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# On the identity involving certain Hardy sums and Kloosterman sums

## Abstract

The main purpose of this paper is, using the properties of Gauss sums and the mean value theorem of Dirichlet L-functions, to study a hybrid mean value problem involving certain Hardy sums and Kloosterman sums and give two exact computational formulae for them.

MSC:11L05, 11M20.

## 1 Introduction

Let c be a natural number and d be an integer prime to c. The classical Dedekind sums

$S\left(d,c\right)=\sum _{j=1}^{c}\left(\left(\frac{j}{c}\right)\right)\left(\left(\frac{dj}{c}\right)\right),$

where

describes the behavior of the logarithm of the eta-function (see [1] and [2]) under modular transformations. Berndt [3] gave an analogous transformation formula for the logarithm of the classical theta-function

$\theta \left(z\right)=\sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}exp\left(\pi i{n}^{2}z\right),\phantom{\rule{1em}{0ex}}Im\left(z\right)>0.$

That is, put $Vz=\left(az+b\right)\left(cz+d\right)$ with $a,b,c,d\in Z$, $c>0$, and $ad-bc=1$. Then we have

$log\theta \left(Vz\right)=log\theta \left(z\right)+\frac{1}{2}log\left(cz+d\right)-\frac{1}{4}\pi i+\frac{1}{4}\pi i{S}_{1}\left(d,c\right),$
(1)

where ${S}_{1}\left(d,c\right)$ are defined as

${S}_{1}\left(d,c\right)=\sum _{j=1}^{c-1}{\left(-1\right)}^{j+1+\left[\frac{dj}{c}\right]}.$

The sums ${S}_{1}\left(d,c\right)$ (and certain related ones) are sometimes called Hardy sums. They are closely connected with Dedekind sums. Some authors had studied the properties of ${S}_{1}\left(d,c\right)$ and related sums and obtained some interesting results, see [48] and [9]. For example, Zhang and Yi [8] proved the following conclusion. Let p be an odd prime. Then, for any fixed positive integer m, we have the asymptotic formula

$\sum _{h=1}^{p-1}|{S}_{1}\left(h,p\right){|}^{2m}={p}^{2m}\cdot \frac{{\zeta }^{2}\left(2m\right)\left(1-\frac{1}{{4}^{m}}\right)}{\zeta \left(4m\right)\left(1+\frac{1}{{4}^{m}}\right)}+O\left({p}^{2m-1}\cdot exp\left(\frac{6lnp}{lnlnp}\right)\right),$

where $\zeta \left(s\right)$ is the Riemann zeta-function and $exp\left(y\right)={e}^{y}$.

On the other hand, we introduce the classical Kloosterman sums $K\left(n,q\right)$ which are defined as follows: For any positive integer $q>1$ and integer n,

$K\left(n,q\right)=\underset{b=1}{\overset{q}{{\sum }^{\prime }}}e\left(\frac{nb+\overline{b}}{q}\right),$

where $\overline{b}$ denotes the solution of the congruence $x\cdot b\equiv 1modq$, ${{\sum }^{\prime }}_{b=1}^{q}$ denotes the summation over all $1\le b\le q$ with $\left(b,q\right)=1$ and $e\left(x\right)={e}^{2\pi ix}$. Some elementary properties of $K\left(n,q\right)$ can be found in [10] and [11].

The main purpose of this paper is, using the properties of Gauss sums and the mean square value theorem of Dirichlet L-functions, to study a hybrid mean value problem involving certain Hardy sums and Kloosterman sums and give two exact computational formulae. That is, we shall prove the following theorem.

Theorem 1 Let p be an odd prime. Then we have the identity

Theorem 2 Let p be an odd prime, then we have the identity

where ${h}_{p}$ denotes the class number of the quadratic field $\mathbf{Q}\left(\sqrt{-p}\right)$.

For general odd number $q\ge 3$, whether there exits a computational formula for the hybrid mean value

$\underset{m=1}{\overset{q}{{\sum }^{\prime }}}\underset{n=1}{\overset{q}{{\sum }^{\prime }}}|K\left(m,q\right){|}^{2}\cdot |K\left(n,q\right){|}^{2}\cdot {S}_{1}\left(2m\cdot \overline{n},q\right)$

is an open problem.

## 2 Several lemmas

In this section, we shall give several lemmas, which are necessary in the proof of our theorems. Hereinafter, we shall use many properties of Gauss sums, all of which can be found in [12], so they will not be repeated here. First we have the following lemma.

Lemma 1 Let p be an odd prime, then we have the identity

$\sum _{n=1}^{p-1}\chi \left(n\right)\cdot |K\left(n,p\right){|}^{2}=\overline{\chi }\left(-1\right)\cdot \frac{{\tau }^{3}\left(\chi \right)\cdot \tau \left({\overline{\chi }}^{2}\right)}{\tau \left(\overline{\chi }\right)}.$

Proof For any non-principal character $\chi modp$, from the properties of Gauss sums we have

$\begin{array}{rcl}\sum _{n=1}^{p-1}\chi \left(n\right)|K\left(n,p\right){|}^{2}& =& \sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\sum _{n=1}^{p-1}\chi \left(n\right)e\left(\frac{n\left(a-b\right)+\left(\overline{a}-\overline{b}\right)}{p}\right)\\ =& \sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\sum _{n=1}^{p-1}\chi \left(n\right)e\left(\frac{nb\left(a-1\right)+\overline{b}\left(\overline{a}-1\right)}{p}\right)\\ =& \tau \left(\chi \right)\cdot \sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\overline{\chi }\left(b\left(a-1\right)\right)e\left(\frac{\overline{b}\left(\overline{a}-1\right)}{p}\right)\\ =& {\tau }^{2}\left(\chi \right)\cdot \sum _{a=1}^{p-1}\overline{\chi }\left(a-1\right)\overline{\chi }\left(\overline{a}-1\right)\\ =& {\tau }^{2}\left(\chi \right)\cdot \sum _{a=1}^{p-1}\chi \left(a\right)\overline{\chi }\left(-{\left(a-1\right)}^{2}\right)=\overline{\chi }\left(-1\right)\cdot {\tau }^{2}\left(\chi \right)\cdot \sum _{a=1}^{p-2}\chi \left(a+1\right)\overline{\chi }\left({a}^{2}\right)\\ =& \overline{\chi }\left(-1\right)\cdot {\tau }^{2}\left(\chi \right)\cdot \sum _{a=1}^{p-2}\chi \left(\overline{a}+{\overline{a}}^{2}\right)=\overline{\chi }\left(-1\right)\cdot {\tau }^{2}\left(\chi \right)\cdot \sum _{a=1}^{p-1}\chi \left({a}^{2}+a\right)\\ =& \overline{\chi }\left(-1\right)\cdot {\tau }^{2}\left(\chi \right)\cdot \frac{1}{\tau \left(\overline{\chi }\right)}\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\overline{\chi }\left(b\right)e\left(\frac{b\left({a}^{2}+a\right)}{p}\right)\\ =& \overline{\chi }\left(-1\right)\cdot {\tau }^{2}\left(\chi \right)\cdot \frac{1}{\tau \left(\overline{\chi }\right)}\cdot \sum _{b=1}^{p-1}\overline{\chi }\left(b\right)e\left(\frac{b}{p}\right)\sum _{a=1}^{p-1}\chi \left(a\right)e\left(\frac{ba}{p}\right)\\ =& \overline{\chi }\left(-1\right)\cdot {\tau }^{3}\left(\chi \right)\cdot \frac{1}{\tau \left(\overline{\chi }\right)}\cdot \sum _{b=1}^{p-1}{\overline{\chi }}^{2}\left(b\right)e\left(\frac{b}{p}\right)\\ =& \overline{\chi }\left(-1\right)\cdot \frac{{\tau }^{3}\left(\chi \right)\cdot \tau \left({\overline{\chi }}^{2}\right)}{\tau \left(\overline{\chi }\right)}.\end{array}$

This proves Lemma 1. □

Lemma 2 Let $q>2$ be an integer, then, for any integer a with $\left(a,q\right)=1$, we have the identity

$S\left(a,q\right)=\frac{1}{{\pi }^{2}q}\sum _{d|q}\frac{{d}^{2}}{\varphi \left(d\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modd}}\chi \left(a\right)|L\left(1,\chi \right){|}^{2},$

where $L\left(1,\chi \right)$ denotes the Dirichlet L-function corresponding to character $\chi modd$.

Proof See Lemma 2 of [9]. □

Lemma 3 Let $q>0$ and $\left(h,q\right)=1$. Then we have the identity

${S}_{1}\left(h,q\right)=-8S\left(h+q,2q\right)+4S\left(h,q\right).$

Proof This formula is an immediate consequence of (5.9) and (5.10) in [7]. □

Lemma 4 Let p be an odd prime and $0. Then we have the identity

${S}_{1}\left(2h,p\right)=-20\cdot S\left(2h,p\right)+8\cdot S\left(4h,p\right)+8\cdot S\left(h,p\right).$

Proof Note that the divisors of 2p are 1, 2, p and 2p. So from Lemma 2 and Lemma 3 we have

$\begin{array}{rcl}{S}_{1}\left(2h,p\right)& =& -8\cdot S\left(2h+p,2p\right)+4\cdot S\left(2h,p\right)\\ =& -\frac{4}{{\pi }^{2}p}\sum _{d|2p}\frac{{d}^{2}}{\varphi \left(d\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modd}}\chi \left(2h+p\right)|L\left(1,\chi \right){|}^{2}\\ +\frac{4}{{\pi }^{2}p}\sum _{d|p}\frac{{d}^{2}}{\varphi \left(d\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modd}}\chi \left(2h\right)|L\left(1,\chi \right){|}^{2}\\ =& -\frac{4}{{\pi }^{2}p}\cdot \frac{{\left(2p\right)}^{2}}{\varphi \left(2p\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi mod2p}}\chi \left(2h+p\right)|L\left(1,\chi \right){|}^{2}\\ =& -\frac{16p}{{\pi }^{2}\left(p-1\right)}\cdot \underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(2h+p\right)\lambda \left(2h+p\right)|L\left(1,\chi \lambda \right){|}^{2}\\ =& -\frac{16p}{{\pi }^{2}\left(p-1\right)}\cdot \underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(2h\right)|L\left(1,\chi \lambda \right){|}^{2},\end{array}$
(2)

where λ denotes the principal character $mod\phantom{\rule{0.25em}{0ex}}2$.

From the Euler infinite product formula we have

$\begin{array}{rcl}|L\left(1,\chi \lambda \right){|}^{2}& =& \prod _{{p}_{1}}|1-\frac{\chi \left({p}_{1}\right)\lambda \left({p}_{1}\right)}{{p}_{1}}{|}^{-2}=\prod _{{p}_{1}>2}|1-\frac{\chi \left({p}_{1}\right)}{{p}_{1}}{|}^{-2}\\ =& |1-\frac{\chi \left(2\right)}{2}{|}^{2}\cdot \prod _{{p}_{1}}|1-\frac{\chi \left({p}_{1}\right)}{{p}_{1}}{|}^{-2}=\left(\frac{5}{4}-\frac{\chi \left(2\right)}{2}-\frac{\overline{\chi }\left(2\right)}{2}\right)\cdot |L\left(1,\chi \right){|}^{2},\end{array}$
(3)

where ${\prod }_{p}$ denotes the product over all primes p.

From Lemma 2 we also have the identity

$S\left(n,p\right)=\frac{1}{{\pi }^{2}}\cdot \frac{p}{p-1}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(n\right)|L\left(1,\chi \right){|}^{2}.$
(4)

Now, combining (2), (3) and (4), we have the identity

$\begin{array}{rcl}{S}_{1}\left(2h,p\right)& =& -\frac{16p}{{\pi }^{2}\left(p-1\right)}\cdot \underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(2h\right)|L\left(1,\chi \lambda \right){|}^{2}\\ =& -16\cdot \frac{p}{{\pi }^{2}\left(p-1\right)}\cdot \underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(2h\right)\left(\frac{5}{4}-\frac{\chi \left(2\right)}{2}-\frac{\overline{\chi }\left(2\right)}{2}\right)\cdot {|L\left(1,\chi \right)|}^{2}\\ =& -20\cdot S\left(2h,p\right)+8\cdot S\left(4h,p\right)+8\cdot S\left(h,p\right).\end{array}$

This proves Lemma 4. □

Lemma 5 Let p be an odd prime. Then we have the identities

Proof From the definition of Dedekind sums we have

$S\left(1,c\right)=\sum _{a=1}^{c-1}{\left(\frac{a}{c}-\frac{1}{2}\right)}^{2}=\frac{\left(c-1\right)\left(c-2\right)}{12c}.$
(5)

If $p\equiv 1modc$, then from (5), and noting the reciprocity theorem of Dedekind sums (see [5]), we have the computational formula

$\begin{array}{rcl}S\left(c,p\right)& =& \frac{{p}^{2}+{c}^{2}+1}{12pc}-\frac{1}{4}-S\left(p,c\right)=\frac{{p}^{2}+{c}^{2}+1}{12pc}-\frac{1}{4}-S\left(1,c\right)\\ =& \frac{{p}^{2}+{c}^{2}+1}{12pc}-\frac{1}{4}-\frac{\left(c-1\right)\left(c-2\right)}{12c}=\frac{\left(p-1\right)\left(p-1-{c}^{2}\right)}{12pc}.\end{array}$
(6)

If $p\equiv 3mod4$, then we also have

$\begin{array}{rcl}S\left(4,p\right)& =& \frac{{p}^{2}+16+1}{48p}-\frac{1}{4}-S\left(p,4\right)=\frac{{p}^{2}+17}{48p}-\frac{1}{4}-S\left(3,4\right)\\ =& \frac{{p}^{2}+17}{48p}-\frac{1}{4}+\frac{1}{8}=\frac{{p}^{2}-6p+17}{48p}.\end{array}$
(7)

Now taking $c=1$ in (6), from (4) we may immediately deduce the identity

$\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}|L\left(1,\chi \right){|}^{2}=\frac{{\pi }^{2}}{12}\cdot \frac{{\left(p-1\right)}^{2}\cdot \left(p-2\right)}{{p}^{2}}.$
(8)

Taking $c=2$ in (6), from (4) we can also deduce the identity

$\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(2\right)\cdot {|L\left(1,\chi \right)|}^{2}=\frac{{\pi }^{2}}{24}\cdot \frac{{\left(p-1\right)}^{2}\cdot \left(p-5\right)}{{p}^{2}}.$
(9)

If $p\equiv 1mod4$, then taking $c=4$ in (6), from (4) we can deduce the identity

$\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(4\right)\cdot {|L\left(1,\chi \right)|}^{2}=\frac{{\pi }^{2}}{48}\cdot \frac{{\left(p-1\right)}^{2}\cdot \left(p-17\right)}{{p}^{2}}.$
(10)

If $p\equiv 3mod4$, then from (4) and (7) we have the identity

$\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(4\right)\cdot |L\left(1,\chi \right){|}^{2}=\frac{{\pi }^{2}}{48}\cdot \frac{\left(p-1\right)\cdot \left({p}^{2}-6p+17\right)}{{p}^{2}}.$
(11)

Now Lemma 5 follows from (8)-(11). □

## 3 Proof of the theorems

In this section, we shall complete the proof of our theorems. Note that if χ is a non-principal character $mod\phantom{\rule{0.25em}{0ex}}p$, then $|\tau \left(\chi \right)|=\sqrt{p}$ and

$|\sum _{m=1}^{p-1}\chi \left(m\right)K\left(m,p\right)|=|\sum _{a=1}^{p-1}\sum _{m=1}^{p-1}\chi \left(m\right)e\left(\frac{ma+\overline{a}}{p}\right)|=|{\tau }^{2}\left(\chi \right)|=p.$
(12)

If $p\equiv 3mod4$, then from (12), Lemma 4 and Lemma 5 we have

$\begin{array}{r}\sum _{m=1}^{p-1}\sum _{n=1}^{p-1}K\left(m,p\right)\cdot K\left(n,p\right)\cdot {S}_{1}\left(2m\cdot \overline{n},p\right)\\ \phantom{\rule{1em}{0ex}}=-\frac{20\cdot p}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(2\right)|\sum _{n=1}^{p-1}\chi \left(n\right)\cdot K\left(n,p\right){|}^{2}\cdot |L\left(1,\chi \right){|}^{2}\\ \phantom{\rule{2em}{0ex}}+\frac{8\cdot p}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(4\right)|\sum _{n=1}^{p-1}\chi \left(n\right)\cdot K\left(n,p\right){|}^{2}\cdot |L\left(1,\chi \right){|}^{2}\\ \phantom{\rule{2em}{0ex}}+\frac{8\cdot p}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}|\sum _{n=1}^{p-1}\chi \left(n\right)\cdot K\left(n,p\right){|}^{2}\cdot |L\left(1,\chi \right){|}^{2}\\ \phantom{\rule{1em}{0ex}}=-\frac{20\cdot {p}^{3}}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(2\right)\cdot |L\left(1,\chi \right){|}^{2}+\frac{8\cdot {p}^{3}}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}|L\left(1,\chi \right){|}^{2}\\ \phantom{\rule{2em}{0ex}}+\frac{8\cdot {p}^{3}}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(4\right)\cdot |L\left(1,\chi \right){|}^{2}\\ \phantom{\rule{1em}{0ex}}=-\frac{5}{6}p\left(p-1\right)\left(p-5\right)+\frac{2}{3}p\left(p-1\right)\left(p-2\right)+\frac{1}{6}p\left({p}^{2}-6p+17\right)=2{p}^{2}.\end{array}$
(13)

If $p\equiv 1mod4$, then from (12), Lemma 4 and Lemma 5 we also have

$\begin{array}{c}\sum _{m=1}^{p-1}\sum _{n=1}^{p-1}K\left(m,p\right)\cdot K\left(n,p\right)\cdot {S}_{1}\left(2m\cdot \overline{n},p\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\frac{20\cdot {p}^{3}}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(2\right)\cdot |L\left(1,\chi \right){|}^{2}+\frac{8\cdot {p}^{3}}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}|L\left(1,\chi \right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{8\cdot {p}^{3}}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(4\right)\cdot |L\left(1,\chi \right){|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=-\frac{5}{6}p\left(p-1\right)\left(p-5\right)+\frac{2}{3}p\left(p-1\right)\left(p-2\right)+\frac{1}{6}p\left(p-1\right)\left(p-17\right)=0.\hfill \end{array}$
(14)

It is clear that Theorem 1 follows from (13) and (14).

Now we prove Theorem 2. If $p\equiv 1mod4$, then from Lemma 1 and the method of proving Theorem 1 we have

$\begin{array}{c}\sum _{m=1}^{p-1}\sum _{n=1}^{p-1}|K\left(m,p\right){|}^{2}\cdot |K\left(n,p\right){|}^{2}\cdot {S}_{1}\left(2m\cdot \overline{n},p\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\frac{20\cdot p}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(2\right)|\sum _{n=1}^{p-1}\chi \left(n\right)\cdot |K\left(n,p\right){|}^{2}{|}^{2}\cdot |L\left(1,\chi \right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{8\cdot p}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(4\right)|\sum _{n=1}^{p-1}\chi \left(n\right)\cdot |K\left(n,p\right){|}^{2}{|}^{2}\cdot |L\left(1,\chi \right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{8\cdot p}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}|\sum _{n=1}^{p-1}\chi \left(n\right)\cdot |K\left(n,p\right){|}^{2}{|}^{2}\cdot |L\left(1,\chi \right){|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=-\frac{20\cdot {p}^{4}}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(2\right)\cdot |L\left(1,\chi \right){|}^{2}+\frac{8\cdot {p}^{4}}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}|L\left(1,\chi \right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{8\cdot {p}^{4}}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(4\right)\cdot |L\left(1,\chi \right){|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=-\frac{5}{6}{p}^{2}\left(p-1\right)\left(p-5\right)+\frac{2}{3}{p}^{2}\left(p-1\right)\left(p-2\right)+\frac{1}{6}{p}^{2}\left(p-1\right)\left(p-17\right)=0.\hfill \end{array}$
(15)

If $p\equiv 3mod4$, then note that the Legendre symbol $\left(\frac{-1}{p}\right)={\chi }_{2}\left(-1\right)=-1$, $L\left(1,{\chi }_{2}\right)=\pi \cdot {h}_{p}/\sqrt{p}$, and

$\sum _{a=1}^{p-1}{\left(\frac{a}{p}\right)}^{2}e\left(\frac{a}{p}\right)=-1,$

so from Lemma 1 and the method of proving Theorem 1 we have

$\begin{array}{c}\sum _{m=1}^{p-1}\sum _{n=1}^{p-1}|K\left(m,p\right){|}^{2}\cdot |K\left(n,p\right){|}^{2}\cdot {S}_{1}\left(2m\cdot \overline{n},p\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\frac{20\cdot p}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(2\right)|\sum _{n=1}^{p-1}\chi \left(n\right)\cdot |K\left(n,p\right){|}^{2}{|}^{2}\cdot |L\left(1,\chi \right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{8\cdot p}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(4\right)|\sum _{n=1}^{p-1}\chi \left(n\right)\cdot |K\left(n,p\right){|}^{2}{|}^{2}\cdot |L\left(1,\chi \right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{8\cdot p}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}|\sum _{n=1}^{p-1}\chi \left(n\right)\cdot |K\left(n,p\right){|}^{2}{|}^{2}\cdot |L\left(1,\chi \right){|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=-\frac{20\cdot {p}^{4}}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(2\right)\cdot |L\left(1,\chi \right){|}^{2}+\frac{8\cdot {p}^{4}}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}|L\left(1,\chi \right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{8\cdot {p}^{4}}{{\pi }^{2}\left(p-1\right)}\underset{\chi \left(-1\right)=-1}{\sum _{\chi modp}}\chi \left(4\right)\cdot |L\left(1,\chi \right){|}^{2}+\frac{20\cdot {p}^{3}}{{\pi }^{2}}\cdot \left(\frac{2}{p}\right)\cdot |L\left(1,{\chi }_{2}\right){|}^{2}\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{8\cdot {p}^{3}}{{\pi }^{2}}\cdot \left(\frac{4}{p}\right)\cdot |L\left(1,{\chi }_{2}\right){|}^{2}-\frac{8\cdot {p}^{3}}{{\pi }^{2}}\cdot |L\left(1,{\chi }_{2}\right){|}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=-\frac{5}{6}{p}^{2}\left(p-1\right)\left(p-5\right)+\frac{2}{3}{p}^{2}\left(p-1\right)\left(p-2\right)+\frac{1}{6}{p}^{2}\left({p}^{2}-6p+17\right)\hfill \\ \phantom{\rule{2em}{0ex}}+20\cdot {p}^{2}\cdot {h}_{p}^{2}\cdot \left(\frac{2}{p}\right)-16\cdot {p}^{2}\cdot {h}_{p}^{2}\hfill \\ \phantom{\rule{1em}{0ex}}=2{p}^{3}+20\cdot \left(\frac{2}{p}\right)\cdot {p}^{2}\cdot {h}_{p}^{2}-16\cdot {p}^{2}\cdot {h}_{p}^{2}.\hfill \end{array}$
(16)

Note that $\left(\frac{2}{p}\right)={\left(-1\right)}^{\frac{{p}^{2}-1}{8}}=-1$ if $p\equiv 3mod8$, and $\left(\frac{2}{p}\right)=1$ if $p\equiv 7mod8$, then from (15) and (16) we may immediately deduce Theorem 2. This completes the proof of our theorems.

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## Acknowledgements

The authors express their gratitude to the referee for very helpful and detailed comments. This work is supported by the N.S.F. (11371291), the S.R.F.D.P. (20136101110014), the N.S.F. (2013JZ001) of Shaanxi Province, P.R. China.

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Correspondence to Han Zhang.

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### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

HZ obtained the theorems and completed the proof. WZ corrected and improved the final version. Both authors read and approved the final manuscript.

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### Cite this article

Zhang, H., Zhang, W. On the identity involving certain Hardy sums and Kloosterman sums. J Inequal Appl 2014, 52 (2014). https://doi.org/10.1186/1029-242X-2014-52

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### Keywords

• Gauss sums
• Kloosterman sums
• identity
• certain Hardy sums
• hybrid mean value
• computational formula