# An hybrid mean value of quadratic Gauss sums and a sum analogous to Kloosterman sums

## Abstract

The main purpose of this paper is, using the analytic methods and the properties of character sums, to study the computational problem of one kind of hybrid mean value involving the quadratic Gauss sums and a new sum analogous to Kloosterman sums, and to give an interesting hybrid mean value formula for it.

MSC:11L03, 11L40.

## 1 Introduction

Let $q\ge 3$ be an integer, and let χ be a Dirichlet character $mod\phantom{\rule{0.25em}{0ex}}q$. Then for any integer n, the famous quadratic Gauss sums $G\left(\chi ,n;q\right)$ is defined as follows:

$G\left(\chi ,n;q\right)=\sum _{a=1}^{q}\chi \left(a\right)\cdot e\left(\frac{n{a}^{2}}{q}\right),$

where $e\left(y\right)={e}^{2\pi iy}$.

This sum plays a very important role in the study of analytic number theory, many famous number theoretic problems are closely related to it. For example, the distribution of primes, the Goldbach problem, the properties of Dirichlet L-functions are some good examples. About the arithmetic properties of $G\left(\chi ,n;q\right)$, some authors had studied it and obtained many interesting results. For example, if $q=p$ is a prime and $\left(p,n\right)=1$, then one can get the estimate $|G\left(\chi ,n;p\right)|\le 2\sqrt{p}$. Some other results can be found in references .

On the other hand, the classical Kloosterman sums $K\left(m,n;q\right)$ is defined as

$K\left(m,n;q\right)=\underset{a=1}{\overset{q-1}{{\sum }^{\prime }}}e\left(\frac{ma+n\overline{a}}{q}\right),$

where ${{\sum }^{\prime }}_{a=1}^{q-1}$ denotes the summation over all $1\le a\le q$ such that $\left(a,q\right)=1$, and $\overline{a}$ denotes the solution of the congruence equation $ax\equiv 1modq$.

Now we define another sum analogous to Kloosterman sums as follows:

$S\left(\chi ,q\right)=\underset{a=1}{\overset{q-1}{{\sum }^{\prime }}}\chi \left(a+\overline{a}\right).$

In fact, this sum is a special case of the general character of polynomials, some related results can be found in [7, 8] and [9, 10].

The main purpose of this paper is using the analytic method and the properties of the character sums to study the hybrid mean value properties of $G\left(\chi ,n;p\right)$ and $S\left(\chi ,p\right)$, and to give an interesting mean value formula. That is, we shall prove the following two conclusions.

Theorem 1 Let p be an odd prime, χ be any non-principal even character (i.e. $\chi \left(-1\right)=1$) $mod\phantom{\rule{0.25em}{0ex}}p$. Then for any integer n with $\left(n,p\right)=1$, we have the identity

$|\sum _{a=1}^{p-1}\chi \left(a\right)\cdot e\left(\frac{n{a}^{2}}{p}\right){|}^{2}=2p+\overline{\chi }\left(2\right)\cdot \left(\frac{n}{p}\right)\cdot \tau \left({\chi }_{2}\right)\cdot \sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right),$

where $\left(\frac{\ast }{p}\right)={\chi }_{2}$ denotes the Legendre symbol, and $\tau \left({\chi }_{2}\right)={\sum }_{a=1}^{p-1}{\chi }_{2}\left(a\right)\cdot e\left(\frac{a}{p}\right)$ denotes the classical Gauss sums with ${\tau }^{2}\left({\chi }_{2}\right)=\left(\frac{-1}{p}\right)\cdot p$.

Theorem 2 Let p be an odd prime with $p\equiv 3mod4$. Then for any integer n with $\left(n,p\right)=1$, we have the identity

$\underset{\chi modp}{{\sum }^{\prime }}|\sum _{a=1}^{p-1}\chi \left(a\right)\cdot e\left(\frac{n{a}^{2}}{p}\right){|}^{2}\cdot |\sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right){|}^{2}=\left(p-1\right)\cdot \left(3{p}^{2}-6p-1\right),$

where ${{\sum }^{\prime }}_{\chi modp}$ denotes the summation over all even character $mod\phantom{\rule{0.25em}{0ex}}p$, i.e. $\chi \left(-1\right)=1$.

Some notes: Theorem 1 tells us that there exists a close relationship between $G\left(\chi ,n;p\right)$ and $S\left(\chi ,p\right)$. That is, ${|G\left(\chi ,n;p\right)|}^{2}$ can be represented by $S\left(\chi ,p\right)$.

Since for any odd character $\chi modp$, we have $G\left(\chi ,n;p\right)=S\left(\chi ,p\right)=0$, we only discussed the summation for all even characters $\chi modp$ in Theorem 2.

If $p\equiv 1mod4$, then we cannot give a computational formula for the hybrid mean value in Theorem 2. In this case, the difficulty is that we cannot obtain an exact value for the behind formula (13). We hope that the interested reader will stay with us as we turn to further study.

For general integer $q\ge 3$, whether there exists a computational formula for the hybrid mean value

$\underset{\chi modq}{{\sum }^{\prime }}|\sum _{a=1}^{q}\chi \left(a\right)\cdot e\left(\frac{n{a}^{2}}{q}\right){|}^{2}\cdot |\underset{a=1}{\overset{q-1}{{\sum }^{\prime }}}\chi \left(a+\overline{a}\right){|}^{2}$

is an interesting open problem, where n is any integer with $\left(n,q\right)=1$.

## 2 Several lemmas

In this section, we shall give two simple lemmas, which are necessary in the proofs of our theorems. Hereinafter, we shall use many properties of character sums and Gauss sums, all of these can be found in references [1, 2] and . First we have the following.

Lemma 1 Let p be an odd prime, χ be any non-principal even character $mod\phantom{\rule{0.25em}{0ex}}p$. Then we have the identity

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

where $\left(\frac{\ast }{p}\right)$ denotes the Legendre symbol.

Proof Let $a+\overline{a}=u$, then we have

$\begin{array}{rcl}\sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right)& =& \sum _{u=1}^{p-1}\chi \left(u\right)\underset{a+\overline{a}\equiv umodp}{\sum _{a=1}^{p-1}}1=\sum _{u=1}^{p-1}\chi \left(u\right)\underset{{a}^{2}-au+1\equiv 0modp}{\sum _{a=1}^{p-1}}1\\ =& \sum _{u=1}^{p-1}\chi \left(u\right)\underset{{\left(2a-u\right)}^{2}\equiv {u}^{2}-4modp}{\sum _{a=0}^{p-1}}1=\sum _{u=1}^{p-1}\chi \left(u\right)\underset{{a}^{2}\equiv {u}^{2}-4modp}{\sum _{a=0}^{p-1}}1.\end{array}$
(1)

Note that for any fixed integer ${u}^{2}-4$, the number of the solutions of the congruence equation ${x}^{2}\equiv {u}^{2}-4modp$ are $1+\left(\frac{{u}^{2}-4}{p}\right)$, so from (1) we have

$\begin{array}{rcl}\sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right)& =& \sum _{u=1}^{p-1}\chi \left(u\right)\left(1+\left(\frac{{u}^{2}-4}{p}\right)\right)\\ =& \sum _{u=1}^{p-1}\chi \left(u\right)\left(\frac{{u}^{2}-4}{p}\right)=\chi \left(2\right)\sum _{u=1}^{p-1}\chi \left(u\right)\left(\frac{{u}^{2}-1}{p}\right).\end{array}$
(2)

Now from (2) and the properties of reduced residue system $mod\phantom{\rule{0.25em}{0ex}}p$ we have

$\begin{array}{rcl}|\sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right){|}^{2}& =& |\sum _{u=1}^{p-1}\chi \left(u\right)\left(\frac{{u}^{2}-1}{p}\right){|}^{2}=\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\chi \left(a\overline{b}\right)\left(\frac{{a}^{2}-1}{p}\right)\left(\frac{{b}^{2}-1}{p}\right)\\ =& \sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{{a}^{2}{b}^{2}-1}{p}\right)\left(\frac{{b}^{2}-1}{p}\right)\\ =& \sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(1+\left(\frac{b}{p}\right)\right)\left(\frac{{a}^{2}b-1}{p}\right)\left(\frac{b-1}{p}\right)\\ =& \sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}b-1\right)\left(b-1\right)}{p}\right)\\ +\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}b-1\right)b\left(b-1\right)}{p}\right).\end{array}$
(3)

Note that $\chi \left(-1\right)=1$, from the properties of the complete residue system $mod\phantom{\rule{0.25em}{0ex}}p$ we also have

$\begin{array}{rcl}\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}b-1\right)\left(b-1\right)}{p}\right)& =& \sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=0}^{p-1}\left(\frac{{\left(2{a}^{2}b-{a}^{2}-1\right)}^{2}-{\left({a}^{2}-1\right)}^{2}}{p}\right)\\ =& \sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=0}^{p-1}\left(\frac{{b}^{2}-{\left({a}^{2}-1\right)}^{2}}{p}\right)\end{array}$
(4)

and

(5)

(This formula can be found in Hua’s book , Section 7.8, Theorem 8.2.)

Combining (4) and (5) we can deduce the identity

$\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}b-1\right)\left(b-1\right)}{p}\right)=2\left(p-1\right)-\sum _{a=2}^{p-2}\chi \left(a\right)=2p.$
(6)

Now Lemma 1 follows from (3) and (6). □

Lemma 2 Let p be an odd prime, χ be any non-principal even character $mod\phantom{\rule{0.25em}{0ex}}p$. Then for any integer m with $\left(m,p\right)=1$, we have the identity

$|\sum _{a=1}^{p-1}\chi \left(a\right)\cdot e\left(\frac{m{a}^{2}}{p}\right){|}^{2}=2p+\left(\frac{m}{p}\right)\cdot \tau \left({\chi }_{2}\right)\cdot \sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right),$

where ${\chi }_{2}=\left(\frac{\ast }{p}\right)$ denotes the Legendre symbol with ${\tau }^{2}\left({\chi }_{2}\right)=\left(\frac{-1}{p}\right)\cdot p$.

Proof If $\left(m,p\right)=1$, then from the properties of Gauss sums and quadratic residue $mod\phantom{\rule{0.25em}{0ex}}p$ we have

$\begin{array}{rcl}\sum _{a=0}^{p-1}e\left(\frac{m{a}^{2}}{p}\right)& =& 1+\sum _{a=1}^{p-1}e\left(\frac{m{a}^{2}}{p}\right)=1+\sum _{a=1}^{p-1}\left(1+\left(\frac{a}{p}\right)\right)\cdot e\left(\frac{ma}{p}\right)\\ =& \sum _{a=0}^{p-1}e\left(\frac{ma}{p}\right)+\sum _{a=1}^{p-1}\left(\frac{a}{p}\right)\cdot e\left(\frac{ma}{p}\right)\\ =& \left(\frac{m}{p}\right)\sum _{a=1}^{p-1}\left(\frac{a}{p}\right)\cdot e\left(\frac{a}{p}\right)=\left(\frac{m}{p}\right)\cdot \tau \left({\chi }_{2}\right).\end{array}$
(7)

Since χ is a non-principal even character $mod\phantom{\rule{0.25em}{0ex}}p$, so from identity (7) and the definition of $G\left(\chi ,m;p\right)$ we have

$\begin{array}{rcl}|G\left(\chi ,m;p\right){|}^{2}& =& \sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\chi \left(a\overline{b}\right)\cdot e\left(\frac{m{a}^{2}-m{b}^{2}}{p}\right)\\ =& \sum _{a=1}^{p-1}\chi \left(a\right)\cdot \sum _{b=1}^{p-1}e\left(\frac{m{b}^{2}\left({a}^{2}-1\right)}{p}\right)=\sum _{a=1}^{p-1}\chi \left(a\right)\cdot \left(\sum _{b=0}^{p-1}e\left(\frac{m{b}^{2}\left({a}^{2}-1\right)}{p}\right)-1\right)\\ =& 2\left(p-1\right)+\sum _{a=2}^{p-2}\chi \left(a\right)\cdot \left(\sum _{b=0}^{p-1}e\left(\frac{m{b}^{2}\left({a}^{2}-1\right)}{p}\right)-1\right)\\ =& 2\left(p-1\right)-\sum _{a=2}^{p-2}\chi \left(a\right)+\tau \left({\chi }_{2}\right)\cdot \sum _{a=2}^{p-2}\chi \left(a\right)\left(\frac{m\left({a}^{2}-1\right)}{p}\right)\\ =& 2p-\sum _{a=1}^{p-1}\chi \left(a\right)+\tau \left({\chi }_{2}\right)\cdot \sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{m\left({a}^{2}-1\right)}{p}\right)\\ =& 2p+\left(\frac{m}{p}\right)\cdot \tau \left({\chi }_{2}\right)\cdot \sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right).\end{array}$

This completes the proof of Lemma 2. □

## 3 Proof of the theorems

In this section, we shall complete the proof of our theorems. First we prove Theorem 1. In fact from (2) and Lemma 2 we may immediately deduce the identity

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

This proves Theorem 1.

Now we prove Theorem 2; from Lemma 1 and Lemma 2 we have

$\begin{array}{c}\underset{\chi modp}{{\sum }^{\prime }}|\sum _{a=1}^{p-1}\chi \left(a\right)\cdot e\left(\frac{n{a}^{2}}{p}\right){|}^{2}\cdot |\sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right){|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=|\sum _{a=1}^{p-1}{\chi }_{0}\left(a\right)\cdot e\left(\frac{n{a}^{2}}{p}\right){|}^{2}\cdot |\sum _{a=1}^{p-1}{\chi }_{0}\left(a+\overline{a}\right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}+\underset{\chi \ne {\chi }_{0}}{\underset{\chi modp}{{\sum }^{\prime }}}\left(2p+\left(\frac{n}{p}\right)\cdot \tau \left({\chi }_{2}\right)\cdot \sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}×\left(2p+\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\right).\hfill \end{array}$
(8)

If $p\equiv 3mod4$, then we have $\left(\frac{-1}{p}\right)=-1$ and

$|\sum _{a=1}^{p-1}{\chi }_{0}\left(a\right)\cdot e\left(\frac{n{a}^{2}}{p}\right){|}^{2}=|\sum _{a=0}^{p-1}e\left(\frac{n{a}^{2}}{p}\right)-1{|}^{2}=|{\chi }_{2}\left(n\right)\tau \left({\chi }_{2}\right)-1{|}^{2}=p+1,$
(9)
$|\sum _{a=1}^{p-1}{\chi }_{0}\left(a+\overline{a}\right){|}^{2}={\left(\sum _{a=1}^{p-1}1\right)}^{2}={\left(p-1\right)}^{2}.$
(10)

Note that the identity

from (5) we have

$\begin{array}{c}\underset{\chi \ne {\chi }_{0}}{\underset{\chi modp}{{\sum }^{\prime }}}\sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right)=\underset{\chi modp}{{\sum }^{\prime }}\sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right)-\sum _{a=1}^{p-1}\left(\frac{{a}^{2}-1}{p}\right)\hfill \\ \phantom{\underset{\chi \ne {\chi }_{0}}{\underset{\chi modp}{{\sum }^{\prime }}}\sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right)}=-\sum _{a=0}^{p-1}\left(\frac{{a}^{2}-1}{p}\right)+\left(\frac{-1}{p}\right)=1+\left(\frac{-1}{p}\right)=0;\hfill \end{array}$
(11)
$\begin{array}{c}\underset{\chi \ne {\chi }_{0}}{\underset{\chi modp}{{\sum }^{\prime }}}\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{\chi modp}{{\sum }^{\prime }}\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)-\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(p-1\right)\cdot \sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left(b-1\right)}{p}\right)-\sum _{b=1}^{p-1}\left(\frac{b-1}{p}\right)\sum _{a=1}^{p-1}\left(\frac{{a}^{2}-\overline{b}}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(p-1\right)\cdot \sum _{b=2}^{p-1}\left(\frac{b}{p}\right)-\sum _{b=1}^{p-1}\left(\frac{b-1}{p}\right)\left(-1-\left(\frac{-\overline{b}}{p}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=1-p-2\left(\frac{-1}{p}\right)=-\left(p-3\right).\hfill \end{array}$
(12)

Note that $\left(\frac{-1}{p}\right)=-1$ and

$\begin{array}{c}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}-1\right)\left({b}^{2}-1\right)\left({a}^{2}{b}^{2}-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}{\overline{b}}^{2}-1\right)\left({b}^{2}-1\right)\left({a}^{2}-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}-{b}^{2}\right)\left({b}^{2}-1\right)\left({a}^{2}-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({b}^{2}-{a}^{2}\right)\left({b}^{2}-1\right)\left({a}^{2}-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({b}^{2}{a}^{2}-{a}^{2}\right)\left({b}^{2}{a}^{2}-1\right)\left({a}^{2}-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}-1\right)\left({b}^{2}-1\right)\left({a}^{2}{b}^{2}-1\right)}{p}\right),\hfill \end{array}$

so that we have the identities

$\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}-1\right)\left({b}^{2}-1\right)\left({a}^{2}{b}^{2}-1\right)}{p}\right)=0$
(13)

and

$\begin{array}{c}\underset{\chi \ne {\chi }_{0}}{\underset{\chi modp}{{\sum }^{\prime }}}\left(\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\right)\left(\sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{\chi modp}{{\sum }^{\prime }}\left(\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\right)\left(\sum _{c=1}^{p-1}\chi \left(c\right)\left(\frac{{c}^{2}-1}{p}\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\left(\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\right)\left(\sum _{a=1}^{p-1}\left(\frac{{a}^{2}-1}{p}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(p-1\right)\cdot \sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\left(\frac{{\overline{a}}^{2}-1}{p}\right)-2-2\left(\frac{-1}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\left(p-1\right)\cdot \sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(1+\left(\frac{b}{p}\right)\right)\left(\frac{\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\left(\frac{{a}^{2}-1}{p}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\left(p-1\right)\cdot \sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\left(\frac{{a}^{2}-1}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\left(p-1\right)\cdot \sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\left(\frac{\left({a}^{2}-1\right)\left({b}^{2}-1\right)\left({a}^{2}{b}^{2}-1\right)}{p}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\left(p-1\right)\cdot \sum _{a=1}^{p-1}\left(\frac{{a}^{2}-1}{p}\right)\sum _{b=0}^{p-1}\left(\frac{{\left(2{a}^{2}b-{a}^{2}-1\right)}^{2}-{\left({a}^{2}-1\right)}^{2}}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(p-1\right)\cdot \sum _{a=1}^{p-1}\left(\frac{{a}^{2}-1}{p}\right)\sum _{b=0}^{p-1}\left(\frac{{b}^{2}-{\left({a}^{2}-1\right)}^{2}}{p}\right)=0.\hfill \end{array}$
(14)

Combining (8)-(12) and (14) we may immediately deduce

$\begin{array}{c}\underset{\chi modp}{{\sum }^{\prime }}|\sum _{a=1}^{p-1}\chi \left(a\right)\cdot e\left(\frac{n{a}^{2}}{p}\right){|}^{2}\cdot |\sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right){|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=|\sum _{a=1}^{p-1}{\chi }_{0}\left(a\right)\cdot e\left(\frac{n{a}^{2}}{p}\right){|}^{2}\cdot |\sum _{a=1}^{p-1}{\chi }_{0}\left(a+\overline{a}\right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}+\underset{\chi \ne {\chi }_{0}}{\underset{\chi modp}{{\sum }^{\prime }}}\left(2p+\left(\frac{n}{p}\right)\cdot \tau \left({\chi }_{2}\right)\cdot \sum _{a=1}^{p-1}\chi \left(a\right)\left(\frac{{a}^{2}-1}{p}\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}×\left(2p+\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}\left(\frac{b\left(b-1\right)\left({a}^{2}b-1\right)}{p}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(p+1\right){\left(p-1\right)}^{2}+4{p}^{2}\left(\frac{p-1}{2}-1\right)-2p\left(p-3\right)\hfill \\ \phantom{\rule{1em}{0ex}}=3{p}^{3}-9{p}^{2}+5p+1=\left(p-1\right)\cdot \left(3{p}^{2}-6p-1\right).\hfill \end{array}$

This completes the proof of our theorems.

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## Acknowledgements

The authors would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper. This work is supported by the P.E.D. (2013JK0561) and N.S.F. (11371291) of P.R. China.

## Author information

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### Corresponding author

Correspondence to Han Zhang.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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Pan, X., Zhang, H. An hybrid mean value of quadratic Gauss sums and a sum analogous to Kloosterman sums. J Inequal Appl 2014, 129 (2014). https://doi.org/10.1186/1029-242X-2014-129

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• DOI: https://doi.org/10.1186/1029-242X-2014-129

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