The Fourth Hybrid Power Mean Involving the Character Sums and Exponential Sums

Abstract

In this paper, we consider the fourth hybrid power mean involving two-term exponential sums and third-order character sum modulo p, a topic of significant importance in analytic number theory. These results generalize prior research, and provide new insights for studying the relationship between character sums and exponential sums.

1. Introduction

Let p be an odd prime, $\chi$ be a non-principal character modulo p, and $\lambda$ be a third-order character defined by $\lambda \left(n\right)=e\left({\displaystyle \frac{{\mathrm{ind}}_{g}n}{3}}\right)$, where g is a primitive root modulo p. For any positive integer k and integer m, we define the two-term exponential sums $H(m,k;p)$ and character sums $S(m,p)$ as follows:

$$H(m,k;p)=\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^k+a}{p}}\right)\mathrm{and}S(m,p)=\sum _{a=1}^{p-1}\chi \left(a+m\overline{a}\right),$$

where $e\left(y\right)=e^{2\pi iy}$, $i^2=-1$ and $a·\overline{a}\equiv 1modp$.

These sums play a very important role in the study of analytic number theory and related problems; many important number theory problems are closely related to them, such as the prime distribution and the Goldbach problem. Therefore, many number theorists and scholars have studied the various properties of $H(m,k;p)$ and $S(m,p)$, and obtained a series of meaningful research results. For example, H. Zhang and W. P. Zhang [1] proved the identity

$$\begin{array}{c}\hfill \sum _{m=1}^{p-1}{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+na}{p}}\right)\right|}^4=\left\{\begin{array}{cc}2p^3-p^2,\hfill & \mathrm{if}3\nmid p-1;\hfill \\ 2p^3-7p^2,\hfill & \mathrm{if}3\mid p-1,\hfill \end{array}\right.\end{array}$$

(1)

where p indicates an odd prime, and n is any integer with $(n,p)=1$.

T. T. Wang and W. P. Zhang [2], by the elementary and analytic methods, proved that

$$\begin{array}{c}\hfill \sum _{m=1}^{p-1}{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+na}{p}}\right)\right|}^8=\left\{\begin{array}{cc}14p^5-21p^4,\hfill & \mathrm{if}p\equiv 5mod6,\hfill \\ 14p^5-75p^4-8p^3{d}^2,\hfill & \mathrm{if}p\equiv 1mod6,\hfill \end{array}\right.\end{array}$$

(2)

where $4p=d^2+27b^2$, and d is uniquely determined by $d\equiv 1mod3$.

N. Bag et al. [3] showed that for any positive integer m, they obtained the asymptotic formula

$$\underset{\chi \ne {\chi }_0}{\sum _{\chi modp}}{\left|\sum _{a=1}^{p-1}\chi \left(a+\overline{a}\right)\right|}^{2m}=\left({\displaystyle \genfrac{}{}{0pt}{}{2m-1}{m}}\right)·p^{m+1}+O\left(p^{m+{\displaystyle \frac{1}{2}}}\right).$$

D. Han [4] studied hybrid power mean involving $H(m,k;p)$ and $S(m,p)$, and proved the following conclusion:

$$\begin{array}{c}\hfill \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\chi \left(ma+\overline{a}\right)\right|}^2·{\left|\sum _{a=1}^{p-1}e\left({\displaystyle \frac{m{a}^k+na}{p}}\right)\right|}^2=\left\{\begin{array}{cc}{\displaystyle 2p^3+O\left(\left|k\right|p^2\right),}\hfill & \mathrm{if}2\mid k;\hfill \\ {\displaystyle 2p^3+O\left(\left|k\right|p^{\frac{5}{2}}\right),}\hfill & \mathrm{if}2\nmid k,\hfill \end{array}\right.\end{array}$$

where $\chi$ denotes any non-principal even character modulo p.

X. X. Lv and W. P. Zhang [5] also studied a similar problem and proved the following two results.

If p is an odd prime with $3\nmid (p-1)$, then for any non-principal character $\chi modp$, the following identity can be obtained:

$$\begin{array}{c}\hfill \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\chi \left(a+b+m\overline{a}\overline{b}\right)\right|}^2·{\left|\sum _{b=1}^{p-1}e\left({\displaystyle \frac{m{b}^3+b}{p}}\right)\right|}^2=p^2·\left(p^2-p-1\right).\end{array}$$

If p is an odd prime with $3\mid (p-1)$, then for any third character $\chi modp$ (i.e., there exists a character ${\chi }_{1}modp$ such that $\chi ={\chi }_1^3$), the following asymptotic formula can be obtained:

$$\begin{array}{c}\hfill \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\sum _{b=1}^{p-1}\chi \left(a+b+m\overline{a}\overline{b}\right)\right|}^2·{\left|\sum _{b=1}^{p-1}e\left({\displaystyle \frac{m{b}^3+b}{p}}\right)\right|}^2=3p^4+E\left(p\right),\end{array}$$

where $E\left(p\right)$ satisfies the estimate $\left|E\left(p\right)\right|\le 15p^3$.

More relevant to this, in [6], J. M. Yu, R. J. Yuan, and T. T. Wang considered the computational problems of the fourth power mean value of one kind of two-term exponential sum through the classification and estimation of Dirichlet characters. In [7], X. Han and T. T. Wang applied the properties of character sums, quadratic character, and classical Gauss sums to study the calculations of the hybrid power mean of the generalized Gauss sums and the generalized two-term exponential sums.

Some related papers can also be found in references [8,9,10,11,12]. Inspired by the above results, in this paper we consider the following calculation problem of the fourth hybrid power mean:

$$\begin{array}{c}\hfill \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\chi \left(a+m\overline{a}\right)\right|}^{2k}·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4,\end{array}$$

(3)

where k is a positive integer.

The remainder of this paper is organized as follows: Section 2 introduces the main results. Section 3 presents several lemmas, and Section 4 provides the proofs of the theorems. Finally, Section 5 concludes the paper.

2. Main Results

This paper investigates the fourth hybrid power mean involving two-term exponential sums and third-order character sum modulo p, including $H(m,k;p)$ and $S(m,p)$. Our results extend prior work by incorporating arithmetic structures tied to $p\equiv 1mod3$, offering a unified framework for hybrid means. In this work, we obtain the result by deriving explicit identities under specific congruence conditions. Using analytic methods and properties of third-order characters, we prove two main theorems. A key innovation lies in linking these results to primitive root modulo p and establishing congruence conditions for $2modp$, which determine the structure of the mean values. These results not only generalize prior research but also provide new tools for studying the interplay between multiplicative characters and additive exponential sums, offering insights applicable to broader problems in number theory. The main results are as follows:

Theorem 1. 

Let p be an odd prime with $p\equiv 1mod3$. Then, for any third-order character $\lambda modp$ and any primitive root $gmodp$, we have the identity

$$\begin{array}{ccc}& & \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\lambda \left(a+m\overline{a}\right)\right|}^2·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ & =& \left\{\begin{array}{cc}{\displaystyle 4p^4-14p^3+p^2\left(d^2-2p\right)·(\delta -3),}\hfill & \mathrm{if}2\equiv g^{3h}modp;\hfill \\ {\displaystyle 4p^4-14p^3+p^2\left(p-\frac{d(d-9b)}{2}\right)·(\delta -3),}\hfill & \mathrm{if}2\equiv g^{3h+1}modp;\hfill \\ {\displaystyle 4p^4-14p^3+p^2\left(p-\frac{d(d+9b)}{2}\right)·(\delta -3),}\hfill & \mathrm{if}2\equiv g^{3h+2}modp,\hfill \end{array}\right.\hfill \end{array}$$

where d and b are the same as in (2), the sign of b is determined by the congruence $9b\equiv \left(2g^{{\displaystyle \frac{p-1}{3}}}+1\right)dmodp$, h is any integer with $0\le h\le {\displaystyle \frac{p-1}{3}}$, and ${\displaystyle \delta =\sum _{a=1}^{p-1}\left(\frac{a-1+\overline{a}}{p}\right)}$ is an integer.

Theorem 2. 

Let p be an odd prime with $p\equiv 1mod3$, and g be any primitive root modulo p. If $2\equiv g^{3h}modp$, then we have

$$\begin{array}{ccc}& & \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\lambda \left(a+m\overline{a}\right)\right|}^4·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ & =& \left(8p^2+d^4-4d^{2}p\right)(2p^3-7p^2)+4p^3(d^2-2p)(\delta -3).\hfill \end{array}$$

If $2\equiv g^{3h+1}(modp)$, then

$$\begin{array}{ccc}& & \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\lambda \left(a+m\overline{a}\right)\right|}^4·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ & =& \left(5p^2+2d^{2}p-9bdp-{\displaystyle \frac{d^3(d-9b)}{2}}\right)(2p^3-7p^2)+4p^3\left(p-{\displaystyle \frac{d(d-9b)}{2}}\right)(\delta -3).\hfill \end{array}$$

If $2\equiv g^{3h+2}modp$, then

$$\begin{array}{ccc}& & \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\lambda \left(a+m\overline{a}\right)\right|}^4·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ & =& \left(5p^2+2d^{2}p+9bdp-{\displaystyle \frac{d^3(d+9b)}{2}}\right)(2p^3-7p^2)+4p^3\left(p-{\displaystyle \frac{d(d-9b)}{2}}\right)(\delta -3),\hfill \end{array}$$

where the sign of b is determined by the congruence $9b\equiv \left(2g^{{\displaystyle \frac{p-1}{3}}}+1\right)dmodp$.

The proofs of these theorems will be presented in Section 4.

Note that $\left|d\right|\ll \sqrt{p}$, $\left|b\right|\ll \sqrt{p}$ and $\left|\delta \right|\ll \sqrt{p}$. From Theorem 1, we may immediately deduce the following:

Corollary 1. 

Let p be an odd prime with $p\equiv 1mod3$. Then, for any third-order character $\lambda modp$, we have the asymptotic formula

$$\begin{array}{c}\hfill \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\lambda \left(a+m\overline{a}\right)\right|}^2·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4=4p^4+O\left(p^{{\displaystyle \frac{7}{2}}}\right).\end{array}$$

3. Several Lemmas

To establish our main theorems, we need several fundamental lemmas. In the following, we shall use some knowledge of elementary number theory and analytic number theory, and the properties of the character sum modulo p, all of which can be found in references [13,14]. Thus, we omit their repetition here. We present the following lemmas:

Lemma 1. 

Let p be an odd prime. Then, for any non-principal character χ modulo p, we have the identity

$$\begin{array}{c}\hfill \tau \left({\chi }^2\right)={\displaystyle \frac{{\chi }^2\left(2\right)·\tau \left(\chi \right)·\tau \left(\chi {\chi }_2\right)}{\sqrt{p}}}.\end{array}$$

Proof. 

The proof follows directly from the Hasse–Davenport product formula for Gauss sums (see H. Davenport and H. Hasse [15]). □

Lemma 2. 

Let p be a prime with $p\equiv 1mod3$, and λ be a three-order character modulo p. Then, for any integer m with $(m,p)=1$, we have the identity

$$\begin{array}{c}\hfill \sum _{a=1}^{p-1}\lambda \left(a+m\overline{a}\right)=\overline{\lambda }\left(m\right)·{\displaystyle \frac{{\tau }^3\left(\lambda \right)}{p}}+\lambda \left(2\right)\overline{\lambda }\left(m\right){\chi }_2\left(m\right)·{\displaystyle \frac{{\tau }^3\left(\overline{\lambda }\right)}{p}}.\end{array}$$

Proof. 

Note that $\overline{\lambda }={\lambda }^2$ and $\tau \left(\lambda \right)·\tau \left(\overline{\lambda }\right)=p$. From the properties of the classical Gauss sums, we have

$$\begin{array}{cc}& \sum _{a=1}^{p-1}\lambda \left(a+m\overline{a}\right)={\displaystyle \frac{1}{\tau \left(\overline{\lambda }\right)}}\sum _{b=1}^{p-1}\overline{\lambda }\left(b\right)\sum _{a=1}^{p-1}e\left({\displaystyle \frac{b\left(a+m\overline{a}\right)}{p}}\right)\hfill \\ =\hfill & {\displaystyle \frac{1}{\tau \left(\overline{\lambda }\right)}}\sum _{b=1}^{p-1}\overline{\lambda }\left(b\right)\sum _{a=1}^{p-1}{\lambda }^2\left(a\right)e\left({\displaystyle \frac{b{a}^2+mb}{p}}\right)\hfill \\ =\hfill & {\displaystyle \frac{1}{\tau \left(\overline{\lambda }\right)}}\sum _{b=1}^{p-1}\overline{\lambda }\left(b\right)e\left({\displaystyle \frac{mb}{p}}\right)\sum _{a=1}^{p-1}\lambda \left(a\right)\left(1+{\chi }_2\left(a\right)\right)e\left({\displaystyle \frac{ba}{p}}\right)\hfill \\ =\hfill & {\displaystyle \frac{1}{\tau \left(\overline{\lambda }\right)}}\sum _{b=1}^{p-1}\overline{\lambda }\left(b\right)e\left({\displaystyle \frac{mb}{p}}\right)\left(\overline{\lambda }\left(b\right)\tau \left(\lambda \right)+\overline{\lambda }\left(b\right){\chi }_2\left(b\right)\tau \left(\lambda {\chi }_2\right)\right)\hfill \\ =\hfill & \overline{\lambda }\left(m\right)·{\displaystyle \frac{{\tau }^2\left(\lambda \right)}{\tau \left(\overline{\lambda }\right)}}+\overline{\lambda }\left(m\right){\chi }_2\left(m\right)·{\displaystyle \frac{{\tau }^2\left(\lambda {\chi }_2\right)}{\tau \left(\overline{\lambda }\right)}}\hfill \\ =\hfill & \overline{\lambda }\left(m\right)·{\displaystyle \frac{{\tau }^3\left(\lambda \right)}{p}}+\overline{\lambda }\left(m\right){\chi }_2\left(m\right)·{\displaystyle \frac{\tau \left(\lambda \right)·{\tau }^2\left(\lambda {\chi }_2\right)}{p}}.\hfill \end{array}$$

(4)

Letting $\chi =\lambda$ in Lemma 1, we have

$$\begin{array}{c}\hfill \tau \left(\lambda {\chi }_2\right)={\displaystyle \frac{\lambda \left(2\right)·{\tau }^2\left(\overline{\lambda }\right)}{\sqrt{p}}}.\end{array}$$

(5)

Combining equations (4) and (5), we find that the result shows:

$$\begin{array}{c}\hfill \sum _{a=1}^{p-1}\lambda \left(a+m\overline{a}\right)=\overline{\lambda }\left(m\right)·{\displaystyle \frac{{\tau }^3\left(\lambda \right)}{p}}+\lambda \left(2\right)\overline{\lambda }\left(m\right){\chi }_2\left(m\right)·{\displaystyle \frac{{\tau }^3\left(\overline{\lambda }\right)}{p}},\end{array}$$

which completes the proof. □

Lemma 3. 

Let p be a prime with $p\equiv 1mod3$, and λ be a three-order character modulo p. Then, we have the identity

$$\begin{array}{c}\hfill {\tau }^3\left(\lambda \right)+{\tau }^3\left(\overline{\lambda }\right)=dp,\end{array}$$

where λ is a three-order character modulo p, and d is uniquely determined by $4p=d^2+27b^2$ and $d\equiv 1mod3$.

Proof. 

See B. C. Berndt and R. J. Evans [16] or W. P. Zhang and J. Y. Hu [17]. □

Lemma 4. 

For any odd prime p, we have the identity

$$\begin{array}{c}\hfill \sum _{m=1}^{p-1}{\chi }_2\left(m\right){\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4=\left\{\begin{array}{cc}{\displaystyle p^2\left(\delta -3\right),}\hfill & \mathrm{if}p\equiv 1mod6;\hfill \\ {\displaystyle p^2\left(\delta +3\right),}\hfill & \mathrm{if}p\equiv -1mod6,\hfill \end{array}\right.\end{array}$$

where ${\displaystyle \delta =\sum _{a=1}^{p-1}\left(\frac{a-1+\overline{a}}{p}\right)}$ is an integer and $\left|\delta \right|\le 2\sqrt{p}$.

Proof. 

See J. Zhang and W. P. Zhang [18]. □

Lemma 5. 

Let p be a prime with $p\equiv 1mod3$ and g be any primitive root modulo p. We define $\lambda \left(n\right)=e\left({\displaystyle \frac{{ind}_{g}n}{3}}\right)$. It is clear that λ is a third-order character modulo p. Then, we have the identity

$$\begin{array}{c}\hfill {\tau }^3\left(\lambda \right)={\displaystyle \frac{p(d+3\sqrt{3}bi)}{2}},\end{array}$$

where $i^2=-1$, ${ind}_{g}n$ denotes the exponent of n under primitive root $gmodp$, i.e., $n\equiv g^{{ind}_{g}n}modp$; d and b are the same as defined in Lemma 3; and the sign of b is determined by congruence $9b\equiv \left(2g^{{\displaystyle \frac{p-1}{3}}}+1\right)dmodp$.

Proof. 

First assume that ${\chi }_1,{\chi }_2,\cdots ,{\chi }_r$ are nontrivial multiplicative character modulo p, and also that ${\chi }_1{\chi }_2\cdots {\chi }_r$ is nontrivial. Then, for Jacobi sums $J({\chi }_1,{\chi }_2,\cdots ,{\chi }_r)$, from Theorem 3 in [19], we have

$$\begin{array}{c}\hfill J({\chi }_1,{\chi }_2,\cdots ,{\chi }_r)={\displaystyle \frac{\tau \left({\chi }_1\right)\tau \left({\chi }_2\right)\cdots \tau \left({\chi }_r\right)}{\tau ({\chi }_1{\chi }_2\cdots {\chi }_r)}},\end{array}$$

(6)

where the Jacobi sum is defined by the formula

$$J({\chi }_1,{\chi }_2,\cdots ,{\chi }_r)=\underset{a_1+a_2+\cdots +a_r=1}{\sum _{a_1=1}^{p-1}\sum _{a_2=1}^{p-1}\cdots \sum _{a_r=1}^{p-1}}{\chi }_1\left(a_1\right){\chi }_2\left(a_2\right)\cdots {\chi }_r\left(a_r\right).$$

As shown in [20], the identity $J(\lambda ,\lambda )={\displaystyle \frac{d+3\sqrt{3}bi}{2}}$ holds.

However, from Equations (6) and (7), and noting that $\tau \left(\lambda \right)·\tau \left(\overline{\lambda }\right)=p$ and ${\lambda }^2=\overline{\lambda }$, we also have

$$\begin{array}{c}\hfill J(\lambda ,\lambda )={\displaystyle \frac{{\tau }^2\left(\lambda \right)}{\tau \left(\overline{\lambda }\right)}}={\displaystyle \frac{{\tau }^3\left(\lambda \right)}{p}}.\end{array}$$

(7)

By combining Equations (7) and (8), we can deduce that

$$\begin{array}{c}\hfill {\tau }^3\left(\lambda \right)={\displaystyle \frac{p(d+3\sqrt{3}bi)}{2}},\end{array}$$

which completes the proof. □

4. Proofs of Theorems

In this section, we provide the proof of our main results. First, we prove Theorem 1. Note that $\left|\tau \left(\lambda \right)\right|=\sqrt{p}$, $\tau \left(\lambda \right)·\tau \left(\overline{\lambda }\right)=p$, ${\lambda }^2=\overline{\lambda }$, and $p\equiv 1mod3$, for any three-order character $\lambda modp$. From Equation (1) and Lemmas 2 and 4, we have

$$\begin{array}{cc}& \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\lambda \left(a+m\overline{a}\right)\right|}^2·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ =\hfill & \sum _{m=1}^{p-1}\left({\left|\overline{\lambda }\left(m\right)·{\displaystyle \frac{{\tau }^3\left(\lambda \right)}{p}}+\lambda \left(2\right)\overline{\lambda }\left(m\right){\chi }_2\left(m\right)·{\displaystyle \frac{{\tau }^3\left(\overline{\lambda }\right)}{p}}\right|}^2\right)·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ =\hfill & \left(\lambda \left(2\right)·{\tau }^6\left(\overline{\lambda }\right)+\overline{\lambda }\left(2\right)·{\tau }^6\left(\lambda \right)\right)\sum _{m=1}^{p-1}{\chi }_2\left(m\right){\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ & +2p·\sum _{m=1}^{p-1}{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ =\hfill & 2p·\left(2p^3-7p^2\right)+\left(\lambda \left(2\right)·{\tau }^6\left(\overline{\lambda }\right)+\overline{\lambda }\left(2\right)·{\tau }^6\left(\lambda \right)\right)·(\delta -3).\hfill \end{array}$$

(8)

For any primitive root g modulo p, taking $\lambda \left(n\right)=e\left({\displaystyle \frac{{\mathrm{ind}}_{g}n}{3}}\right)$, then

$$\begin{array}{c}\hfill \lambda \left(2\right)=\left\{\begin{array}{cc}{\displaystyle 1,}\hfill & \mathrm{if}2\equiv g^{3h}modp;\hfill \\ {\displaystyle \frac{-1+\sqrt{3}i}{2},}\hfill & \mathrm{if}2\equiv g^{3h+1}modp;\hfill \\ {\displaystyle \frac{-1-\sqrt{3}i}{2},}\hfill & \mathrm{if}2\equiv g^{3h+2}modp.\hfill \end{array}\right.\end{array}$$

(9)

From Equation (10) and Lemma 5, we obtain:

$$\begin{array}{cc}& \lambda \left(2\right)·{\tau }^3\left(\overline{\lambda }\right)+\overline{\lambda }\left(2\right)·{\tau }^3\left(\lambda \right)=\lambda \left(2\right)·{\displaystyle \frac{p(d-3\sqrt{3}bi)}{2}}+\overline{\lambda }\left(2\right)·{\displaystyle \frac{p(d+3\sqrt{3}bi)}{2}}\hfill \\ =\hfill & \left(\lambda \left(2\right)+\overline{\lambda }\left(2\right)\right)·{\displaystyle \frac{dp}{2}}-\left(\lambda \left(2\right)-\overline{\lambda }\left(2\right)\right)·{\displaystyle \frac{3\sqrt{3}bpi}{2}}\hfill \\ =\hfill & \left\{\begin{array}{cc}{\displaystyle dp,}\hfill & \mathrm{if}2\equiv g^{3h}modp;\hfill \\ {\displaystyle -p·\frac{d-9b}{2},}\hfill & \mathrm{if}2\equiv g^{3h+1}modp;\hfill \\ {\displaystyle -p·\frac{d+9b}{2},}\hfill & \mathrm{if}2\equiv g^{3h+2}modp.\hfill \end{array}\right.\hfill \end{array}$$

(10)

From Equation (11) and Lemma 1, we obtain:

$$\begin{array}{cc}& dp·\left(\lambda \left(2\right)·{\tau }^3\left(\overline{\lambda }\right)+\overline{\lambda }\left(2\right)·{\tau }^3\left(\lambda \right)\right)=\left({\tau }^3\left(\overline{\lambda }\right)+{\tau }^3\left(\lambda \right)\right)\left(\lambda \left(2\right)·{\tau }^3\left(\overline{\lambda }\right)+\overline{\lambda }\left(2\right)·{\tau }^3\left(\lambda \right)\right)\hfill \\ =\hfill & \left(\lambda \left(2\right)+\overline{\lambda }\left(2\right)\right)·p^3+\lambda \left(2\right)·{\tau }^6\left(\overline{\lambda }\right)+\overline{\lambda }\left(2\right)·{\tau }^6\left(\lambda \right).\hfill \end{array}$$

(11)

Combining Equations (11) and (12), we obtain:

$$\begin{array}{cc}& \lambda \left(2\right)·{\tau }^6\left(\overline{\lambda }\right)+\overline{\lambda }\left(2\right)·{\tau }^6\left(\lambda \right)=dp·\left(\lambda \left(2\right)·{\tau }^3\left(\overline{\lambda }\right)+\overline{\lambda }\left(2\right)·{\tau }^3\left(\lambda \right)\right)-\left(\lambda \left(2\right)+\overline{\lambda }\left(2\right)\right)·p^3\hfill \\ =\hfill & \left\{\begin{array}{cc}{\displaystyle p^2·\left(d^2-2p\right),}\hfill & \mathrm{if}2\equiv g^{3h}modp;\hfill \\ {\displaystyle p^2·\left(p-\frac{d(d-9b)}{2}\right),}\hfill & \mathrm{if}2\equiv g^{3h+1}modp;\hfill \\ {\displaystyle p^2·\left(p-\frac{d(d+9b)}{2}\right),}\hfill & \mathrm{if}2\equiv g^{3h+2}modp.\hfill \end{array}\right.\hfill \end{array}$$

(12)

Combining Equations (9) and (13), we establish Theorem 1.

Similarly, from Lemma 1 we also have

$$\begin{array}{cc}& \left({\tau }^6\left(\overline{\lambda }\right)+{\tau }^6\left(\lambda \right)\right)·\left(\lambda \left(2\right)·{\tau }^6\left(\overline{\lambda }\right)+\overline{\lambda }\left(2\right)·{\tau }^6\left(\lambda \right)\right)\hfill \\ =\hfill & \lambda \left(2\right)·{\tau }^{12}\left(\overline{\lambda }\right)+\overline{\lambda }\left(2\right)·{\tau }^{12}\left(\lambda \right)+\left(\lambda \left(2\right)+\overline{\lambda }\left(2\right)\right)·p^6\hfill \\ =\hfill & \left(d^2{p}^2-2p^3\right)·\left(\lambda \left(2\right)·{\tau }^6\left(\overline{\lambda }\right)+\overline{\lambda }\left(2\right)·{\tau }^6\left(\lambda \right)\right),\hfill \end{array}$$

and

$$\begin{array}{cc}& \lambda \left(2\right)·{\tau }^{12}\left(\overline{\lambda }\right)+\overline{\lambda }\left(2\right)·{\tau }^{12}\left(\lambda \right)\hfill \\ =\hfill & \left\{\begin{array}{cc}{\displaystyle p^4·\left(d^4-4d^{2}p+2p^2\right),}\hfill & \mathrm{if}2\equiv g^{3h}modp;\hfill \\ {\displaystyle p^4·\left(-p^2+2d^{2}p-9bdp-\frac{d^3(d-9b)}{2}\right),}\hfill & \mathrm{if}2\equiv g^{3h+1}modp;\hfill \\ {\displaystyle p^4·\left(-p^2+2d^{2}p+9bdp-\frac{d^3(d+9b)}{2}\right),}\hfill & \mathrm{if}2\equiv g^{3h+2}modp.\hfill \end{array}\right.\hfill \end{array}$$

(13)

From Equation (1) and Lemmas 2 and 4, we derive:

$$\begin{array}{cc}& \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\lambda \left(a+m\overline{a}\right)\right|}^4·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ =\hfill & \sum _{m=1}^{p-1}{\left({\left|\overline{\lambda }\left(m\right)·{\displaystyle \frac{{\tau }^3\left(\lambda \right)}{p}}+\lambda \left(2\right)\overline{\lambda }\left(m\right){\chi }_2\left(m\right)·{\displaystyle \frac{{\tau }^3\left(\overline{\lambda }\right)}{p}}\right|}^2\right)}^2·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ =\hfill & {\displaystyle \frac{1}{p^4}}·\sum _{m=1}^{p-1}{\left(2p^3+{\chi }_2\left(m\right)\lambda \left(2\right)·{\tau }^6\left(\overline{\lambda }\right)+{\chi }_2\left(m\right)\overline{\lambda }\left(2\right)·{\tau }^6\left(\lambda \right)\right)}^2·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ =\hfill & \left(6p^2+{\displaystyle \frac{\overline{\lambda }\left(2\right){\tau }^{12}\left(\overline{\lambda }\right)+\lambda \left(2\right){\tau }^{12}\left(\lambda \right)}{p^4}}\right)\sum _{m=1}^{p-1}{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ & +{\displaystyle \frac{4}{p}}\left(\lambda \left(2\right)·{\tau }^6\left(\overline{\lambda }\right)+\overline{\lambda }\left(2\right)·{\tau }^6\left(\lambda \right)\right)\sum _{m=1}^{p-1}{\chi }_2\left(m\right){\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ =\hfill & \left(6p^2+{\displaystyle \frac{\overline{\lambda }\left(2\right){\tau }^{12}\left(\overline{\lambda }\right)+\lambda \left(2\right){\tau }^{12}\left(\lambda \right)}{p^4}}\right)·\left(2p^3-7p^2\right)\hfill \\ & +4p·\left(\lambda \left(2\right)·{\tau }^6\left(\overline{\lambda }\right)+\overline{\lambda }\left(2\right)·{\tau }^6\left(\lambda \right)\right)·(\delta -3).\hfill \end{array}$$

(14)

Combining Equations (13)–(15), we know that if $2\equiv g^{3h}modp$, then we have

$$\begin{array}{cc}& \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\lambda \left(a+m\overline{a}\right)\right|}^4·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ =\hfill & \left(8p^2+d^4-4d^{2}p\right)(2p^3-7p^2)+4p^3(d^2-2p)(\delta -3).\hfill \end{array}$$

(15)

If $2\equiv g^{3h+1}(modp)$, then

$$\begin{array}{cc}& \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\lambda \left(a+m\overline{a}\right)\right|}^4·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ =\hfill & \left(5p^2+2d^{2}p-9bdp-{\displaystyle \frac{d^3(d-9b)}{2}}\right)(2p^3-7p^2)+4p^3\left(p-{\displaystyle \frac{d(d-9b)}{2}}\right)(\delta -3).\hfill \end{array}$$

(16)

If $2\equiv g^{3h+2}(modp)$, then

$$\begin{array}{cc}& \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\lambda \left(a+m\overline{a}\right)\right|}^4·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ =\hfill & \left(5p^2+2d^{2}p+9bdp-{\displaystyle \frac{d^3(d+9b)}{2}}\right)(2p^3-7p^2)+4p^3\left(p-{\displaystyle \frac{d(d-9b)}{2}}\right)(\delta -3).\hfill \end{array}$$

(17)

Combining Equations (16) with (17) and (18), we immediately derive Theorem 2. This completes the proofs of all our results.

5. Conclusions

The main contribution of this paper is establishing a novel computational formula for a fourth-order hybrid power mean that incorporates character sums and two-term exponential sums. Specifically, we establish the following result:

Let p be an odd prime with $p\equiv 1mod3$, and let g be a primitive root modulo p. Then, for any third-order character $\lambda modp$, the following identity holds:

$$\begin{array}{cc}& \sum _{m=1}^{p-1}{\left|\sum _{a=1}^{p-1}\lambda \left(a+m\overline{a}\right)\right|}^2·{\left|\sum _{a=0}^{p-1}e\left({\displaystyle \frac{m{a}^3+a}{p}}\right)\right|}^4\hfill \\ =\hfill & \left\{\begin{array}{cc}{\displaystyle 4p^4-14p^3+p^2\left(d^2-2p\right)·(\delta -3),}\hfill & \mathrm{if}2\equiv g^{3h}modp;\hfill \\ {\displaystyle 4p^4-14p^3+p^2\left(p-\frac{d(d-9b)}{2}\right)·(\delta -3),}\hfill & \mathrm{if}2\equiv g^{3h+1}modp;\hfill \\ {\displaystyle 4p^4-14p^3+p^2\left(p-\frac{d(d+9b)}{2}\right)·(\delta -3),}\hfill & \mathrm{if}2\equiv g^{3h+2}modp,\hfill \end{array}\right.\hfill \end{array}$$

where d and b are the same as in (2), the sign of b is determined by the congruence $9b\equiv \left(2g^{{\displaystyle \frac{p-1}{3}}}+1\right)dmodp$, h is any integer with $0\le h\le {\displaystyle \frac{p-1}{3}}$, and ${\displaystyle \delta =\sum _{a=1}^{p-1}\left(\frac{a-1+\overline{a}}{p}\right)}$ is an integer.

The proof methods proposed in this paper are not only novel but also provide a valuable framework for further research-related problems.