\chi }_2(a)\lambda \left(a-\overline{a}\right)\\ =& {\displaystyle \sum _{a=1}^{p-1}}\lambda \left(-a+\overline{a}\right)+{\displaystyle \sum _{a=1}^{p-1}}{\lambda }^2(a)\overline{\lambda }(a)\lambda \left(a^2-1\right)\\ =& -{\displaystyle \sum _{a=1}^{p-1}}\lambda \left(a-\overline{a}\right)+{\displaystyle \sum _{a=1}^{p-1}}\lambda (a)\lambda \left(a^2-1\right)={\displaystyle \sum _{a=1}^{p-1}}\lambda (-a)\lambda \left({(-a)}^2-1\right)\\ =& -{\displaystyle \sum _{a=1}^{p-1}}\lambda (a)\lambda \left(a^2-1\right)=0.\end{array}$$

(1)

Similarly, we can also deduce that

$$\begin{array}{c}\sum _{a=1}^{p-1}\overline{\lambda }\left(a^2-{\overline{a}}^2\right)=0.\end{array}$$

(2)

It is clear that the first formula in Lemma 2 follows from (1) and (2).

Now, we prove the second one. Note that $\tau ({\chi }_2)=\sqrt{p}$, $\lambda (-1)=-1$, ${\lambda }^2={\overline{\lambda }}^2={\chi }_2$, from Lemma 1, and the properties of Gauss sums, we have

$$\begin{array}{ll}& {\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a-\overline{a}}{p}\right)={\displaystyle \sum _{a=1}^{p-1}}{\lambda }^2(a){\chi }_2\left(a^2-1\right)\\ =& {\displaystyle \sum _{a=1}^{p-1}}\lambda (a){\chi }_2\left(a-1\right)+{\displaystyle \sum _{a=1}^{p-1}}\lambda (a){\chi }_2(a){\chi }_2\left(a-1\right)\\ =& \frac{1}{\sqrt{p}}{\displaystyle \sum _{b=1}^{p-1}}{\chi }_2(b){\displaystyle \sum _{a=1}^{p-1}}\lambda (a)e\left(\frac{b(a-1)}{p}\right)+\frac{1}{\sqrt{p}}{\displaystyle \sum _{b=1}^{p-1}}{\chi }_2(b){\displaystyle \sum _{a=1}^{p-1}}\overline{\lambda }(a)e\left(\frac{b(a-1)}{p}\right)\\ =& -\frac{1}{\sqrt{p}}\left({\tau }^2(\lambda )+{\tau }^2\left(\overline{\lambda }\right)\right)=-2{\displaystyle \sum _{a=1}^{\frac{p-1}{2}}}\left(\frac{a+\overline{a}}{p}\right)=-{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a+\overline{a}}{p}\right),\end{array}$$

or

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

This proves the second formula in Lemma 2. □

Lemma 3.

Let p be an odd prime with $p\equiv 5mod8$. Then, we have the identity

$$\begin{array}{c}\hfill \sum _{a=1}^{p-1}\left(\frac{a^2-{\overline{a}}^2}{p}\right)=-2\sum _{a=1}^{\frac{p-1}{2}}\left(\frac{a+\overline{a}}{p}\right)-2.\end{array}$$

Proof. 

Note that ${\lambda }^2={\overline{\lambda }}^2={\chi }_2$, from the method of proving Lemma 2, the properties of the Gauss sums and the Legendre’s symbol $modp$ we have

$$\begin{array}{ll}& {\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a^2-{\overline{a}}^2}{p}\right)={\displaystyle \sum _{a=1}^{p-1}}{\chi }_2\left(a-\overline{a}\right)+{\displaystyle \sum _{a=1}^{p-1}}{\chi }_2(a){\chi }_2\left(a-\overline{a}\right)\\ =& {\displaystyle \sum _{a=1}^{p-1}}{\lambda }^2(a){\chi }_2\left(a^2-1\right)+{\displaystyle \sum _{a=1}^{p-1}}{\chi }_2\left(a^2-1\right)\\ =& {\displaystyle \sum _{a=1}^{p-1}}\lambda (a){\chi }_2\left(a-1\right)+{\displaystyle \sum _{a=1}^{p-1}}\lambda (a){\chi }_2(a){\chi }_2\left(a-1\right)+{\displaystyle \sum _{a=0}^{p-1}}{\chi }_2\left({(a+1)}^2-1\right)-1\\ =& -\frac{{\tau }^2(\lambda )}{\sqrt{p}}-\frac{{\tau }^2\left(\overline{\lambda }\right)}{\sqrt{p}}+{\displaystyle \sum _{a=1}^{p-1}}{\chi }_2\left(a^2+2a\right)-1\\ =& -2{\displaystyle \sum _{a=1}^{\frac{p-1}{2}}}\left(\frac{a+\overline{a}}{p}\right)+{\displaystyle \sum _{a=1}^{p-1}}{\chi }_2\left(1+2\overline{a}\right)-1=-2{\displaystyle \sum _{a=1}^{\frac{p-1}{2}}}\left(\frac{a+\overline{a}}{p}\right)-2.\end{array}$$

This proves Lemma 3. □

Lemma 4.

Let p be an odd prime with $p\equiv 5mod8$. Then, we have the identity

$$\begin{array}{c}\hfill \sum _{a=1}^{p-1}{\chi }_2\left(a+\overline{a}\right)\lambda \left(a-\overline{a}\right)=\sum _{a=1}^{p-1}{\chi }_2\left(a-\overline{a}\right)\lambda \left(a+\overline{a}\right)=0.\end{array}$$

Proof. 

Note that $\lambda (-1)=-1$, ${\lambda }^2={\overline{\lambda }}^2={\chi }_2$, and for any integer a with $(a,p)=1$, we have ${\chi }_2(a)={\chi }_2\left(\overline{a}\right)$. Thus,

$$\begin{array}{ll}& {\displaystyle \sum _{a=1}^{p-1}}{\chi }_2\left(a+\overline{a}\right)\lambda \left(a-\overline{a}\right)={\displaystyle \sum _{a=1}^{p-1}}{\lambda }^2(a){\chi }_2\left(a^2+1\right)\overline{\lambda }(a)\lambda \left(a^2-1\right)\\ =& {\displaystyle \sum _{a=1}^{p-1}}\lambda (a){\chi }_2\left(a^2+1\right)\lambda \left(a^2-1\right)={\displaystyle \sum _{a=1}^{p-1}}\lambda (-a){\chi }_2\left({(-a)}^2+1\right)\lambda \left({(-a)}^2-1\right)\\ =& -{\displaystyle \sum _{a=1}^{p-1}}\lambda (a){\chi }_2\left(a^2+1\right)\lambda \left(a^2-1\right)=0.\end{array}$$

(3)

Similarly, we can also deduce the identity

$$\begin{array}{c}\hfill \sum _{a=1}^{p-1}{\chi }_2\left(a-\overline{a}\right)\lambda \left(a+\overline{a}\right)=0.\end{array}$$

(4)

From (3) and (4), we can deduce Lemma 4. □

Lemma 5.

Let p be an odd prime with $p\equiv 5mod8$. Then, for any fourth-order character $\lambda modp$, we have the identities

$$\begin{array}{c}\hfill \sum _{a=1}^{p-1}\lambda \left(a-\overline{a}\right)\overline{\lambda }\left(a+\overline{a}\right)=\sum _{a=1}^{p-1}\lambda \left(a+\overline{a}\right)\overline{\lambda }\left(a-\overline{a}\right)=0.\end{array}$$

Proof. 

Note that $\lambda (-1)=-1$, from the properties of the reduced residue system modulo p, we have the identity

$$\begin{array}{ll}& {\displaystyle \sum _{a=1}^{p-1}}\lambda \left(a^2+1\right)\overline{\lambda }\left(a^2-1\right)={\displaystyle \sum _{a=1}^{p-1}}\lambda \left(1+{\overline{a}}^2\right)\overline{\lambda }\left(1-{\overline{a}}^2\right)\\ =& {\displaystyle \sum _{a=1}^{p-1}}\lambda \left(1+a^2\right)\overline{\lambda }\left(1-a^2\right)=-{\displaystyle \sum _{a=1}^{p-1}}\lambda \left(a^2+1\right)\overline{\lambda }\left(a^2-1\right),\end{array}$$

which implies that

$$\begin{array}{c}\hfill \sum _{a=1}^{p-1}\lambda \left(a^2+1\right)\overline{\lambda }\left(a^2-1\right)=0.\end{array}$$

(5)

From (5), we may immediately get

$$\begin{array}{c}\sum _{a=1}^{p-1}\lambda \left(a+\overline{a}\right)\overline{\lambda }\left(a-\overline{a}\right)=\sum _{a=1}^{p-1}\lambda \left(a^2+1\right)\overline{\lambda }\left(a^2-1\right)=0.\hfill \end{array}$$

This proves Lemma 5. □

Lemma 6.

Let p be an odd prime with $p\equiv 5mod8$. For any fourth-order character $\lambda modp$, if $r+\overline{r}\equiv 0modp$, then

$$\begin{array}{c}\hfill \left(1+\lambda (2r)+{\lambda }^2(2r)+\overline{\lambda }(2r)\right)+\left(1+\lambda (-2r)+{\lambda }^2(-2r)+\overline{\lambda }(-2r)\right)=4.\end{array}$$

Proof. 

Since ${\lambda }^3=\overline{\lambda }$, $\lambda (-1)=-1$ and $\left(\frac{2}{p}\right)={\lambda }^2(2)={\chi }_2(2)={(-1)}^{\frac{p^2-1}{8}}=-1$, note that $r^2\equiv -1modp$ or $r^3\equiv -rmodp$, we have

$$\begin{array}{l}\lambda (2r)=\lambda (2){\overline{\lambda }}^3(r)=\lambda (2)\overline{\lambda }\left(r^3\right)=\lambda (2)\overline{\lambda }\left(-r\right)\\ =\lambda (-1){\lambda }^2(2)\overline{\lambda }\left(2r\right)=-{\chi }_2(2)·\overline{\lambda }\left(2r\right)=\overline{\lambda }\left(2r\right).\end{array}$$

Therefore, $\lambda (2r)$ is a real number. However, $|\lambda (2r)|=1$, and we must have $\lambda (2r)=1$ or $-1$.

If $\lambda (2r)=1$, then we have $1+\lambda (2r)+{\lambda }^2(2r)+\overline{\lambda }(2r)=1+1+1+1=4$ and

$$1+\lambda (-2r)+{\lambda }^2(-2r)+\overline{\lambda }(-2r)=1-1+1-1=0.$$

If $\lambda (2r)=-1$, then we have $1+\lambda (2r)+{\lambda }^2(2r)+\overline{\lambda }(2r)=1-1+1-1=0$ and

$$1+\lambda (-2r)+{\lambda }^2(-2r)+\overline{\lambda }(-2r)=1+1+1+1=4.$$

This proves Lemma 6. □

3. Proofs of the Theorems

In this section, we shall complete the proofs of our main results. First, we prove Theorem 1. For any prime p with $p\equiv 5mod8$, there must exist an integer $1<r<p-1$ such that $r^2\equiv -1modp$ or $r+\overline{r}\equiv 0modp$. Let $\lambda$ denote any fourth-order character modulo p. Then, for any integer n with $(n,p)=1$, we have the characteristic function:

$$1+\lambda (n)+{\lambda }^2(n)+{\lambda }^3(n)=\left\{\begin{array}{ll}4,& \mathrm{if}n\mathrm{is}\mathrm{a}\mathrm{quartic}\mathrm{residue}\mathrm{modulo}p;\\ 0,& \mathrm{if}n\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{quartic}\mathrm{residue}\mathrm{modulo}p.\end{array}\right.$$

Since $\lambda (2)={\overline{\lambda }}^3(2)={\chi }_2(2)\overline{\lambda }(2)=-\overline{\lambda }(2)$ and ${\chi }_2(\pm 2)=-1$, from Lemma 6 and the definition of $N(p)$, we have

$$\begin{array}{ll}& N(p)=\frac{1}{16}{\displaystyle \underset{(a^2+1,p)=1}{{\displaystyle \sum _{a=2}^{p-2}}}}\left(1+\lambda \left(a+\overline{a}\right)+{\chi }_2\left(a+\overline{a}\right)+\overline{\lambda }\left(a+\overline{a}\right)\right)\\ & \times \left(1+\lambda \left(a-\overline{a}\right)+{\chi }_2\left(a-\overline{a}\right)+\overline{\lambda }\left(a-\overline{a}\right)\right)\\ =& \frac{1}{16}{\displaystyle \sum _{a=1}^{p-1}}\left(1+\lambda \left(a+\overline{a}\right)+{\chi }_2\left(a+\overline{a}\right)+\overline{\lambda }\left(a+\overline{a}\right)\right)\\ & \times \left(1+\lambda \left(a-\overline{a}\right)+{\chi }_2\left(a-\overline{a}\right)+\overline{\lambda }\left(a-\overline{a}\right)\right)-\frac{4}{16}\\ =& \frac{1}{16}{\displaystyle \sum _{a=1}^{p-1}}\left(\lambda \left(a+\overline{a}\right)+\overline{\lambda }\left(a+\overline{a}\right)+\overline{\lambda }\left(a-\overline{a}\right)+\lambda \left(a-\overline{a}\right)+1\right)\\ & +\frac{1}{16}{\displaystyle \sum _{a=1}^{p-1}}\left(\lambda \left(a+\overline{a}\right)\lambda \left(a-\overline{a}\right)+\overline{\lambda }\left(a+\overline{a}\right)\overline{\lambda }\left(a-\overline{a}\right)+{\chi }_2\left(a^2-{\overline{a}}^2\right)\right)\\ & +\frac{1}{16}{\displaystyle \sum _{a=1}^{p-1}}\left(\lambda \left(a+\overline{a}\right)\overline{\lambda }\left(a-\overline{a}\right)+\overline{\lambda }\left(a+\overline{a}\right)\lambda \left(a-\overline{a}\right)\right)\\ & +\frac{1}{16}{\displaystyle \sum _{a=1}^{p-1}}\left({\chi }_2\left(a+\overline{a}\right)\lambda \left(a-\overline{a}\right)+{\chi }_2\left(a+\overline{a}\right)\overline{\lambda }\left(a-\overline{a}\right)\right)\\ & +\frac{1}{16}{\displaystyle \sum _{a=1}^{p-1}}\left(\lambda \left(a+\overline{a}\right){\chi }_2\left(a-\overline{a}\right)+\overline{\lambda }\left(a+\overline{a}\right){\chi }_2\left(a-\overline{a}\right)\right)\\ & +\frac{1}{16}{\displaystyle \sum _{a=1}^{p-1}}\left({\chi }_2\left(a+\overline{a}\right)+{\chi }_2\left(a-\overline{a}\right)\right)-\frac{4}{16}.\end{array}$$

(6)

If $p\equiv 5mod8$, then note that

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

from (6), Lemma 2–Lemma 5, we have

$$\begin{array}{lll}N(p)& =& \frac{p-1}{16}-\frac{1}{16}\left(2+2{\displaystyle \sum _{a=1}^{p-1}}\left(\frac{a+\overline{a}}{p}\right)\right)-\frac{4}{16}\\ & =& \frac{1}{16}·\left(p-7-2{\displaystyle \sum _{a=1}^{\frac{p-1}{2}}}\left(\frac{a+\overline{a}}{p}\right)\right).\end{array}$$

This proves Theorem 1.

If p is a prime with $p\equiv 1mod8$, then note that ${\chi }_2(\pm 2)=1$, $\lambda (\pm 1)=1$, $\lambda (r)=\overline{\lambda }(r)$ and $\lambda (2)=\overline{\lambda }(2)$, from (6), we have

$$\begin{array}{ll}& N(p)=\frac{1}{16}{\displaystyle \underset{(a^2+1,p)=1}{{\displaystyle \sum _{a=2}^{p-2}}}}\left(1+\lambda \left(a+\overline{a}\right)+{\chi }_2\left(a+\overline{a}\right)+\overline{\lambda }\left(a+\overline{a}\right)\right)\\ & \times \left(1+\lambda \left(a-\overline{a}\right)+{\chi }_2\left(a-\overline{a}\right)+\overline{\lambda }\left(a-\overline{a}\right)\right)\\ =& \frac{1}{16}{\displaystyle \sum _{a=1}^{p-1}}\left(1+\lambda \left(a+\overline{a}\right)+{\chi }_2\left(a+\overline{a}\right)+\overline{\lambda }\left(a+\overline{a}\right)\right)\\ & \times \left(1+\lambda \left(a-\overline{a}\right)+{\chi }_2\left(a-\overline{a}\right)+\overline{\lambda }\left(a-\overline{a}\right)\right)-\frac{1+\lambda (2)}{4}-\frac{1+2\lambda (2r)+{\chi }_2(r)}{8}.\end{array}$$

(7)

Since, in this case, we can not get those formulas like Lemma 3–Lemma 5. In this time, we only use the trivial estimates (see Weil’s work [22] or Bourgain et al. [23]):

$$\begin{array}{c}\hfill \left|\sum _{a=1}^{p-1}\lambda \left(a^2-{\overline{a}}^2\right)\right|=\left|\sum _{a=1}^{p-1}{\chi }_2(a)\lambda \left(a^4-1\right)\right|\le 5·\sqrt{p},\end{array}$$

(8)

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

(9)

$$\begin{array}{c}\hfill \left|\sum _{a=1}^{p-1}{\chi }_2\left(a+\overline{a}\right)\lambda \left(a-\overline{a}\right)\right|=\left|\sum _{a=1}^{p-1}\lambda (a){\chi }_2\left(a^2+1\right)\lambda \left(a^2-1\right)\right|\le 5·\sqrt{p},\end{array}$$

(10)

$$\begin{array}{c}\hfill \left|\sum _{a=1}^{p-1}{\chi }_2\left(a-\overline{a}\right)\lambda \left(a+\overline{a}\right)\right|=\left|\sum _{a=1}^{p-1}\lambda (a){\chi }_2\left(a^2-1\right)\lambda \left(a^2+1\right)\right|\le 5·\sqrt{p},\end{array}$$

(11)

$$\begin{array}{c}\hfill \left|\sum _{a=1}^{p-1}\lambda \left(a\pm \overline{a}\right)\right|=\left|\sum _{a=1}^{p-1}\overline{\lambda }(a)\lambda \left(a^2\pm 1\right)\right|\le 3·\sqrt{p},\end{array}$$

(12)

$$\begin{array}{c}\hfill \left|\sum _{a=1}^{p-1}{\chi }_2\left(a^2-{\overline{a}}^2\right)\right|\le 4·\sqrt{p},\end{array}$$

(13)

$$\begin{array}{c}\hfill \left|\sum _{a=1}^{p-1}{\chi }_2\left(a\pm \overline{a}\right)\right|\le 3·\sqrt{p},\end{array}$$

(14)

$$\begin{array}{c}\hfill 0\le 1+\lambda \left(2\right)\le 2,\end{array}$$

(15)

$$\begin{array}{c}\hfill 0\le 1+2\lambda \left(2r\right)+{\chi }_2(r)\le 4.\end{array}$$

(16)

Combining (7)–(16) we have asymptotic formula

$$\begin{array}{c}\hfill N(p)=\frac{1}{16}·(p-17)+E(p),\end{array}$$

where $E(p)$ satisfies the estimate $\left|E(p)\right|\le \frac{15}{4}·\sqrt{p}$.

Now, we prove Corollary 1. Since $N(p)\ge 0$ is an integer, from Theorem 1, we have

$$p-7-2\sum _{a=1}^{\frac{p-1}{2}}\left(\frac{a+\overline{a}}{p}\right)\equiv 0mod16$$

or

$$p\equiv 7+2\sum _{a=1}^{\frac{p-1}{2}}\left(\frac{a+\overline{a}}{p}\right)mod16.$$

In order to prove Corollary 2, we just have to make sure that $N(p)>0$. From Theorem 1 and Theorem 2, we just have the inequalities

$$\begin{array}{c}\hfill N(p)=\frac{1}{16}\left(p-7-2\sum _{a=1}^{\frac{p-1}{2}}\left(\frac{a+\overline{a}}{2}\right)\right)>0\end{array}$$

(17)

and

$$\begin{array}{c}\hfill N(p)=\frac{1}{16}\left(p-17\right)+E(p)>\frac{1}{16}\left(p-17\right)-\frac{15}{4}·\sqrt{p}>0.\end{array}$$

(18)

Note that, from the estimate

$$\left|\sum _{a=1}^{\frac{p-1}{2}}\left(\frac{a+\overline{a}}{2}\right)\right|\le \sqrt{p},$$

we can get $p>3700$ for solving inequalities (17) and (18).

This completes the proofs of our all results.

4. Conclusions

The main results of this paper are two theorems and two corollaries. For prime p with $p\equiv 5mod8$, Theorem 1 of this paper gives an exact calculation formula for $N(p)$, which contains an interesting constant

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

This $\alpha (p)$ is closely related to prime p. In fact, we have (see [20]: Theorem 4–11)

$$\begin{array}{c}\hfill p={\left(\frac{1}{2}\sum _{a=1}^{p-1}\left(\frac{a+\overline{a}}{p}\right)\right)}^2+{\left(\frac{1}{2}\sum _{a=1}^{p-1}\left(\frac{a+s\overline{a}}{p}\right)\right)}^2={\alpha }^2(p)+{\beta }^2(p),\end{array}$$

where s denotes any quadratic non-residue modulo p.

We know very little about the arithmetic properties of $\alpha (p)$, or even its parity. Obviously, for any prime p with $p\equiv 5mod8$, Corollary 1 gives for the first time a nontrivial congruence property of $\alpha (p)$.

For any prime p with $p\equiv 1mod8$, we naturally ask the following two problems:

(1).

Whether there exists an exact computing formula for $N(p)$?

(2).

What is the residue of $\alpha (p)mod8$?

Interested readers are advised to study them with us.