• Research
• Open Access

# Retracted Article: On the fourth power mean of the two-term exponential sums

Journal of Inequalities and Applications20142014:116

https://doi.org/10.1186/1029-242X-2014-116

• Received: 7 February 2014
• Accepted: 6 March 2014
• Published:

The Retraction Note to this article has been published in Journal of Inequalities and Applications 2014 2014:344

## Abstract

### Abstract

The main purpose of this paper is using the analytic methods and the properties of Gauss sums to study the computational problem of one kind of fourth power mean of two-term exponential sums, and to give an interesting identity and asymptotic formula for it.

MSC: 11L05.

## Keywords

• the two-term exponential sums
• fourth power mean
• Gauss sums
• identity
• asymptotic formula

## 1 Introduction

Let $q\ge 3$ be a positive integer. For any integers m and n, the two-term exponential sum $C\left(m,n,k;q\right)$ is defined as follows:
$C\left(m,n,k;q\right)=\sum _{a=1}^{q}e\left(\frac{m{a}^{k}+na}{q}\right),$

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

About the various properties of $C\left(m,n,k;q\right)$, some authors had studied it, and they obtained a series of results, some related works can be found in references . For example, Gauss’ classical work proved the remarkable formula (see )

where ${i}^{2}=-1$.

Generally, for any odd number q and $\left(2m,q\right)=1$, the exact value of $|C\left(m,n,2;q\right)|$ is (e.g. see Berndt, Evans, Williams and Apostol’s related works). Cochrane and Zheng  show for the general sum that
$|C\left(m,n,k;q\right)|\le {k}^{\omega \left(q\right)}{q}^{\frac{1}{2}},$

where $\omega \left(q\right)$ denotes the number of all distinct prime divisors of q.

In this paper, we study the fourth power mean of the two-term exponential sum $C\left(m,n,k;q\right)$ as follows:
$\sum _{m=1}^{q}|C\left(m,n,k;q\right){|}^{4},$
(1.1)

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

As regards this problem, it seems that none has yet studied it, at least we have not seen any related result before. The problem is interesting, because it can reflect that the mean value of $C\left(m,n,k;q\right)$ is well behaved. The main purpose of this paper is to show this point. That is, we shall prove the following conclusion.

Theorem Let $p>3$ be a prime. Then for any integer n with $\left(n,p\right)=1$, we have the identity

where ${\chi }_{1}$ is any 3-order character $mod\phantom{\rule{0.25em}{0ex}}p$.

Note that for any non-principal character $\chi modp$, we have $|\tau \left(\chi \right)|=\sqrt{p}$, so from our theorem we may immediately deduce the following.

Corollary Let $p>3$ be a prime with $3|p-1$. Then for any integer n with $\left(n,p\right)=1$, we have the asymptotic formula
$\sum _{m=1}^{p}|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{4}=2{p}^{3}-5{p}^{2}+O\left({p}^{\frac{3}{2}}\right).$

It seems that our method can also be used to deal with (1.1) for all prime p and integer $k\ge 4$. But this time, the computing is very complex.

For any integer $h\ge 3$, whether there exists an exact computational formula for
$\sum _{m=1}^{p}|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{2h},$

where p is an odd prime and $\left(n,p\right)=1$, is an open problem.

## 2 Several lemmas

In this section, we will give several lemmas which are necessary in the proof of our theorem. In the proving process of all lemmas, we used many properties of Gauss sums; all these can be found in , we will not be repeat them here. First we have the following.

Lemma 1 Let p be an odd prime, χ be any non-principal character $mod\phantom{\rule{0.25em}{0ex}}p$. Then for any integer n with $\left(n,p\right)=1$, we have the identity
Proof Note that χ is a non-principal character $mod\phantom{\rule{0.25em}{0ex}}p$, so if χ is not a 3-order character $mod\phantom{\rule{0.25em}{0ex}}p$ (that is, ${\chi }^{3}\ne {\chi }_{0}$, the principal character $mod\phantom{\rule{0.25em}{0ex}}p$), then from the properties of Gauss sums we have
$\begin{array}{c}\sum _{m=1}^{p-1}\chi \left(m\right)|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\sum _{m=1}^{p-1}\chi \left(m\right)e\left(\frac{m\left({a}^{3}-{b}^{3}\right)+n\left(a-b\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\sum _{m=1}^{p-1}\chi \left(m\right)e\left(\frac{m{b}^{3}\left({a}^{3}-1\right)+nb\left(a-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\tau \left(\chi \right)\sum _{a=1}^{p-1}\overline{\chi }\left({a}^{3}-1\right)\sum _{b=1}^{p-1}{\overline{\chi }}^{3}\left(b\right)e\left(\frac{nb\left(a-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\tau \left(\chi \right)\tau \left({\overline{\chi }}^{3}\right)\sum _{a=1}^{p-1}\overline{\chi }\left({a}^{3}-1\right){\chi }^{3}\left(n\left(a-1\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}={\chi }^{3}\left(n\right)\tau \left(\chi \right)\tau \left({\overline{\chi }}^{3}\right)\sum _{a=1}^{p-2}\overline{\chi }\left({\left(a+1\right)}^{3}-1\right)\chi \left({a}^{3}\right)\hfill \\ \phantom{\rule{1em}{0ex}}={\chi }^{3}\left(n\right)\tau \left(\chi \right)\tau \left({\overline{\chi }}^{3}\right)\sum _{a=1}^{p-2}\overline{\chi }\left({\left(\overline{a}+1\right)}^{3}-{\overline{a}}^{3}\right)\hfill \\ \phantom{\rule{1em}{0ex}}={\chi }^{3}\left(n\right)\tau \left(\chi \right)\tau \left({\overline{\chi }}^{3}\right)\sum _{a=1}^{p-2}\overline{\chi }\left(3{a}^{2}+3a+1\right),\hfill \end{array}$
(2.1)

where $\tau \left(\chi \right)={\sum }_{a=1}^{p-1}\chi \left(a\right)e\left(\frac{a}{p}\right)$ denotes the classical Gauss sum.

If χ is a 3-order character $mod\phantom{\rule{0.25em}{0ex}}p$, then ${\chi }^{3}={\chi }_{0}$; note that for any integer a with $\left(a,p\right)=1$, we have
From the method of proving (2.1) we have the identity
$\begin{array}{c}\sum _{m=1}^{p-1}\chi \left(m\right)|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=\tau \left(\chi \right)\sum _{a=1}^{p-1}\overline{\chi }\left({a}^{3}-1\right)\sum _{b=1}^{p-1}{\overline{\chi }}^{3}\left(b\right)e\left(\frac{nb\left(a-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\tau \left(\chi \right)\sum _{a=1}^{p-1}\overline{\chi }\left({a}^{3}-1\right)\sum _{b=1}^{p-1}e\left(\frac{nb\left(a-1\right)}{p}\right)=-\tau \left(\chi \right)\sum _{a=1}^{p-1}\overline{\chi }\left({a}^{3}-1\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\tau \left(\chi \right)\sum _{a=1}^{p-1}\left({\chi }^{2}\left(a\right)+\chi \left(a\right)+1\right)\overline{\chi }\left(a-1\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\tau \left(\chi \right)\left(\sum _{a=1}^{p-1}\overline{\chi }\left(a-1\right)+\sum _{a=1}^{p-1}\overline{\chi }\left(\overline{a}\left(1-\overline{a}\right)\right)+\sum _{a=1}^{p-1}\overline{\chi }\left(1-\overline{a}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\tau \left(\chi \right)\left(-2+\sum _{a=1}^{p-1}\overline{\chi }\left(a\left(1-a\right)\right)\right).\hfill \end{array}$
(2.2)

Now note that $|\tau \left(\chi \right)|=\sqrt{p}$, if $\chi \ne {\chi }_{0}$. From (2.1) and (2.2) we may immediately deduce Lemma 1. □

Lemma 2 Let p be an odd prime, χ be any non-real character $mod\phantom{\rule{0.25em}{0ex}}p$. Then we have the identity
$\sum _{a=1}^{p-1}\chi \left(a\left(a-1\right)\right)=\frac{{\tau }^{2}\left(\chi \right)}{\tau \left({\chi }^{2}\right)}.$
Therefore,
$|\sum _{a=1}^{p-1}\chi \left(a\left(a-1\right)\right)|=\sqrt{p}.$
Proof From the definition and properties of the classical Gauss sums we have
$\begin{array}{rcl}{\tau }^{2}\left(\chi \right)& =& \sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\chi \left(a\right)\chi \left(b\right)e\left(\frac{a+b}{p}\right)=\sum _{a=1}^{p-1}\chi \left(a\right)\sum _{b=1}^{p-1}{\chi }^{2}\left(b\right)e\left(\frac{b\left(a+1\right)}{p}\right)\\ =& \tau \left({\chi }^{2}\right)\sum _{a=1}^{p-1}\chi \left(a\right)\overline{\chi }\left({\left(a+1\right)}^{2}\right)=\tau \left({\chi }^{2}\right)\sum _{a=2}^{p}\chi \left(a-1\right)\overline{\chi }\left({a}^{2}\right)\\ =& \tau \left({\chi }^{2}\right)\sum _{a=2}^{p-1}\chi \left(\left(1-\overline{a}\right)\overline{a}\right)=\tau \left({\chi }^{2}\right)\sum _{a=1}^{p-1}\chi \left(a\left(1-a\right)\right)\end{array}$
or
$\sum _{a=1}^{p-1}\chi \left(a\left(a-1\right)\right)=\frac{{\tau }^{2}\left(\chi \right)}{\tau \left({\chi }^{2}\right)}.$

This proves Lemma 2. □

## 3 Proof of the theorem

In this section, we shall complete the proof of our theorem. First from the orthogonality of characters $mod\phantom{\rule{0.25em}{0ex}}p$ we have
$\sum _{\chi modp}|\sum _{m=1}^{p-1}\chi \left(m\right)|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{2}{|}^{2}=\left(p-1\right)\sum _{m=1}^{p-1}|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{4}.$
(3.1)
On the other hand, if $3\nmid p-1$, then any non-principal character χ is not a 3-order character $mod\phantom{\rule{0.25em}{0ex}}p$. Note that
$\begin{array}{c}\sum _{m=1}^{p-1}{\chi }_{0}\left(m\right)|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\sum _{m=1}^{p-1}e\left(\frac{m{b}^{3}\left({a}^{3}-1\right)+nb\left(a-1\right)}{p}\right)\hfill \\ \phantom{\rule{1em}{0ex}}={\left(p-1\right)}^{2}+p-2={p}^{2}-p-1.\hfill \end{array}$
(3.2)
From (3.2) and Lemma 1 we have
$\begin{array}{c}\sum _{\chi modp}|\sum _{m=1}^{p-1}\chi \left(m\right)|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{2}{|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=|\sum _{m=1}^{p-1}{\chi }_{0}\left(m\right)|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{2}{|}^{2}+\underset{\chi \ne {\chi }_{0}}{\sum _{\chi modp}}|\sum _{m=1}^{p-1}\chi \left(m\right)|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{2}{|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\left({p}^{2}-p-1\right)}^{2}+\underset{\chi \ne {\chi }_{0}}{\sum _{\chi modp}}{p}^{2}\cdot |\sum _{a=1}^{p-2}\overline{\chi }\left(3{a}^{2}+3a+1\right){|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\left({p}^{2}-p-1\right)}^{2}+{p}^{2}\sum _{\chi modp}|\sum _{a=1}^{p-2}\overline{\chi }\left(3{a}^{2}+3a+1\right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}-{p}^{2}{\left(\sum _{a=1}^{p-2}{\chi }_{0}\left(3{a}^{2}+3a+1\right)\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\left({p}^{2}-p-1\right)}^{2}+{p}^{2}\left(p-1\right)\underset{{a}^{2}+a\equiv {b}^{2}+bmodp}{\sum _{a=1}^{p-2}\sum _{b=1}^{p-2}}1-{p}^{2}{\left(p-2\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\left({p}^{2}-p-1\right)}^{2}+{p}^{2}\left(p-1\right)\underset{\left(a-b\right)\left(a+b+1\right)\equiv 0modp}{\sum _{a=1}^{p-2}\sum _{b=1}^{p-2}}1-{p}^{2}{\left(p-2\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\left({p}^{2}-p-1\right)}^{2}+{p}^{2}\left(p-1\right)\left(p-2+p-2-1\right)-{p}^{2}{\left(p-2\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=\left(p-1\right)\left(2{p}^{3}-3{p}^{2}-3p-1\right).\hfill \end{array}$
(3.3)
If $3\nmid p-1$, then combining (3.1) and (3.3) we may immediately deduce the identity
$\sum _{m=1}^{p-1}|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{4}=2{p}^{3}-3{p}^{2}-3p-1$
or
$\sum _{m=1}^{p}|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{4}=2{p}^{3}-3{p}^{2}-3p.$
(3.4)
If $3|p-1$, let ${\chi }_{1}\ne {\chi }_{0}$ be a 3-order character $mod\phantom{\rule{0.25em}{0ex}}p$, then ${\overline{\chi }}_{1}={\chi }_{1}^{2}$ is also a 3-order character $mod\phantom{\rule{0.25em}{0ex}}p$; this time note that
$\sum _{m=1}^{p-1}{\chi }_{0}\left(m\right)|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{2}=p\left(p-1\right)-2p-1={p}^{2}-3p-1$
and
$\begin{array}{rcl}\sum _{a=1}^{p-2}{\chi }_{1}\left({\left(a+1\right)}^{3}-{a}^{3}\right)& =& \sum _{a=1}^{p-2}{\chi }_{1}\left({\left(\overline{a}+1\right)}^{3}-1\right)=\sum _{a=1}^{p-1}{\chi }_{1}\left({a}^{3}-1\right)\\ =& \sum _{a=1}^{p-1}\left({\overline{\chi }}_{1}^{2}\left(a\right)+{\overline{\chi }}_{1}\left(a\right)+1\right){\chi }_{1}\left(a-1\right)\\ =& -2+\sum _{a=1}^{p-1}{\chi }_{1}\left(a\left(1-a\right)\right),\end{array}$
and from Lemma 2 and the method of proving (3.3) we have
$\begin{array}{c}\sum _{\chi modp}|\sum _{m=1}^{p-1}\chi \left(m\right)|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{2}{|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\left({p}^{2}-3p-1\right)}^{2}+\underset{\chi \ne {\chi }_{0},{\chi }_{1},{\chi }_{1}^{2}}{\sum _{\chi modp}}{p}^{2}|\sum _{a=1}^{p-2}\overline{\chi }\left(3{a}^{2}+3a+1\right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}+2p|-2+\sum _{a=1}^{p-1}{\chi }_{1}\left(a\left(1-a\right)\right){|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\left({p}^{2}-3p-1\right)}^{2}+{p}^{2}\left(p-1\right)\left(2p-5\right)-2{p}^{2}|\sum _{a=1}^{p-2}{\chi }_{1}\left({\left(a+1\right)}^{3}-{a}^{3}\right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}+2p\left(p+4-2\sum _{a=1}^{p-1}{\chi }_{1}\left(a\left(1-a\right)\right)-2\sum _{a=1}^{p-1}{\overline{\chi }}_{1}\left(a\left(1-a\right)\right)\right)-{p}^{2}{\left(p-4\right)}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}={\left({p}^{2}-3p-1\right)}^{2}+{p}^{2}\left(p-1\right)\left(2p-5\right)-{p}^{2}{\left(p-4\right)}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}-\left(2{p}^{2}-2p\right)\left(p+4-2\sum _{a=1}^{p-1}{\chi }_{1}\left(a\left(1-a\right)\right)-2\sum _{a=1}^{p-1}{\overline{\chi }}_{1}\left(a\left(1-a\right)\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(p-1\right)\left(2{p}^{3}-5{p}^{2}-15p-1\right)+4\left(p-1\right)\left({\tau }^{3}\left({\chi }_{1}\right)+{\tau }^{3}\left({\overline{\chi }}_{1}\right)\right).\hfill \end{array}$
(3.5)
So if $3|p-1$, then combining (3.1) and (3.5) we can deduce the asymptotic formula
$\begin{array}{rcl}\sum _{m=1}^{p}|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{4}& =& 1+\sum _{m=1}^{p-1}|\sum _{a=1}^{p-1}e\left(\frac{m{a}^{3}+na}{p}\right){|}^{4}\\ =& 2{p}^{3}-5{p}^{2}-15p+4\left({\tau }^{3}\left({\chi }_{1}\right)+{\tau }^{3}\left({\overline{\chi }}_{1}\right)\right).\end{array}$
(3.6)

Now our theorem follows from (3.4) and (3.6).

## Authors’ Affiliations

(1)
School of Science, Xi’an Polytechnic University, Xi’an, Shaanxi, P.R. China
(2)
Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R. China

## References 