Theory and Modern Applications

# On the fourth power mean of the general k th Kloosterman sums

## Abstract

Let $q>2$ be an integer, and let χ be a Dirichlet character modulo q. For integers m, n and k, the general k th Kloosterman sum $S\left(m,n,k,\chi ;q\right)$ is defined by $S\left(m,n,k,\chi ;q\right)={{\sum }^{\prime }}_{a=1}^{q}\chi \left(a\right)\mathrm{e}\left(\frac{m{a}^{k}+n{\overline{a}}^{k}}{q}\right)$, where ∑ denotes the summation over all a with $\left(a,q\right)=1$, $\mathrm{e}\left(y\right)={\mathrm{e}}^{2\pi iy}$, and $\overline{a}$ is the inverse of a modulo q such that $a\overline{a}\equiv 1modq$ and $1\le \overline{a}\le q$. In this paper we further study the fourth power mean ${\sum }_{\chi modq}{\sum }_{m=1}^{q}|S\left(m,n,k,\chi ;q\right){|}^{4}$, and we give some identities.

MSC:11F20.

## 1 Introduction

Let $q>2$ be an integer, and let χ be a Dirichlet character modulo q. For arbitrary integers m and n, the general Kloosterman sum $S\left(m,n,\chi ;q\right)$ is defined by

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

where ∑ denotes the summation over all a with $\left(a,q\right)=1$, $\mathrm{e}\left(y\right)={\mathrm{e}}^{2\pi iy}$, and $\overline{a}$ is the inverse of a modulo q such that $a\overline{a}\equiv 1modq$ and $1\le \overline{a}\le q$. For $q=p$ a prime, Chowla [1] and Malyshev [2] proved an upper bound:

$|S\left(m,n,\chi ;p\right)|\ll {\left(m,n,p\right)}^{1/2}{p}^{1/2+ϵ},$

where $\left(m,n,p\right)$ is the great common divisor of m, n and p.

For general integer $q>2$, we do not know how large $|S\left(m,n,\chi ;q\right)|$ is. However, $|S\left(m,n,\chi ;q\right)|$ enjoys good value distribution properties. For fixed integer n with $\left(n,q\right)=1$, Zhang [3] showed the identity

$\sum _{\chi modq}\sum _{m=1}^{q}|S\left(m,n,\chi ;q\right){|}^{4}={\varphi }^{2}\left(q\right){q}^{2}d\left(q\right)\prod _{{p}^{\alpha }\parallel q}\left(1-\frac{2}{\alpha +1}\cdot \frac{{p}^{\alpha -1}-1}{{p}^{\alpha }\left(p-1\right)}+\frac{\alpha -4{p}^{\alpha -1}}{\left(\alpha +1\right){p}^{\alpha }}\right),$

where $d\left(q\right)$ is the divisor function, $\varphi \left(q\right)$ is the Euler function, and ${\prod }_{{p}^{\alpha }\parallel q}$ denotes the product over all prime divisors p of q with ${p}^{\alpha }\mid q$ and ${p}^{\alpha +1}\nmid q$.

For integers m, n and k, the general k th Kloosterman sum $S\left(m,n,k,\chi ;q\right)$ is defined by

$S\left(m,n,k,\chi ;q\right)=\underset{a=1}{\overset{q}{{\sum }^{\prime }}}\chi \left(a\right)\mathrm{e}\left(\frac{m{a}^{k}+n{\overline{a}}^{k}}{q}\right).$

Suppose that $\left(nk,q\right)=1$. Liu and Zhang [4] proved the identity:

$\begin{array}{c}\sum _{\chi modq}\sum _{m=1}^{q}|S\left(m,n,k,\chi ;q\right){|}^{4}\hfill \\ \phantom{\rule{1em}{0ex}}={\varphi }^{2}\left(q\right){q}^{2}\prod _{{p}^{\alpha }\parallel q}{\left(k,p-1\right)}^{2}\left(\alpha -1+\frac{2\left(p-1\right)}{\left(k,p-1\right)p}+\frac{\alpha }{{p}^{\alpha }}-\frac{2\left({p}^{\alpha }-1\right)}{{p}^{\alpha }\left(p-1\right)}\right).\hfill \end{array}$

In this paper we further consider the situation $\left(k,q\right)>1$. Our result is a generalization of [4].

Theorem 1.1 Let p be an odd prime, n be any integer. Let α and k be positive integers with $d=\left(k,p-1\right)$ and $\left(k,{p}^{\alpha -1}\right)={p}^{\delta }$. Then we have

$\begin{array}{c}\sum _{\chi mod{p}^{\alpha }}\sum _{m=1}^{{p}^{\alpha }}|S\left(m,n,k,\chi ;{p}^{\alpha }\right){|}^{4}\hfill \\ \phantom{\rule{1em}{0ex}}=\left\{\begin{array}{ll}2d{\varphi }^{3}\left({p}^{\alpha }\right){p}^{\alpha +\delta }\left(1-\frac{d}{2\varphi \left(p\right)}\right)+{d}^{2}{\varphi }^{2}\left({p}^{\alpha }\right){p}^{2\alpha +\delta -1}\left({p}^{\alpha -\delta -1}-1\right),& \frac{\alpha -2}{2}\le \delta \le \alpha -1;\\ 2d{\varphi }^{3}\left({p}^{\alpha }\right){p}^{\alpha +\delta }\left(1-\frac{d}{2\varphi \left(p\right)}\right)+{d}^{2}{\varphi }^{2}\left({p}^{\alpha }\right){p}^{2\alpha +\delta -1}\left({p}^{\delta +1}-1\right)\\ \phantom{\rule{1em}{0ex}}+{d}^{2}{\varphi }^{2}\left({p}^{\alpha }\right){p}^{2\alpha +2\delta -1}\left(p-1\right)\left(\alpha -2\delta -2\right),& 0\le \delta \le \frac{\alpha -3}{2}.\end{array}\hfill \end{array}$

From Theorem 1.1 we immediately get the following corollary.

Corollary 1.1 Let p be an odd prime, n be any integer. Let α and k be positive integers. Assume that one of the following conditions holds:

1. (1)

$\alpha =1$;

2. (2)

$\alpha =2$ and $\left(k,{p}^{\alpha -1}\right)>1$;

3. (3)

$\alpha \ge 2$ and $\left(k,{p}^{\alpha -1}\right)={p}^{\alpha -1}$.

Then we have

$\sum _{\chi mod{p}^{\alpha }}\sum _{m=1}^{{p}^{\alpha }}|S\left(m,n,k,\chi ;{p}^{\alpha }\right){|}^{4}=2{\varphi }^{3}\left({p}^{\alpha }\right){p}^{\alpha }\left(k,\varphi \left({p}^{\alpha }\right)\right)\left(1-\frac{d}{2\varphi \left(p\right)}\right).$

Theorem 1.2 Let $q>2$ be an odd number, n be an integer with $\left(n,q\right)=1$, and let k be a positive integer. Then we have

$\begin{array}{c}\sum _{\chi modq}\sum _{m=1}^{q}|S\left(m,n,k,\chi ;q\right){|}^{4}\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{\frac{\alpha -2}{2}\le \delta \le \alpha -1}{\underset{\left(k,{p}^{\alpha -1}\right)={p}^{\delta }}{\prod _{{p}^{\alpha }\parallel q}}}\left(2\left(k,p-1\right){\varphi }^{3}\left({p}^{\alpha }\right){p}^{\alpha +\delta }\left(1-\frac{\left(k,p-1\right)}{2\varphi \left(p\right)}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{\left(k,p-1\right)}^{2}{\varphi }^{2}\left({p}^{\alpha }\right){p}^{2\alpha +\delta -1}\left({p}^{\alpha -\delta -1}-1\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}×\underset{0\le \delta \le \frac{\alpha -3}{2}}{\underset{\left(k,{p}^{\alpha -1}\right)={p}^{\delta }}{\prod _{{p}^{\alpha }\parallel q}}}\left(2\left(k,p-1\right){\varphi }^{3}\left({p}^{\alpha }\right){p}^{\alpha +\delta }\left(1-\frac{\left(k,p-1\right)}{2\varphi \left(p\right)}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{\left(k,p-1\right)}^{2}{\varphi }^{2}\left({p}^{\alpha }\right){p}^{2\alpha +\delta -1}\left({p}^{\delta +1}-1\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{\left(k,p-1\right)}^{2}{\varphi }^{2}\left({p}^{\alpha }\right){p}^{2\alpha +2\delta -1}\left(p-1\right)\left(\alpha -2\delta -2\right)\right).\hfill \end{array}$

From Theorem 1.2 we can get the following corollary.

Corollary 1.2 Let $q>2$ be an odd number, n be an integer with $\left(n,q\right)=1$, and let k be a positive integer. Assume that one of the following conditions holds:

1. (1)

q is square-free;

2. (2)

$q={p}_{1}^{2}\cdots {p}_{r}^{2}$ and ${p}_{1}\cdots {p}_{r}\mid k$;

3. (3)

$q={\prod }_{\begin{array}{c}i=1\\ {\alpha }_{i}\ge 2\end{array}}^{r}{p}_{i}^{{\alpha }_{i}}$ and $q\mid k{\prod }_{i=1}^{r}{p}_{i}$.

Then we have

$\sum _{\chi modq}\sum _{m=1}^{q}|S\left(m,n,k,\chi ;q\right){|}^{4}={2}^{\omega \left(q\right)}{\varphi }^{3}\left(q\right)q\left(k,\varphi \left(q\right)\right)\prod _{{p}^{\alpha }\parallel q}\left(1-\frac{\left(k,p-1\right)}{2\varphi \left(p\right)}\right).$

## 2 Some lemmas

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

Lemma 2.1 Let p be an odd prime, and let α and k be positive integers with $\left(k,{p}^{\alpha -1}\right)={p}^{\delta }$. Write $d=\left(k,p-1\right)$. Then we have

$\underset{{p}^{\alpha }\mid \left({a}^{k}-1\right)\left({b}^{k}-1\right)}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1=\left\{\begin{array}{ll}2d\varphi \left({p}^{\alpha }\right){p}^{\delta }\left(1-\frac{d}{2\varphi \left(p\right)}\right)+{d}^{2}{p}^{\alpha +\delta -1}\left({p}^{\alpha -\delta -1}-1\right),& \frac{\alpha -2}{2}\le \delta \le \alpha -1;\\ 2d\varphi \left({p}^{\alpha }\right){p}^{\delta }\left(1-\frac{d}{2\varphi \left(p\right)}\right)+{d}^{2}{p}^{\alpha +\delta -1}\left({p}^{\delta +1}-1\right)\\ \phantom{\rule{1em}{0ex}}+{d}^{2}{p}^{\alpha +2\delta -1}\left(p-1\right)\left(\alpha -2\delta -2\right),& 0\le \delta \le \frac{\alpha -3}{2}.\end{array}$

Proof For $\alpha =1$, we get

$\underset{{p}^{\alpha }\mid \left({a}^{k}-1\right)\left({b}^{k}-1\right)}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1=\underset{p\mid \left({a}^{k}-1\right)\left({b}^{k}-1\right)}{\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}}1=2d\left(p-1\right)-{d}^{2}=2d\varphi \left(p\right)\left(1-\frac{d}{2\varphi \left(p\right)}\right).$
(2.1)

Now we assume that $\alpha \ge 2$. It is not hard to show that

$\begin{array}{rcl}\underset{{p}^{\alpha }\mid \left({a}^{k}-1\right)\left({b}^{k}-1\right)}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1& =& \underset{\beta +\gamma \ge \alpha }{\sum _{\beta =0}^{\alpha }\sum _{\gamma =0}^{\alpha }}\underset{{p}^{\beta }\parallel {a}^{k}-1}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}\underset{{p}^{\gamma }\parallel {b}^{k}-1}{\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1\\ =& \sum _{\beta =0}^{\alpha }\underset{{p}^{\beta }\parallel {a}^{k}-1}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}\sum _{\gamma =\alpha -\beta }^{\alpha }\underset{{p}^{\gamma }\parallel {b}^{k}-1}{\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1=\sum _{\beta =0}^{\alpha }\underset{{p}^{\beta }\parallel {a}^{k}-1}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}\underset{{p}^{\alpha -\beta }\mid {b}^{k}-1}{\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1\\ =& \underset{p\nmid {a}^{k}-1}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}\underset{{p}^{\alpha }\mid {b}^{k}-1}{\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1+\sum _{\beta =1}^{\alpha -1}\underset{{p}^{\beta }\parallel {a}^{k}-1}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}\underset{{p}^{\alpha -\beta }\mid {b}^{k}-1}{\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1+\underset{{p}^{\alpha }\mid {a}^{k}-1}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}1\\ =& \left(\varphi \left({p}^{\alpha }\right)-{p}^{\alpha -1}\left(k,\varphi \left(p\right)\right)\right)\cdot \left(k,\varphi \left({p}^{\alpha }\right)\right)+\left(k,\varphi \left({p}^{\alpha }\right)\right)\varphi \left({p}^{\alpha }\right)\\ +\sum _{\beta =1}^{\alpha -1}\left({p}^{\alpha -\beta }\left(k,\varphi \left({p}^{\beta }\right)\right)-{p}^{\alpha -\beta -1}\left(k,\varphi \left({p}^{\beta +1}\right)\right)\right)\cdot {p}^{\beta }\left(k,\varphi \left({p}^{\alpha -\beta }\right)\right)\\ =& {p}^{\alpha -1}\left(2\varphi \left(p\right)-\left(k,\varphi \left(p\right)\right)\right)\cdot \left(k,{p}^{\alpha -1}\right)\cdot \left(k,\varphi \left(p\right)\right)\\ +{p}^{\alpha -1}{\left(k,\varphi \left(p\right)\right)}^{2}\sum _{\beta =1}^{\alpha -1}\left(p\left(k,{p}^{\beta -1}\right)-\left(k,{p}^{\beta }\right)\right)\cdot \left(k,{p}^{\alpha -\beta -1}\right)\\ =& 2d\varphi \left({p}^{\alpha }\right)\left(k,{p}^{\alpha -1}\right)\left(1-\frac{d}{2\varphi \left(p\right)}\right)\\ +{d}^{2}{p}^{\alpha -1}\sum _{\beta =1}^{\alpha -1}\left(p\left(k,{p}^{\beta -1}\right)-\left(k,{p}^{\beta }\right)\right)\cdot \left(k,{p}^{\alpha -\beta -1}\right).\end{array}$
(2.2)

If $\left(k,{p}^{\alpha -1}\right)=1$, by (2.2) we have

$\underset{{p}^{\alpha }\mid \left({a}^{k}-1\right)\left({b}^{k}-1\right)}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1=2d\varphi \left({p}^{\alpha }\right)\left(1-\frac{d}{2\varphi \left(p\right)}\right)+{d}^{2}\varphi \left({p}^{\alpha }\right)\left(\alpha -1\right).$
(2.3)

If $\left(k,{p}^{\alpha -1}\right)={p}^{\alpha -1}$, then

$\underset{{p}^{\alpha }\mid \left({a}^{k}-1\right)\left({b}^{k}-1\right)}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1=2d\varphi \left({p}^{\alpha }\right){p}^{\alpha -1}\left(1-\frac{d}{2\varphi \left(p\right)}\right).$
(2.4)

On the other hand, for $\alpha =2$ and $\left(k,{p}^{\alpha -1}\right)>1$, we have

$\underset{{p}^{\alpha }\mid \left({a}^{k}-1\right)\left({b}^{k}-1\right)}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1=2d\varphi \left({p}^{\alpha }\right)\left(k,{p}^{\alpha -1}\right)\left(1-\frac{d}{2\varphi \left(p\right)}\right).$
(2.5)

Next we suppose that $\alpha \ge 3$ and $\left(k,{p}^{\alpha -1}\right)={p}^{\delta }$ with $1\le \delta \le \alpha -2$. Then

$\begin{array}{c}\underset{{p}^{\alpha }\mid \left({a}^{k}-1\right)\left({b}^{k}-1\right)}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1\hfill \\ \phantom{\rule{1em}{0ex}}=2d\varphi \left({p}^{\alpha }\right)\left(k,{p}^{\alpha -1}\right)\left(1-\frac{d}{2\varphi \left(p\right)}\right)+{d}^{2}{p}^{\alpha -1}\sum _{\beta =1}^{\alpha -1}\left(p\left(k,{p}^{\beta -1}\right)-\left(k,{p}^{\beta }\right)\right)\cdot \left(k,{p}^{\alpha -\beta -1}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=2d\varphi \left({p}^{\alpha }\right){p}^{\delta }\left(1-\frac{d}{2\varphi \left(p\right)}\right)+{d}^{2}\varphi \left({p}^{\alpha }\right){p}^{\delta }\sum _{\beta =\delta +1}^{\alpha -1}\left(k,{p}^{\alpha -\beta -1}\right).\hfill \end{array}$

Noting that

$\begin{array}{rcl}\sum _{\beta =\delta +1}^{\alpha -1}\left(k,{p}^{\alpha -\beta -1}\right)& =& \sum _{u=0}^{\alpha -\delta -2}\left(k,{p}^{u}\right)\\ =& \left\{\begin{array}{ll}{\sum }_{u=0}^{\alpha -\delta -2}{p}^{u},& \alpha -\delta -2\le \delta ,\\ {\sum }_{u=0}^{\delta }{p}^{u}+{\sum }_{u=\delta +1}^{\alpha -\delta -2}{p}^{\delta },& \alpha -\delta -2\ge \delta +1\end{array}\\ =& \left\{\begin{array}{ll}\frac{{p}^{\alpha -\delta -1}-1}{p-1},& \delta \ge \frac{\alpha -2}{2},\\ \frac{{p}^{\delta +1}-1}{p-1}+\left(\alpha -2\delta -2\right){p}^{\delta },& \delta \le \frac{\alpha -3}{2}.\end{array}\end{array}$

Therefore

$\begin{array}{c}\underset{{p}^{\alpha }\mid \left({a}^{k}-1\right)\left({b}^{k}-1\right)}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1\hfill \\ \phantom{\rule{1em}{0ex}}=\left\{\begin{array}{ll}2d\varphi \left({p}^{\alpha }\right){p}^{\delta }\left(1-\frac{d}{2\varphi \left(p\right)}\right)+{d}^{2}{p}^{\alpha +\delta -1}\left({p}^{\alpha -\delta -1}-1\right),& \frac{\alpha -2}{2}\le \delta \le \alpha -2;\\ 2d\varphi \left({p}^{\alpha }\right){p}^{\delta }\left(1-\frac{d}{2\varphi \left(p\right)}\right)\\ \phantom{\rule{1em}{0ex}}+{d}^{2}{p}^{\alpha +\delta -1}\left({p}^{\delta +1}-1\right)+{d}^{2}{p}^{\alpha +2\delta -1}\left(p-1\right)\left(\alpha -2\delta -2\right),& 1\le \delta \le \frac{\alpha -3}{2}.\end{array}\hfill \end{array}$
(2.6)

Now combining (2.1), (2.3)-(2.6) we immediately get

$\begin{array}{c}\underset{{p}^{\alpha }\mid \left({a}^{k}-1\right)\left({b}^{k}-1\right)}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1\hfill \\ \phantom{\rule{1em}{0ex}}=\left\{\begin{array}{ll}2d\varphi \left({p}^{\alpha }\right){p}^{\delta }\left(1-\frac{d}{2\varphi \left(p\right)}\right)+{d}^{2}{p}^{\alpha +\delta -1}\left({p}^{\alpha -\delta -1}-1\right),& \frac{\alpha -2}{2}\le \delta \le \alpha -1;\\ 2d\varphi \left({p}^{\alpha }\right){p}^{\delta }\left(1-\frac{d}{2\varphi \left(p\right)}\right)\\ \phantom{\rule{1em}{0ex}}+{d}^{2}{p}^{\alpha +\delta -1}\left({p}^{\delta +1}-1\right)+{d}^{2}{p}^{\alpha +2\delta -1}\left(p-1\right)\left(\alpha -2\delta -2\right),& 0\le \delta \le \frac{\alpha -3}{2},\end{array}\hfill \end{array}$

for $\alpha \ge 1$. □

Lemma 2.2 Let n, k, ${k}_{1}$, ${k}_{2}$ be integers with $\left(n,{k}_{1}{k}_{2}\right)=\left({k}_{1},{k}_{2}\right)=1$. Then for any character $\chi mod{k}_{1}{k}_{2}$, there exist integers ${n}_{1}$ and ${n}_{2}$ with $\left({n}_{1},{k}_{1}\right)=\left({n}_{2},{k}_{2}\right)=1$ such that

$n\equiv {n}_{1}{k}_{2}^{2}+{n}_{2}{k}_{1}^{2}\phantom{\rule{0.25em}{0ex}}\left(mod\phantom{\rule{0.25em}{0ex}}{k}_{1}{k}_{2}\right),$

and for these integers we have

$|S\left(m,n,k,\chi ;{k}_{1}{k}_{2}\right)|=|S\left(m{\overline{k}}_{2},{n}_{1}{k}_{2},k,{\chi }_{1};{k}_{1}\right)|\cdot |S\left(m{\overline{k}}_{1},{n}_{2}{k}_{1},k,{\chi }_{2};{k}_{2}\right)|,$

where $\chi ={\chi }_{1}{\chi }_{2}$ with ${\chi }_{1}mod{k}_{1}$ and ${\chi }_{2}mod{k}_{2}$.

Proof This is Lemma 2.2 of [4]. □

## 3 Proof of the theorems

First we prove Theorem 1.1. Let p be an odd prime, and let α and k be positive integers. Assume that n is any integer. We have

$\begin{array}{rcl}|S\left(m,n,k,\chi ;{p}^{\alpha }\right){|}^{2}& =& \sum _{a=1}^{{p}^{\alpha }}\chi \left(a\right)\mathrm{e}\left(\frac{m{a}^{k}+n{\overline{a}}^{k}}{{p}^{\alpha }}\right)\sum _{b=1}^{{p}^{\alpha }}\overline{\chi }\left(b\right)\mathrm{e}\left(-\frac{m{b}^{k}+n{\overline{b}}^{k}}{{p}^{\alpha }}\right)\\ =& \underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\chi \left(a\right)\mathrm{e}\left(\frac{m{b}^{k}\left({a}^{k}-1\right)+n{\overline{b}}^{k}\left({\overline{a}}^{k}-1\right)}{{p}^{\alpha }}\right).\end{array}$

Then from the orthogonality relation for characters and the trigonometric identity we get

$\begin{array}{c}\sum _{\chi mod{p}^{\alpha }}\sum _{m=1}^{{p}^{\alpha }}|S\left(m,n,k,\chi ;{p}^{\alpha }\right){|}^{4}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{\chi mod{p}^{\alpha }}\sum _{m=1}^{{p}^{\alpha }}\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\chi \left(a\right)\mathrm{e}\left(\frac{m{b}^{k}\left({a}^{k}-1\right)+n{\overline{b}}^{k}\left({\overline{a}}^{k}-1\right)}{{p}^{\alpha }}\right)\hfill \\ \phantom{\rule{2em}{0ex}}×\underset{c=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{d=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\overline{\chi }\left(c\right)\mathrm{e}\left(-\frac{m{d}^{k}\left({c}^{k}-1\right)+n{\overline{d}}^{k}\left({\overline{c}}^{k}-1\right)}{{p}^{\alpha }}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\varphi \left({p}^{\alpha }\right)\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{d=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\sum _{m=1}^{{p}^{\alpha }}\mathrm{e}\left(\frac{m\left({a}^{k}-1\right)\left({b}^{k}-{d}^{k}\right)+n\left({\overline{a}}^{k}-1\right)\left({\overline{b}}^{k}-{\overline{d}}^{k}\right)}{{p}^{\alpha }}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\varphi \left({p}^{\alpha }\right)\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{d=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\sum _{m=1}^{{p}^{\alpha }}\mathrm{e}\left(\frac{m{d}^{k}\left({a}^{k}-1\right)\left({b}^{k}-1\right)+n{\overline{d}}^{k}\left({\overline{a}}^{k}-1\right)\left({\overline{b}}^{k}-1\right)}{{p}^{\alpha }}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\varphi \left({p}^{\alpha }\right)\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{d=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\sum _{m=1}^{{p}^{\alpha }}\mathrm{e}\left(\frac{m{d}^{k}\left({a}^{k}-1\right)\left({b}^{k}-1\right)+n{\overline{d}}^{k}{\overline{a}}^{k}{\overline{b}}^{k}\left({a}^{k}-1\right)\left({b}^{k}-1\right)}{{p}^{\alpha }}\right)\hfill \\ \phantom{\rule{1em}{0ex}}={\varphi }^{2}\left({p}^{\alpha }\right){p}^{\alpha }\underset{{p}^{\alpha }\mid \left({a}^{k}-1\right)\left({b}^{k}-1\right)}{\underset{a=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}\underset{b=1}{\overset{{p}^{\alpha }}{{\sum }^{\prime }}}}1.\hfill \end{array}$

Write $d=\left(k,p-1\right)$ and $\left(k,{p}^{\alpha -1}\right)={p}^{\delta }$. By Lemma 2.1 we immediately have

$\begin{array}{c}\sum _{\chi mod{p}^{\alpha }}\sum _{m=1}^{{p}^{\alpha }}|S\left(m,n,k,\chi ;{p}^{\alpha }\right){|}^{4}\hfill \\ \phantom{\rule{1em}{0ex}}=\left\{\begin{array}{ll}2d{\varphi }^{3}\left({p}^{\alpha }\right){p}^{\alpha +\delta }\left(1-\frac{d}{2\varphi \left(p\right)}\right)+{d}^{2}{\varphi }^{2}\left({p}^{\alpha }\right){p}^{2\alpha +\delta -1}\left({p}^{\alpha -\delta -1}-1\right),& \frac{\alpha -2}{2}\le \delta \le \alpha -1;\\ 2d{\varphi }^{3}\left({p}^{\alpha }\right){p}^{\alpha +\delta }\left(1-\frac{d}{2\varphi \left(p\right)}\right)+{d}^{2}{\varphi }^{2}\left({p}^{\alpha }\right){p}^{2\alpha +\delta -1}\left({p}^{\delta +1}-1\right)\\ \phantom{\rule{1em}{0ex}}+{d}^{2}{\varphi }^{2}\left({p}^{\alpha }\right){p}^{2\alpha +2\delta -1}\left(p-1\right)\left(\alpha -2\delta -2\right),& 0\le \delta \le \frac{\alpha -3}{2}.\end{array}\hfill \end{array}$

This completes the proof of Theorem 1.1.

Now we prove Theorem 1.2. Let $q>2$ be an odd number, n be an integer with $\left(n,q\right)=1$, and let k be a positive integer. Write

$q=\prod _{i=1}^{r}{p}_{i}^{{\alpha }_{i}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}m=\sum _{i=1}^{r}\frac{{m}_{i}q}{{p}_{i}^{{\alpha }_{i}}}.$

By Lemma 2.2 and Theorem 1.1 we have

$\begin{array}{c}\sum _{\chi modq}\sum _{m=1}^{q}|S\left(m,n,k,\chi ;q\right){|}^{4}\hfill \\ \phantom{\rule{1em}{0ex}}=\prod _{i=1}^{r}\left(\sum _{{\chi }_{i}mod{p}_{i}^{{\alpha }_{i}}}\sum _{{m}_{i}=1}^{{p}_{i}^{{\alpha }_{i}}}|S\left({m}_{i}\frac{q}{{p}_{i}^{{\alpha }_{i}}}\overline{\left(\frac{q}{{p}_{i}^{{\alpha }_{i}}}\right)},{n}_{i}\frac{q}{{p}_{i}^{{\alpha }_{i}}},k,{\chi }_{i};{p}_{i}^{{\alpha }_{i}}\right){|}^{4}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\prod _{i=1}^{r}\left(\sum _{{\chi }_{i}mod{p}_{i}^{{\alpha }_{i}}}\sum _{{m}_{i}=1}^{{p}_{i}^{{\alpha }_{i}}}|S\left({m}_{i},{n}_{i}\frac{q}{{p}_{i}^{{\alpha }_{i}}},k,{\chi }_{i};{p}_{i}^{{\alpha }_{i}}\right){|}^{4}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\underset{\frac{\alpha -2}{2}\le \delta \le \alpha -1}{\underset{\left(k,{p}^{\alpha -1}\right)={p}^{\delta }}{\prod _{{p}^{\alpha }\parallel q}}}\left(2\left(k,p-1\right){\varphi }^{3}\left({p}^{\alpha }\right){p}^{\alpha +\delta }\left(1-\frac{\left(k,p-1\right)}{2\varphi \left(p\right)}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{\left(k,p-1\right)}^{2}{\varphi }^{2}\left({p}^{\alpha }\right){p}^{2\alpha +\delta -1}\left({p}^{\alpha -\delta -1}-1\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}×\underset{0\le \delta \le \frac{\alpha -3}{2}}{\underset{\left(k,{p}^{\alpha -1}\right)={p}^{\delta }}{\prod _{{p}^{\alpha }\parallel q}}}\left(2\left(k,p-1\right){\varphi }^{3}\left({p}^{\alpha }\right){p}^{\alpha +\delta }\left(1-\frac{\left(k,p-1\right)}{2\varphi \left(p\right)}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{\left(k,p-1\right)}^{2}{\varphi }^{2}\left({p}^{\alpha }\right){p}^{2\alpha +\delta -1}\left({p}^{\delta +1}-1\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{\left(k,p-1\right)}^{2}{\varphi }^{2}\left({p}^{\alpha }\right){p}^{2\alpha +2\delta -1}\left(p-1\right)\left(\alpha -2\delta -2\right)\right).\hfill \end{array}$

This proves Theorem 1.2.

## References

1. Chowla S: On Kloosterman’s sum. Norske Vid. Selsk. Forhdl. 1967, 40: 70–72.

2. Malyshev AV: A generalization of Kloosterman sums and their estimates. Vestn. Leningr. Univ. 1960, 15: 59–75. (Russian)

3. Zhang W: On the general Kloosterman sum and its fourth power mean. J. Number Theory 2004, 104: 156–161. 10.1016/S0022-314X(03)00154-9

4. Liu HY, Zhang WP: On the general k -th Kloosterman sums and its fourth power mean. Chin. Ann. Math., Ser. B 2004, 25: 97–102. 10.1142/S0252959904000093

## Acknowledgements

This work is supported by the P.E.D. (2013JK0561) and N.S.F. (11371291) of P.R. China.

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Correspondence to Xiaoyan Guo.

### 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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Guo, X., Geng, G. & Pan, X. On the fourth power mean of the general k th Kloosterman sums. Adv Differ Equ 2014, 103 (2014). https://doi.org/10.1186/1687-1847-2014-103