A sum analogous to the high-dimensional Kloosterman sums and its upper bound estimate

Abstract

The main purpose of this paper is, using the properties of Gauss sums and the estimate for the generalized exponential sums, to study the upper bound estimate problem of one kind sums analogous to the high-dimensional Kloosterman sums and to give some interesting mean value formula and an upper bound estimate for it.

MSC:11L05.

1 Introduction

For any integer $q\ge 3$, the high-dimensional Kloosterman sums $K(c_1,c_2,\dots ,c_k,m;q)$ are defined as follows:

$$K(c_1,c_2,\dots ,c_k,m;q)=\underset{a_1=1}{\overset{q}{{\sum }^{\prime }}}\cdots \underset{a_k=1}{\overset{q}{{\sum }^{\prime }}}e\left(\frac{c_1{a}_1+\cdots +c_k{a}_k+m{\overline{a}}_1\cdots {\overline{a}}_k}{q}\right),$$

where $e(x)=e^{2\pi ix}$, ${{\sum }^{\prime }}_{a_i=1}^q$ denotes the summation over all integers $1\le a_i\le q$ such that $(a_i,q)=1$, $c_i$ and m are integers with $(m,q)=1$, ${\overline{a}}_i$ denotes the solution of the congruent equation $x\cdot a_i\equiv 1modq$ ($i=1,2,\dots ,k$).

There are several results on the properties of the Kloosterman sums $K(c_1,c_2,\dots ,c_k,m;q)$. For example, see [1, 2] and [3]. Related works can also be found in [4–8] and [9].

In this paper, we consider a sum analogous to the high-dimensional Kloosterman sums as follows:

$$S(c_1,c_2,\dots ,c_k,m,\chi ;q)=\underset{a_1=1}{\overset{q}{{\sum }^{\prime }}}\cdots \underset{a_k=1}{\overset{q}{{\sum }^{\prime }}}\chi (c_1{a}_1+\cdots +c_k{a}_k+m{\overline{a}}_1\cdots {\overline{a}}_k),$$

(1.1)

where χ is a Dirichlet character modq.

If $k=1$ and $q=p$ (an odd prime), then for any integer a with $(a,p)=1$, applying the Fermat little theorem, one can deduce $a^{p-2}\equiv \overline{a}modp$. So, the sum (1.1) becomes

$$\sum _{a=1}^{p-1}\chi (ca+m\overline{a}).$$

It is a special case of the general polynomial character sums

$$\sum _{a=N+1}^{N+M}\chi (f(a)),$$

where M and N are any positive integers and $f(x)$ is a polynomial. Let χ be a q th-order character modp. If $f(x)$ is not a perfect q th power modp, then from Weil’s classical work (see [10]), we can deduce the estimate

$$\sum _{x=N+1}^{N+M}\chi (f(x))\ll p^{\frac{1}{2}}lnp,$$

where ‘≪’ constant depends only on the degree of $f(x)$. Some related results can also be found in [11–13] and [14].

Now we are concerned with the upper bound estimate problem of (1.1). Regarding this contents, it seems that nobody has yet studied it, at least we have not seen any related result before. The problem is interesting because it can reflect some new properties of character sums. The main purpose of this paper is, using the analytic methods and the properties of Gauss sums, to study this problem and give a sharp upper bound estimate for (1.1). That is, we prove the following conclusions.

Theorem 1 Let p be an odd prime, let k be a positive integer with $k\ge 2$, and let χ be any non-principal character modp. Then for any integers $c_1,c_2,\dots ,c_k$ and m with $(c_1{c}_2\cdots c_{k}m,p)=1$, we have the identity

$$|S(c_1,c_2,\dots ,c_k,m,\chi ;p)|=\{\begin{array}{cc}{p}^{\frac{k}{2}}\hfill & \mathit{\text{if}}(k+1,p-1)=1,\hfill \\ 0\hfill & \mathit{\text{if}}(k+1)|(p-1)\mathit{\text{and}}{\chi }^{\frac{p-1}{k+1}}\ne {\chi }_0,\hfill \end{array}$$

where ${\chi }_0$ denotes the principal character modp.

Theorem 2 Let p be an odd prime, let k be a positive integer with $k\ge 2$, and let χ be any non-principal character modp. Then for any integers $c_1,c_2,\dots ,c_k$ and m with $(c_1{c}_2\cdots c_{k}m,p)=1$, we have the estimate

$$|S(c_1,c_2,\dots ,c_k,m,\chi ;p)|\le (k+1)\cdot p^{\frac{k}{2}}.$$

Theorem 3 Let p and q be two odd primes, let r be any qth non-residue modp. Then for any integers $c_1,c_2,\dots ,c_{q-1}$ with $(c_1{c}_2\cdots c_{q-1},p)=1$, we have the identity

$$\begin{array}{c}\sum _{i=0}^{q-1}{[\sum _{a_1=1}^{p-1}\sum _{a_2=1}^{p-1}\cdots \sum _{a_{q-1}=1}^{p-1}\left(\frac{c_1{a}_1+c_2{a}_2+\cdots +c_{q-1}{a}_{q-1}+r^i{\overline{a}}_1{\overline{a}}_2\cdots {\overline{a}}_{q-1}}{p}\right)]}^2\hfill \\ =\{\begin{array}{cc}{q}^2\cdot p^{q-1}\hfill & \mathit{\text{if}}q|(p-1),\hfill \\ q\cdot p^{q-1}\hfill & \mathit{\text{if}}(q,p-1)=1.\hfill \end{array}\hfill \end{array}$$

If $p\equiv 1mod4$, then the above formula also holds for $q=2$, where $(\frac{\ast }{p})$ denotes the Legendre symbol.

Taking $p=4m+1$ and $q=2$ in Theorem 3, note that $2|(p-1)$, we may immediately deduce the following.

Corollary Let p be an odd prime with $p\equiv 1mod4$, then we have the identity

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

where r and s are any two integers such that $(\frac{r}{p})\cdot (\frac{s}{p})=-1$.

This gives another proof for a classical work in elementary number theory (i.e., see [15] Theorems 4-11): For any prime p with $p\equiv 1mod4$, there exist two positive integers x and y such that $p=x^2+y^2$.

2 Several lemmas

To complete the proof of our theorems, we need the following basic lemmas.

Lemma 1 Let p be an odd prime, let χ be any non-principal character modp, and let k be any positive integer such that $(k,p-1)=1$ or $k|p-1$. Then for any integer m with $(m,p)=1$, we have the identity

$$\sum _{a=1}^{p-1}\chi (a)e\left(\frac{m{a}^k}{p}\right)=\{\begin{array}{cc}{\overline{\chi }}^r(m)\cdot \tau ({\chi }^r)\hfill & \mathit{\text{if}}(k,p-1)=1,\hfill \\ 0\hfill & \mathit{\text{if}}k|(p-1)\mathit{\text{and}}{\chi }^{\frac{p-1}{k}}\ne {\chi }_0,\hfill \\ {\overline{\chi }}_1(m)\cdot {\sum }_{i=0}^{k-1}{\overline{\chi }}_k^i(m)\tau ({\chi }_1{\chi }_k^i)\hfill & \mathit{\text{if}}k|(p-1)\mathit{\text{and}}{\chi }^{\frac{p-1}{k}}={\chi }_0,\hfill \end{array}$$

where $r\cdot k\equiv 1mod(p-1)$, ${\chi }_0$ denotes the principal character modp, ${\chi }_k$ denotes any k-order character modp and ${\chi }_1^k=\chi$.

Proof If $(k,p-1)=1$, then there exists one integer r with $(r,p-1)=1$ such that $r\cdot k\equiv 1mod(p-1)$. This time, for any integer a with $(a,p)=1$, we have $a^{rk}\equiv amodp$. If a passes through a reduced residue system modp, then $a^r$ also passes through a reduced residue system modp. Therefore, we have

$$\begin{array}{rcl}\sum _{a=1}^{p-1}\chi (a)e\left(\frac{m{a}^k}{p}\right)& =& \sum _{a=1}^{p-1}\chi \left(a^r\right)e\left(\frac{m{a}^{rk}}{p}\right)\\ =& \sum _{a=1}^{p-1}{\chi }^r(a)e\left(\frac{ma}{p}\right)={\overline{\chi }}^r(m)\cdot \tau \left({\chi }^r\right).\end{array}$$

(2.1)

If $k>1$ and $k|(p-1)$ with ${\chi }^{\frac{p-1}{k}}\ne {\chi }_0$, then there must exist an integer n with $(n,p)=1$ such that ${\chi }^{\frac{p-1}{k}}(n)\ne 1$. For this n, we have

$$\begin{array}{rcl}\sum _{a=1}^{p-1}\chi (a)e\left(\frac{m{a}^k}{p}\right)& =& \sum _{a=1}^{p-1}\chi (a\cdot n^{\frac{p-1}{k}})e\left(\frac{m{(a\cdot n^{\frac{p-1}{k}})}^k}{p}\right)\\ =& \chi \left(n^{\frac{p-1}{k}}\right)\sum _{a=1}^{p-1}\chi (a)e\left(\frac{m{a}^k{n}^{p-1}}{p}\right)={\chi }^{\frac{p-1}{k}}(n)\sum _{a=1}^{p-1}\chi (a)e\left(\frac{m{a}^k}{p}\right)\end{array}$$

or

$$(1-{\chi }^{\frac{p-1}{k}}(n))\sum _{a=1}^{p-1}\chi (a)e\left(\frac{m{a}^k}{p}\right)=0.$$

Since ${\chi }^{\frac{p-1}{k}}(n)\ne 1$, from the above identity, we have

$$\sum _{a=1}^{p-1}\chi (a)e\left(\frac{m{a}^k}{p}\right)=0.$$

(2.2)

If ${\chi }^{\frac{p-1}{k}}={\chi }_0$, then χ must be a k th character modp, so there exists one character ${\chi }_{1}modp$ such that $\chi ={\chi }_1^k$. Let ${\chi }_k$ be a k-order character modp (i.e., ${\chi }_k^k={\chi }_0$), then for any integer a with $(a,p)=1$, note that

$$1+{\chi }_k(a)+{\chi }_k^2(a)+\cdots +{\chi }_k^{k-1}(a)=\{\begin{array}{cc}k\hfill & \text{if }a\text{ is a }k\text{th residue }modp,\hfill \\ 0,\hfill & \text{otherwise}.\hfill \end{array}$$

From the properties of Gauss sums, we have

$$\begin{array}{rcl}\sum _{a=1}^{p-1}\chi (a)e\left(\frac{m{a}^k}{p}\right)& =& \sum _{a=1}^{p-1}{\chi }_1^k(a)e\left(\frac{m{a}^k}{p}\right)=\sum _{a=1}^{p-1}{\chi }_1\left(a^k\right)e\left(\frac{m{a}^k}{p}\right)\\ =& \sum _{a=1}^{p-1}{\chi }_1(a)(1+{\chi }_k(a)+{\chi }_k^2(a)+\cdots +{\chi }_k^{k-1}(a))e\left(\frac{ma}{p}\right)\\ =& {\overline{\chi }}_1(m)\cdot \sum _{i=0}^{k-1}{\overline{\chi }}_k^i(m)\tau \left({\chi }_1{\chi }_k^i\right).\end{array}$$

(2.3)

Now Lemma 1 follows from (2.1), (2.2) and (2.3). □

Lemma 2 Let p and q be two odd primes with $q|(p-1)$, and let ${\chi }_q$ be any q-order character modp. Then for any integers $c_1,c_2,\dots ,c_{q-1}$ and m with $(m{c}_1{c}_2\cdots c_{q-1},p)=1$, we have the identities

$$\begin{array}{rl}(\mathrm{I})& \sum _{a_1=1}^{p-1}\sum _{a_2=1}^{p-1}\cdots \sum _{a_{q-1}=1}^{p-1}\left(\frac{c_1{a}_1+c_2{a}_2+\cdots +c_{q-1}{a}_{q-1}+m{\overline{a}}_1{\overline{a}}_2\cdots {\overline{a}}_{q-1}}{p}\right)\\ =\left(\frac{n}{p}\right)\cdot {\tau }^{q-1}({\chi }_2)(\sum _{i=0}^{q-1}{\chi }_q^i(n)\frac{{\tau }^q({\chi }_2{\chi }_q^i)}{{\tau }^q({\chi }_2)});\\ (\mathrm{II})& \sum _{a=1}^{p-1}\left(\frac{a+m\overline{a}}{p}\right)=\{\begin{array}{cc}\frac{{\overline{\chi }}_1(m)}{\tau ({\chi }_2)}({\tau }^2({\chi }_1)+(\frac{m}{p}){\tau }^2({\overline{\chi }}_1))\hfill & \mathit{\text{if}}p\equiv 1mod4,\hfill \\ 0\hfill & \mathit{\text{if}}p\equiv 3mod4,\hfill \end{array}\end{array}$$

where $(\frac{\ast }{p})={\chi }_2$ denotes the Legendre symbol, $n=m{c}_1{c}_2\cdots c_{q-1}$ and ${\chi }_1^2={\chi }_2$.

Proof If q is an odd prime, then ${\chi }_2^q={\chi }_2$ and ${\overline{\chi }}_2={\chi }_2$, so applying (2.3) and the properties of Gauss sums, we have

$$\begin{array}{c}\sum _{a_1=1}^{p-1}\sum _{a_2=1}^{p-1}\cdots \sum _{a_{q-1}=1}^{p-1}\left(\frac{c_1{a}_1+c_2{a}_2+\cdots +c_{q-1}{a}_{q-1}+m{\overline{a}}_1{\overline{a}}_2\cdots {\overline{a}}_{q-1}}{p}\right)\hfill \\ =\sum _{a_1=1}^{p-1}\sum _{a_2=1}^{p-1}\cdots \sum _{a_{q-1}=1}^{p-1}\left(\frac{a_1+a_2+\cdots +a_{q-1}+n{\overline{a}}_1{\overline{a}}_2\cdots {\overline{a}}_{q-1}}{p}\right)\hfill \\ =\frac{1}{\tau ({\chi }_2)}\sum _{a_1=1}^{p-1}\sum _{a_2=1}^{p-1}\cdots \sum _{a_{q-1}=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_2(b)e\left(\frac{b(a_1+\cdots +a_{q-1})+bn{\overline{a}}_1\cdots {\overline{a}}_{q-1}}{p}\right)\hfill \\ =\frac{1}{\tau ({\chi }_2)}\sum _{a_1=1}^{p-1}\sum _{a_2=1}^{p-1}\cdots \sum _{a_{q-2}=1}^{p-1}\sum _{b=1}^{p-1}\chi (b)e\left(\frac{b(1+a_1+\cdots +a_{q-2})}{p}\right)\hfill \\ \times \sum _{a_{q-1}=1}^{p-1}{\chi }_2(a_{q-1})e\left(\frac{bn{\overline{a}}_{q-1}^q{\overline{a}}_1\cdots {\overline{a}}_{q-2}}{p}\right)\hfill \\ =\frac{1}{\tau ({\chi }_2)}\sum _{a_1=1}^{p-1}\sum _{a_2=1}^{p-1}\cdots \sum _{a_{q-2}=1}^{p-1}\sum _{b=1}^{p-1}\chi (b)e\left(\frac{b(1+a_1+\cdots +a_{q-2})}{p}\right)\hfill \\ \times \sum _{c=1}^{p-1}{\chi }_2(c)(1+{\chi }_q(c)+\cdots +{\chi }_q^{q-1}(c))e\left(\frac{bnc{\overline{a}}_1\cdots {\overline{a}}_{q-2}}{p}\right)\hfill \\ =\left(\frac{n}{p}\right)\cdot {\tau }^{q-1}({\chi }_2)(\sum _{i=0}^{q-1}{\chi }_q^i(n)\frac{{\tau }^q({\chi }_2{\chi }_q^i)}{{\tau }^q({\chi }_2)}).\hfill \end{array}$$

This proves formula (I).

To prove formula (II), note that if $p\equiv 3mod4$, then ${\chi }_2$ must be an odd character modp (i.e., ${\chi }_2(-1)=-1$) so that

$$\begin{array}{rcl}\sum _{a=1}^{p-1}\left(\frac{a+m\overline{a}}{p}\right)& =& \sum _{a=1}^{p-1}\left(\frac{p-a+m\overline{p-a}}{p}\right)\\ =& \left(\frac{-1}{p}\right)\sum _{a=1}^{p-1}\left(\frac{a+m\overline{a}}{p}\right)=-\sum _{a=1}^{p-1}\left(\frac{a+m\overline{a}}{p}\right)\end{array}$$

or

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

If $p\equiv 1mod4$, then there exists one character ${\chi }_{1}modp$ such that ${\chi }_1^2={\chi }_2$. Note that ${\chi }_1^3={\overline{\chi }}_1$ and ${\chi }_2{\overline{\chi }}_1={\chi }_1$; from the properties of Gauss sums, we have

$$\begin{array}{rcl}\sum _{a=1}^{p-1}\left(\frac{a+m\overline{a}}{p}\right)& =& \frac{1}{\tau ({\chi }_2)}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}{\chi }_2(b)e\left(\frac{b(a+m\overline{a})}{p}\right)\\ =& \frac{1}{\tau ({\chi }_2)}\sum _{b=1}^{p-1}{\chi }_2(b)\sum _{a=1}^{p-1}{\chi }_2(\overline{a})e\left(\frac{b+mb{\overline{a}}^2}{p}\right)\\ =& \frac{1}{\tau ({\chi }_2)}\sum _{b=1}^{p-1}{\chi }_2(b)e\left(\frac{b}{p}\right)\sum _{a=1}^{p-1}{\chi }_1^2(a)e\left(\frac{mb{a}^2}{p}\right)\\ =& \frac{1}{\tau ({\chi }_2)}\sum _{b=1}^{p-1}{\chi }_2(b)e\left(\frac{b}{p}\right)\sum _{a=1}^{p-1}{\chi }_1(a)(1+{\chi }_2(a))e\left(\frac{mba}{p}\right)\\ =& \frac{{\overline{\chi }}_1(m)}{\tau ({\chi }_2)}({\tau }^2({\chi }_1)+\left(\frac{m}{p}\right){\tau }^2({\overline{\chi }}_1)).\end{array}$$

This proves Lemma 2. □

3 Proof of the theorems

In this section, we complete the proof of our theorems. First we prove Theorems 1 and 2. Let $n=m{c}_1{c}_2\cdots c_k$, $(k+1,p-1)=d$. If $d=1$, we can assume $r(k+1)\equiv 1mod(p-1)$, then from Lemma 1, the properties of a reduced residue system modp and Gauss sums, we have

$$\begin{array}{c}S(c_1,c_2,\dots ,c_k,m,\chi ;p)\hfill \\ =\sum _{a_1=1}^{p-1}\cdots \sum _{a_k=1}^{p-1}\chi (a_1+\cdots +a_k+m{c}_1{c}_2\cdots c_k\cdot {\overline{a}}_1\cdots {\overline{a}}_k)\hfill \\ =\frac{1}{\tau (\overline{\chi })}\sum _{a_1=1}^{p-1}\cdots \sum _{a_k=1}^{p-1}\sum _{b=1}^{p-1}\overline{\chi }(b)e\left(\frac{b(a_1+\cdots +a_k+n{\overline{a}}_1\cdots {\overline{a}}_k)}{p}\right)\hfill \\ =\frac{1}{\tau (\overline{\chi })}\sum _{a_2=1}^{p-1}\cdots \sum _{a_k=1}^{p-1}\sum _{a_1=1}^{p-1}\sum _{b=1}^{p-1}\overline{\chi }(b{\overline{a}}_1)e\left(\frac{b(1+\cdots +a_k)+nb{\overline{a}}_1^{k+1}{\overline{a}}_2\cdots {\overline{a}}_k}{p}\right)\hfill \\ =\frac{1}{\tau (\overline{\chi })}\sum _{a_2=1}^{p-1}\cdots \sum _{a_k=1}^{p-1}\sum _{b=1}^{p-1}\overline{\chi }(b)e\left(\frac{b+b(a_2+\cdots +a_k)}{p}\right)\hfill \\ \times \sum _{a_1=1}^{p-1}\chi (a_1)e\left(\frac{nb{\overline{a}}_1^{k+1}{\overline{a}}_2\cdots {\overline{a}}_k}{p}\right)\hfill \\ =\frac{1}{\tau (\overline{\chi })}\sum _{a_2=1}^{p-1}\cdots \sum _{a_k=1}^{p-1}\sum _{b=1}^{p-1}\overline{\chi }(b)e\left(\frac{b+b(a_2+\cdots +a_k)}{p}\right)\hfill \\ \times \sum _{a_1=1}^{p-1}{\overline{\chi }}^r(a_1)e\left(\frac{nb{a}_1{\overline{a}}_2\cdots {\overline{a}}_k}{p}\right)\hfill \\ =\frac{\tau ({\overline{\chi }}^r)}{\tau (\overline{\chi })}\sum _{b=1}^{p-1}\overline{\chi }(b)e\left(\frac{b}{p}\right)\sum _{a_2=1}^{p-1}\cdots \sum _{a_k=1}^{p-1}{\chi }^r(nb{\overline{a}}_2\cdots {\overline{a}}_k)e\left(\frac{b(a_2+\cdots +a_k)}{p}\right)\hfill \\ ={\chi }^r(n)\cdot \frac{{\tau }^{k+1}({\overline{\chi }}^r)}{\tau (\overline{\chi })}\hfill \\ ={\chi }^r(m{c}_1{c}_2\cdots c_k)\cdot \frac{{\tau }^{k+1}({\overline{\chi }}^r)}{\tau (\overline{\chi })}.\hfill \end{array}$$

(3.1)

If $d>1$ and ${\chi }^{\frac{p-1}{d}}\ne {\chi }_0$, then from the method of proving (2.2), we have

$$\sum _{a_1=1}^{p-1}\chi (a_1)e\left(\frac{nb{\overline{a}}_1^{k+1}{\overline{a}}_2\cdots {\overline{a}}_k}{p}\right)=0.$$

From this identity and the method of proving (3.1), we may immediately deduce that if ${\chi }^{\frac{p-1}{d}}\ne {\chi }_0$, then

$$S(c_1,c_2,\dots ,c_k,m,\chi ;p)=0.$$

(3.2)

If $d>1$ and ${\chi }^{\frac{p-1}{d}}={\chi }_0$, then χ must be a d th character modp, so there exists a character ${\chi }_{1}modp$ such that $\chi ={\chi }_1^d$. Let ${\chi }_d$ be a d-order character modp, then we have

$$\begin{array}{c}\sum _{a_1=1}^{p-1}\chi (a_1)e\left(\frac{nb{\overline{a}}_1^{k+1}{\overline{a}}_2\cdots {\overline{a}}_k}{p}\right)\hfill \\ =\sum _{a=1}^{p-1}{\chi }_1\left(a^d\right)e\left(\frac{nb{\overline{a}}^{d\cdot \frac{k+1}{d}}{\overline{a}}_2\cdots {\overline{a}}_k}{p}\right)\hfill \\ =\sum _{a=1}^{p-1}{\overline{\chi }}_1(a)(1+{\chi }_d(a)+{\chi }_d^2(a)+\cdots +{\chi }_d^{d-1}(a))e\left(\frac{nb{a}^{\frac{k+1}{d}}{\overline{a}}_2\cdots {\overline{a}}_k}{p}\right)\hfill \\ =\sum _{i=0}^{d-1}\sum _{a=1}^{p-1}{\overline{\chi }}_1(a){\chi }_d^i(a)e\left(\frac{nb{a}^{\frac{k+1}{d}}{\overline{a}}_2\cdots {\overline{a}}_k}{p}\right).\hfill \end{array}$$

(3.3)

Let $(\frac{k+1}{d},p-1)=d_1$, then repeat the process of proving (2.1), (2.2) and (2.3). Combining (3.1), (3.2) and (3.3), we may immediately deduce the estimate

$$|S(c_1,c_2,\dots ,c_k,m,\chi ;p)|\le (k+1)\cdot p^{\frac{k}{2}}.$$

(3.4)

Now note that $|\tau (\chi )|=|\tau ({\chi }^r)|=\sqrt{p}$, Theorems 1 and 2 follow from (3.1), (3.2) and (3.4).

Now we prove Theorem 3. If $q\ge 3$, we separate q into two cases $(q,p-1)=1$ and $(q,p-1)=q$. If $(q,p-1)=q$, then note that for any q th non-residue $rmodp$, we have

$$\sum _{h=0}^{q-1}{\chi }_q^l\left(r^h\right)=\{\begin{array}{cc}q\hfill & \text{if }q|l,\hfill \\ 0\hfill & \text{if }(q,l)=1.\hfill \end{array}$$

From (I) of Lemma 2, we can deduce that

$$\begin{array}{c}\sum _{i=0}^{q-1}{[\sum _{a_1=1}^{p-1}\sum _{a_2=1}^{p-1}\cdots \sum _{a_{q-1}=1}^{p-1}\left(\frac{c_1{a}_1+c_2{a}_2+\cdots +c_{q-1}{a}_{q-1}+r^i{\overline{a}}_1{\overline{a}}_2\cdots {\overline{a}}_{q-1}}{p}\right)]}^2\hfill \\ =p^{q-1}\sum _{i=0}^{q-1}\sum _{j=0}^{q-1}\sum _{h=0}^{q-1}{\chi }_q^{i+j}(c_1{c}_2\cdots c_{q-1}{r}^h)\frac{{\tau }^q({\chi }_2{\chi }_q^i)}{{\tau }^q({\chi }_2)}\cdot \frac{{\tau }^q({\chi }_2{\chi }_q^j)}{{\tau }^q({\chi }_2)}\hfill \\ =q\cdot p^{q-1}\underset{i+j=0,q}{\sum _{i=0}^{q-1}\sum _{j=0}^{q-1}}\frac{{\tau }^q({\chi }_2{\chi }_q^i)}{{\tau }^q({\chi }_2)}\cdot \frac{{\tau }^q({\overline{\chi }}_2{\overline{\chi }}_q^i)}{{\tau }^q({\chi }_2)}\hfill \\ =q^2\cdot p^{q-1}.\hfill \end{array}$$

If $(q,p-1)=1$, then from the method of proving (2.1) and the properties of Gauss sums, we can deduce that

$$\begin{array}{c}\sum _{i=0}^{q-1}{[\sum _{a_1=1}^{p-1}\sum _{a_2=1}^{p-1}\cdots \sum _{a_{q-1}=1}^{p-1}\left(\frac{c_1{a}_1+c_2{a}_2+\cdots +c_{q-1}{a}_{q-1}+r^i{\overline{a}}_1{\overline{a}}_2\cdots {\overline{a}}_{q-1}}{p}\right)]}^2\hfill \\ =\sum _{i=0}^{q-1}{p}^{q-1}=q\cdot p^{q-1}.\hfill \end{array}$$

If $q=2$ and $p\equiv 1mod4$, then applying (II) of Lemma 2, we have

$$\begin{array}{rcl}{\left(\sum _{a=1}^{p-1}\left(\frac{a+m\overline{a}}{p}\right)\right)}^2& =& p\cdot |1+\left(\frac{m}{p}\right)\cdot \frac{{\tau }^2({\overline{\chi }}_1)}{{\tau }^2({\chi }_1)}{|}^2\\ =& 2p+\left(\frac{m}{p}\right)\cdot \frac{{\tau }^2({\overline{\chi }}_1)}{{\tau }^2({\chi }_1)}+\left(\frac{m}{p}\right)\cdot \frac{{\tau }^2({\chi }_1)}{{\tau }^2({\overline{\chi }}_1)}.\end{array}$$

(3.5)

Therefore, from (3.5) we can deduce that

$${\left(\sum _{a=1}^{p-1}\left(\frac{a+\overline{a}}{p}\right)\right)}^2+{\left(\sum _{a=1}^{p-1}\left(\frac{a+r\overline{a}}{p}\right)\right)}^2=4p.$$

This proves Theorem 3.

To prove the corollary, note that

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

From (3.5) we may immediately deduce the identity

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

where r and s are any two integers such that $(\frac{r}{p})\cdot (\frac{s}{p})=-1$.

This completes the proof of our corollary.