The hybrid power mean of the generalized Gauss sums

Abstract

The main purpose of this paper is using the analytic method and the orthogonal properties of the character sums to study the computational problem of one kind hybrid power mean of the generalized Gauss sums $\operatorname{mod}\ p$, an odd prime with $p\equiv 1\operatorname{mod}\ 3$, and give some exact computational formulas for this kind hybrid power mean.

1 Introduction

Let $q \ge 3$ be a positive integer, χ be any Dirichlet character $\operatorname{mod}\ q$. For any positive integer $k\geq 1$ and integer m, the generalized Gauss sum $G(m, k,\chi; q)$ is defined as

$$G(m, k, \chi; q) = \sum_{a=1}^{q} \chi (a) e \biggl( \frac{ma^{k} }{q} \biggr) $$

with $e(y) = e^{2\pi i y}$.

If $k=m=1$, then $G(m, k, \chi; q)$ becomes the classical Gauss sums $\tau (\chi)$; see [1] for its definition and some basic properties.

About the various arithmetical properties of $G(m, k, \chi; q)$, many authors had studied it, and obtained a series of interesting results; see [2–9]. For example, from Weil’s classical result [2] one can obtain the upper bound estimate

$$\bigl\vert G(m, k, \chi; p)\bigr\vert \leq (k+1)\sqrt{p}. $$

Zhang Wenpeng and Liu Huaning [3] studied the fourth power mean of the generalized third Gauss sums with $3\vert (p-1)$, and they obtained a complex but exact computational formula. That is, they proved that

$$\sum_{\chi \operatorname{mod}\ p}\Biggl\vert \sum _{a=1}^{p-1}\chi (a) e \biggl( \frac{a^{3}}{p} \biggr) \Biggr\vert ^{4}=5p^{3}-18p^{2}+20p+1+ \frac{U^{5}}{p}+5pU-5U^{3}-4U^{2}+4U, $$

where $U=\sum_{a=1}^{p}e ( \frac{a^{3}}{p} )$ is a real constant.

In this paper, we are concerned with the computational problem of the following hybrid power mean:

$$\begin{aligned} \sum_{m=1}^{p-1}\Biggl\vert \sum _{a=1}^{p-1}\psi (a) e \biggl( \frac{ma^{2}}{p} \biggr) \Biggr\vert ^{2k}\cdot \Biggl\vert \sum _{b=1}^{p-1} \biggl( \frac{b}{p} \biggr) e \biggl( \frac{mb ^{3}}{p} \biggr) \Biggr\vert ^{2h} , \end{aligned}$$

(1)

where k and h are positive integers, ψ is any three-order character $\operatorname{mod}\ p$ and $( \frac{*}{p} )$ denotes the Legendre symbol $\operatorname{mod}\ p$.

About the hybrid power mean (1), it seems that no one had studied it, at least we have not seen such a related paper before. The characteristic of this hybrid power mean is symmetrical. That is, the two-order character corresponds to the third power, and the three-order character corresponds to the second power. And the interesting thing is that, for any positive integers k and h, one can give an exact computational formula for (1). The main purpose of this paper is to illustrate this point. That is, we shall use the analytic method and the orthogonal properties of the character sums to prove the following main results.

Theorem 1

Let p be an odd prime with $p\equiv 1\operatorname{mod}\ 3$, k be any positive integer. For any three-order character $\psi \operatorname{mod}\ p$, if $p\equiv 1\operatorname{mod}\ 12$, then we have

$$\begin{aligned}& \sum_{m=1}^{p-1}\Biggl\vert \sum _{a=1}^{p-1}\psi (a)e \biggl( \frac{ma^{2}}{p} \biggr) \Biggr\vert ^{2k}\cdot \Biggl\vert \sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma ^{3}}{p} \biggr) \Biggr\vert ^{2} \\& \quad =3p(p-1)\sum_{i=0}^{ [ \frac{k}{2} ] } \binom{k}{2i}2^{k-2i}p ^{k-i} \Biggl( \sum_{a=0}^{p-1} \biggl( \frac{a^{3}-1}{p} \biggr) \Biggr) ^{2i}. \end{aligned}$$

If $p\equiv 7\operatorname{mod}\ 12$, then we have

$$\begin{aligned}& \sum_{m=1}^{p-1}\Biggl\vert \sum _{a=1}^{p-1}\psi (a)e \biggl( \frac{ma^{2}}{p} \biggr) \Biggr\vert ^{2k}\cdot \Biggl\vert \sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma ^{3}}{p} \biggr) \Biggr\vert ^{2} \\& \quad =3p(p-1)\sum_{i=0}^{ [ \frac{k}{2} ] }\binom{k}{2i}2^{k-2i}p ^{k-i}\Biggl\vert \sum _{a=1}^{p-1}\psi (a) \biggl( \frac{a^{2}-1}{p} \biggr) \Biggr\vert ^{2i}. \end{aligned}$$

Theorem 2

Let p be an odd prime with $p\equiv 1\operatorname{mod}\ 3$. For any positive integer k, if $p\equiv 1\operatorname{mod}\ 12$, then we have the identity

$$\begin{aligned}& \sum_{m=1}^{p-1}\Biggl\vert \sum _{a=1}^{p-1}\psi (a)e \biggl( \frac{ma^{2}}{p} \biggr) \Biggr\vert ^{2k}\cdot \Biggl\vert \sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma ^{3}}{p} \biggr) \Biggr\vert ^{4} \\& \quad =p(p-1) \Biggl( 9p+2\Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \psi \bigl( a^{3}-1 \bigr) \Biggr\vert ^{2} \Biggr) \\& \qquad {}\times \sum_{i=0}^{ [ \frac{k}{2} ] } \binom{k}{2i}2^{k-2i}p^{k-i} \Biggl( \sum _{a=0}^{p-1} \biggl( \frac{a^{3}-1}{p} \biggr) \Biggr) ^{2i}. \end{aligned}$$

If $p\equiv 7\operatorname{mod}\ 12$, then we have

$$\begin{aligned}& \sum_{m=1}^{p-1}\Biggl\vert \sum _{a=1}^{p-1}\psi (a)e \biggl( \frac{ma^{2}}{p} \biggr) \Biggr\vert ^{2k}\cdot \Biggl\vert \sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma ^{3}}{p} \biggr) \Biggr\vert ^{4} \\& \quad =p(p-1) \Biggl( 9p+2\Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \psi \bigl( a^{3}-1 \bigr) \Biggr\vert ^{2} \Biggr) \sum_{i=0}^{ [ \frac{k}{2} ] } \binom{k}{2i}2^{k-2i}p^{k-i}\Biggl\vert \sum _{a=1}^{p-1}\psi (a) \biggl( \frac{a ^{2}-1}{p} \biggr) \Biggr\vert ^{2i}. \end{aligned}$$

From Theorem 1 and Theorem 2 we may immediately deduce the following corollaries.

Corollary 1

For any odd prime p with $p\equiv 1\operatorname{mod}\ 3$, we have the identity

$$\sum_{m=1}^{p-1}\Biggl\vert \sum _{a=1}^{p-1}\psi (a)e \biggl( \frac{ma^{2}}{p} \biggr) \Biggr\vert ^{2}\cdot \Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma ^{3}}{p} \biggr) \Biggr\vert ^{2} =6p^{2}(p-1). $$

Corollary 2

Let p be an odd prime with $p\equiv 1\operatorname{mod}\ 3$, then we have

$$\begin{aligned}& \sum_{m=1}^{p-1}\Biggl\vert \sum _{a=1}^{p-1}\psi (a)e \biggl( \frac{ma^{2}}{p} \biggr) \Biggr\vert ^{2}\cdot \Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma ^{3}}{p} \biggr) \Biggr\vert ^{4} \\& \quad =2p^{2}(p-1) \Biggl( 9p+2\Biggl\vert \sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \psi \bigl( a^{3}-1 \bigr) \Biggr\vert ^{2} \Biggr). \end{aligned}$$

Corollary 3

Let p be an odd prime with $p\equiv 1\operatorname{mod}\ 3$, then we have

$$\begin{aligned}& \sum_{m=1}^{p-1}\Biggl\vert \sum _{a=1}^{p-1}\psi (a)e \biggl( \frac{ma^{2}}{p} \biggr) \Biggr\vert ^{4}\cdot \Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma ^{3}}{p} \biggr) \Biggr\vert ^{4} \\& \quad =p^{2}(p-1) \Biggl( 9p+2\Biggl\vert \sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \psi \bigl( a^{3}-1 \bigr) \Biggr\vert ^{2} \Biggr) \bigl( 4p + \bigl\vert C(p)\bigr\vert ^{2} \bigr), \end{aligned}$$

where $C(p)=\sum_{a=0}^{p-1} ( \frac{a^{3}-1}{p} )$ or $\sum_{a=1}^{p-1}\psi (a) ( \frac{a^{2}-1}{p} )$ according to $p\equiv 1$ or $7\operatorname{mod}\ 12$.

Corollary 4

Let p be an odd prime with $p\equiv 1\operatorname{mod}\ 3$, then we have

$$\begin{aligned} \sum_{m=1}^{p-1}\Biggl\vert \sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma ^{3}}{p} \biggr) \Biggr\vert ^{4} =9p^{2}(p-1)+ 2p(p-1)\Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \psi \bigl( a^{3}-1 \bigr) \Biggr\vert ^{2}. \end{aligned}$$

In fact for any positive integers k and h, by using our method we can give an exact computational formula for (1). But when k and h are larger, the formula becomes more complex, therefore it is not listed here.

It is very regrettable that we cannot get the exact value for $\vert \sum_{a=1}^{p-1} ( \frac{a}{p} ) \psi ( a^{3}-1) \vert$, $\vert \sum_{a=1}^{p-1}\psi (a) ( \frac{a^{2}-1}{p} ) \vert$ with $p\equiv 7\operatorname{mod}\ 12$ and $\sum_{a=0}^{p-1} ( \frac{a^{3}-1}{p} )$ with $p\equiv 1\operatorname{mod}\ 12$, which makes our theorems and corollaries look less beautiful.

Whether there exists exact values for these sums are three interesting open problems.

2 Some lemmas

In this section, we will give several simple lemmas, which are necessary in the proofs of our theorems. Hereinafter, we shall use many elementary number theory knowledge and the properties of the classical Gauss sums and Dirichlet characters, all of them can be found in reference [1], so we do not repeat them here. First we have the following.

Lemma 1

Let p be an odd prime with $p\equiv 1\operatorname{mod}\ 3$, ψ be any three-order Dirichlet character $\operatorname{mod}\ p$. Then, for any integer m with $(m, p)=1$, we have the identity

$$\Biggl\vert \sum_{a=1}^{p-1}\psi (a)e \biggl( \frac{ma^{2}}{p} \biggr) \Biggr\vert ^{2}=\textstyle\begin{cases} 2p+ ( \frac{m}{p} ) \sqrt{p}\sum_{a=0}^{p-1} ( \frac{a ^{3}-1}{p} ) &\textit{if }p\equiv 1\operatorname{mod}\ 12; \\ 2p + ( \frac{m}{p} ) \tau (\chi_{2})\sum_{a=1}^{p-1}\psi (a) ( \frac{a^{2}-1}{p} ) &\textit{if }p\equiv 7\operatorname{mod}\ 12. \end{cases} $$

Proof

For any integer m with $(m,p)=1$, note that the trigonometric identity

$$\begin{aligned} \sum_{n=0}^{p-1}e \biggl( \frac{mn^{2}}{p} \biggr) = \biggl( \frac{m}{p} \biggr) \cdot \tau ( \chi_{2} ), \end{aligned}$$

(2)

where $\chi_{2}= ( \frac{*}{p} )$ denotes the Legendre symbol $\operatorname{mod}\ p$.

From (2) and the properties of the reduced residue system $\operatorname{mod}\ p$ we have

$$\begin{aligned}& \Biggl\vert \sum_{a=1}^{p-1}\psi (a) e \biggl( \frac{ma^{2}}{p} \biggr) \Biggr\vert ^{2} \\& \quad = \sum _{a=1}^{p-1}\sum_{b=1}^{p-1} \psi (a)\overline{\psi }(b)e \biggl( \frac{m ( a^{2}-b^{2} ) }{p} \biggr) \\& \quad =\sum_{a=1}^{p-1}\psi (a)\sum _{b=1}^{p-1}e \biggl( \frac{mb^{2} ( a ^{2}-1) }{p} \biggr) \\& \quad =2(p-1)+\mathop{\sum_{a=1}^{p-1}}_{a\neq \pm 1} \psi (a)\sum_{b=1}^{p-1}e \biggl( \frac{mb^{2} ( a^{2}-1) }{p} \biggr) \\& \quad =2p + \mathop{\sum_{a=1}^{p-1}}_{a\neq \pm 1} \psi (a)\sum_{b=0} ^{p-1}e \biggl( \frac{mb^{2} ( a^{2}-1) }{p} \biggr) - \sum_{a=1} ^{p-1} \psi (a) \\& \quad = 2p + \biggl( \frac{m}{p} \biggr) \tau (\chi_{2})\sum _{a=1}^{p-1} \psi (a) \biggl( \frac{a^{2}-1}{p} \biggr). \end{aligned}$$

(3)

If $p\equiv 1\operatorname{mod}\ 12$, then note that $( \frac{-1}{p} ) =1$, $( \frac{a}{p} ) = ( \frac{\overline{a}}{p} )$, $1+\psi (a)+\overline{\psi }(a)=3$ or 0 according to $a\equiv b^{3} \operatorname{mod}\ p$ or $p\nmid ( a-b^{3} )$ for some $1\leq b\leq p-1$, we have

$$\begin{aligned}& \sum_{a=1}^{p-1}\psi (a) \biggl( \frac{a^{2}-1}{p} \biggr) \\& \quad =\sum_{a=1} ^{p-1} \psi (\overline{a}) \biggl( \frac{\overline{a}^{2}-1}{p} \biggr) \\& \quad = \sum _{a=1}^{p-1}\overline{\psi }(a) \biggl( \frac{1-a^{2}}{p} \biggr) \\& \quad =\sum_{a=1}^{p-1}\psi (\overline{a}) \biggl( \frac{\overline{a}^{2}-1}{p} \biggr) \\& \quad = \sum_{a=1}^{p-1} \overline{\psi }(a) \biggl( \frac{a^{2}-1}{p} \biggr) \\& \quad =\frac{1}{2}\sum_{a=1}^{p-1} \bigl( 1+\psi (a)+\psi (\overline{a}) \bigr) \biggl( \frac{a^{2}-1}{p} \biggr) - \frac{1}{2}\sum_{a=1}^{p-1} \biggl( \frac{a ^{2}-1}{p} \biggr) \\& \quad =\frac{1}{2}\sum_{a=1}^{p-1} \biggl( \frac{a^{6}-1}{p} \biggr) - \frac{1}{2}\sum_{a=1}^{p-1} \biggl( \frac{a-1}{p} \biggr) -\frac{1}{2} \sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \biggl( \frac{a-1}{p} \biggr) \\& \quad =\frac{1}{2}\sum_{a=1}^{p-1} \biggl( \frac{a^{3}-1}{p} \biggr) + \frac{1}{2}\sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \biggl( \frac{a^{3}-1}{p} \biggr) - \frac{1}{2}\sum_{a=1}^{p-1} \biggl( \frac{a-1}{p} \biggr) -\frac{1}{2} \sum_{a=1}^{p-1} \biggl( \frac{1-\overline{a}}{p} \biggr) \\& \quad =\frac{1}{2}\sum_{a=1}^{p-1} \biggl( \frac{a^{3}-1}{p} \biggr) + \frac{1}{2}\sum_{a=1}^{p-1} \biggl( \frac{1-\overline{a}^{3}}{p} \biggr) +1= \sum_{a=0}^{p-1} \biggl( \frac{a^{3}-1}{p} \biggr). \end{aligned}$$

(4)

Note that $\tau (\chi_{2})=\sqrt{p}$, if $p\equiv 1 \operatorname{mod}\ 4$. From (3) and (4) we may immediately deduce Lemma 1. □

Lemma 2

Let p be an odd prime with $p\equiv 1\operatorname{mod}\ 3$. Then we have the identities

$$\begin{aligned}& \sum_{m=1}^{p-1}\Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma^{3}}{p} \biggr) \Biggr\vert ^{2}=3p(p-1); \end{aligned}$$

(A)

$$\begin{aligned}& \sum_{m=1}^{p-1} \biggl( \frac{m}{p} \biggr) \Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma^{3}}{p} \biggr) \Biggr\vert ^{2}=0. \end{aligned}$$

(B)

Proof

First note the trigonometric identity

$$\begin{aligned} \sum_{m=1}^{q}e \biggl( \frac{nm}{q} \biggr) =\textstyle\begin{cases} q &\mbox{if }q\mid n, \\ 0 &\mbox{if }q\nmid n. \end{cases}\displaystyle \end{aligned}$$

(5)

Since $p\equiv 1\operatorname{mod}\ 3$, the congruence equation $x^{3}\equiv 1 \operatorname{mod}\ p$ has three solutions. Let these three solutions be $x=1$, g and $g^{2}$, respectively. Note that $( \frac{g}{p} ) = ( \frac{g}{p} ) ^{3}= ( \frac{g^{3}}{p} ) =1$, so from (5) we have

$$\begin{aligned}& \sum_{m=1}^{p-1}\Biggl\vert \sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma^{3}}{p} \biggr) \Biggr\vert ^{2} \\& \quad = \sum _{m=1}^{p-1}\sum_{a=1} ^{p-1}\sum_{b=1}^{p-1} \biggl( \frac{ab}{p} \biggr) e \biggl( \frac{m ( a ^{3}-b^{3} ) }{p} \biggr) \\& \quad =\sum_{m=1}^{p-1}\sum _{a=1}^{p-1}\sum_{b=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{mb^{3} ( a^{3}-1) }{p} \biggr) \\& \quad = \sum _{b=1}^{p-1} \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \sum_{m=0}^{p-1}e \biggl( \frac{m ( a^{3}-1) }{p} \biggr) \\& \quad =(p-1) \biggl( \biggl( \frac{1}{p} \biggr) + \biggl( \frac{g}{p} \biggr) + \biggl( \frac{g^{2}}{p} \biggr) \biggr) p+ \mathop{\sum _{a=1}^{p-1}}_{p \nmid (a^{3}-1)} \biggl( \frac{a}{p} \biggr) \sum_{m=0}^{p-1}e \biggl( \frac{m ( a^{3}-1) }{p} \biggr) \\& \quad =3p(p-1). \end{aligned}$$

This proves equation (A) of Lemma 2.

On the other hand, from the properties of Gauss sums we have

$$\begin{aligned}& \sum_{m=1}^{p-1} \biggl( \frac{m}{p} \biggr) \Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma^{3}}{p} \biggr) \Biggr\vert ^{2} \\& \quad = \sum_{m=1}^{p-1} \biggl( \frac{m}{p} \biggr) \sum_{a=1}^{p-1}\sum _{b=1}^{p-1} \biggl( \frac{ab}{p} \biggr) e \biggl( \frac{m ( a^{3}-b^{3} ) }{p} \biggr) \\& \quad =\sum_{a=1}^{p-1}\sum _{b=1}^{p-1} \biggl( \frac{a}{p} \biggr) \sum _{m=1} ^{p-1} \biggl( \frac{m}{p} \biggr) e \biggl( \frac{mb^{3} ( a^{3}-1) }{p} \biggr) \\& \quad = \sum_{b=1}^{p-1} \biggl( \frac{b^{3}}{p} \biggr) \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \sum_{m=1}^{p-1} \biggl( \frac{m}{p} \biggr) e \biggl( \frac{m ( a^{3}-1) }{p} \biggr) \\& \quad = \Biggl( \sum_{b=1}^{p-1} \biggl( \frac{b}{p} \biggr) \Biggr) \sum_{a=1} ^{p-1} \biggl( \frac{a}{p} \biggr) \sum _{m=1}^{p-1} \biggl( \frac{m}{p} \biggr) e \biggl( \frac{m ( a^{3}-1) }{p} \biggr) =0. \end{aligned}$$

This proves equation (B) of Lemma 2. □

Lemma 3

Let p be an odd prime with $p\equiv 1\operatorname{mod}\ 3$. Then, for any integer m with $(m,p)=1$, we have the identity

$$\begin{aligned} \Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma^{3}}{p} \biggr) \Biggr\vert ^{2} =&3p+ \overline{\psi }(m)\tau (\psi)\sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \overline{ \psi } \bigl( a^{3}-1 \bigr) \\ &{}+ \psi (m)\tau ( \overline{\psi } ) \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \psi \bigl( a^{3}-1 \bigr). \end{aligned}$$

Proof

From the properties of three-order character $\operatorname{mod}\ p$ and Gauss sums we have

$$\begin{aligned}& \Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma^{3}}{p} \biggr) \Biggr\vert ^{2} \\& \quad =\sum_{a=1}^{p-1}\sum _{b=1}^{p-1} \biggl( \frac{ab}{p} \biggr) e \biggl( \frac{m ( a^{3}-b^{3} ) }{p} \biggr) \\& \quad =\sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \sum_{b=0}^{p-1}e \biggl( \frac{mb ^{3} ( a^{3}-1) }{p} \biggr) -\sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \\& \quad =3p+\mathop{\sum_{a=1}^{p-1}}_{p\nmid (a^{3}-1)} \biggl( \frac{a}{p} \biggr) \Biggl( 1+\sum_{b=1}^{p-1} \bigl( 1+\psi (b)+\overline{\psi }(b) \bigr) e \biggl( \frac{mb ( a^{3}-1) }{p} \biggr) \Biggr) \\& \quad =3p+ \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \bigl( \overline{\psi } \bigl( m\bigl(a^{3}-1 \bigr) \bigr) \tau (\psi)+ \psi \bigl( m\bigl(a^{3}-1\bigr) \bigr) \tau ( \overline{\psi } ) \bigr) \\& \quad =3p+ \overline{\psi }(m)\tau (\psi)\sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \overline{ \psi } \bigl( a^{3}-1 \bigr) + \psi (m)\tau ( \overline{\psi } ) \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \psi \bigl( a^{3}-1 \bigr). \end{aligned}$$

This proves Lemma 3. □

3 Proofs of the theorems

Now we will complete the proofs of our theorems. First we prove Theorem 1. Let

$$A(p)=\sqrt{p}\sum_{a=0}^{p-1} \biggl( \frac{a^{3}-1}{p} \biggr)\quad \text{or}\quad \tau (\chi_{2})\sum _{a=1}^{p-1}\psi (a) \biggl( \frac{a ^{2}-1}{p} \biggr). $$

Then, for any integer $k\geq 1$, from Lemma 1 and the binomial theorem we have

$$\begin{aligned} \Biggl\vert \sum_{a=1}^{p-1}\psi (a)e \biggl( \frac{ma^{2}}{p} \biggr) \Biggr\vert ^{2k} =& \biggl( 2p+ \biggl( \frac{m}{p} \biggr) A(p) \biggr) ^{k} \\ =&\sum _{i=0} ^{k}\binom{k}{i}(2p)^{k-i}A^{i}(p) \biggl( \frac{m}{p} \biggr) ^{i}. \end{aligned}$$

(6)

If i is an odd number, then from (B) of Lemma 2 we know that

$$\begin{aligned} \sum_{m=1}^{p-1} \biggl( \frac{m}{p} \biggr) ^{i}\Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma^{3}}{p} \biggr) \Biggr\vert ^{2}= \sum_{m=1}^{p-1} \biggl( \frac{m}{p} \biggr) \Biggl\vert \sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma^{3}}{p} \biggr) \Biggr\vert ^{2}=0. \end{aligned}$$

(7)

Now combining (6), (7) and (A) of Lemma 2 we have

$$\begin{aligned}& \sum_{m=1}^{p-1}\Biggl\vert \sum _{a=1}^{p-1}\psi (a)e \biggl( \frac{ma^{2}}{p} \biggr) \Biggr\vert ^{2k}\cdot \Biggl\vert \sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma ^{3}}{p} \biggr) \Biggr\vert ^{2} \\& \quad =\sum_{i=0}^{k} \binom{k}{i}(2p)^{k-i}A^{i}(p)\sum_{m=1}^{p-1} \biggl( \frac{m}{p} \biggr) ^{i}\Biggl\vert \sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma^{3}}{p} \biggr) \Biggr\vert ^{2} \\& \quad =\sum_{i=0}^{ [ \frac{k}{2} ] } \binom{k}{2i}(2p)^{k-2i}A ^{2i}(p)\sum_{m=1}^{p-1} \Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma^{3}}{p} \biggr) \Biggr\vert ^{2} \\& \quad =3p(p-1)\sum_{i=0}^{ [ \frac{k}{2} ] } \binom{k}{2i}(2p)^{k-2i}A ^{2i}(p). \end{aligned}$$

(8)

Note that if $p\equiv 7\operatorname{mod}\ 12$, then $\tau^{2}(\chi_{2})=-p$ and $A(p)$ is a pure imaginary number, so from (8) and the definition of $A(p)$ we may immediately deduce Theorem 1.

Now we prove Theorem 2. From Lemma 3 we have

$$\begin{aligned}& \Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma^{3}}{p} \biggr) \Biggr\vert ^{4} \\& \quad =9p^{2}+6p \Biggl( \overline{\psi }(m)\tau (\psi)\sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \overline{\psi } \bigl( a^{3}-1 \bigr) + \psi (m) \tau ( \overline{\psi } ) \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \psi \bigl( a^{3}-1 \bigr) \Biggr) \\& \qquad {}+\psi (m)\tau^{2}(\psi) \Biggl( \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \overline{ \psi } \bigl( a^{3}-1 \bigr) \Biggr) ^{2}+\overline{\psi }(m)\tau^{2} ( \overline{ \psi } ) \Biggl( \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \psi \bigl( a^{3}-1 \bigr) \Biggr) ^{2} \\& \qquad {}+2p\Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \overline{\psi } \bigl( a^{3}-1 \bigr) \Biggr\vert ^{2}. \end{aligned}$$

(9)

Note that, for any integer i, we have

$$\sum_{m=1}^{p-1} \biggl( \frac{m}{p} \biggr) ^{i}\psi (m)=\sum_{m=1}^{p-1} \biggl( \frac{m}{p} \biggr) ^{i}\overline{\psi }(m)=0. $$

So from (6), (7), (8) and (A) of Lemma 2 we have

$$\begin{aligned}& \sum_{m=1}^{p-1}\Biggl\vert \sum _{a=1}^{p-1}\psi (a)e \biggl( \frac{ma^{2}}{p} \biggr) \Biggr\vert ^{2k}\cdot \Biggl\vert \sum _{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) e \biggl( \frac{ma ^{3}}{p} \biggr) \Biggr\vert ^{4} \\& \quad = \Biggl( 9p^{2}+2p\Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \overline{ \psi } \bigl( a^{3}-1 \bigr) \Biggr\vert ^{2} \Biggr) \sum_{i=0}^{ [ \frac{k}{2} ] } \binom{k}{2i}(2p)^{k-2i}A^{2i}(p)\sum _{m=1}^{p-1}1 \\& \quad =p(p-1) \Biggl( 9p+2\Biggl\vert \sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr) \psi \bigl( a^{3}-1 \bigr) \Biggr\vert ^{2} \Biggr) \sum_{i=0}^{ [ \frac{k}{2} ] } \binom{k}{2i}(2p)^{k-2i}A^{2i}(p). \end{aligned}$$

(10)

Now Theorem 2 follows from (10) and the definition of $A(p)$. This completes the proofs of all our results.