On the $k$-th power mean of one kind generalized cubic Gauss sums

1.   Introduction

Let $p$ be an odd prime, $\chi$ denotes a Dirichlet character modulo $p$. For any integers $k > h\geq 1$ and integers $m$ and $n$, the generalized two-term exponential sums $S(m, n, k, h, \chi; p)$ is defined as follows:

where $e(y) = e^{2\pi i y}$ and $i$ is the imaginary unit, i.e., $i^2 = -1$.

This sum plays a very important role in the study of analytic number theory and additive number theory, many important problems in number theory are closely related to it, such as prime distribution and Waring's problems. And because of that, many number theorists and scholars had studied the various properties of $S(m, n, k, h, \chi; p)$, and obtained a series of meaningful research results. For example, R. Duan and W. P. Zhang ^[1] proved that for any prime $p$ with $3\nmid (p-1)$, and any Dirichlet character $\lambda \bmod p$, one has the identity

where $\left(\frac{*}{p}\right)$ denotes the Legendre symbol, $\chi_0$ is the principal character modulo $p$.

L. Chen and X. Wang ^[2] studied the fourth power mean of $S(m, 1, 4, 1, \chi_0; p)$, and proved the identities

where $\alpha = \alpha(p) = \sum\limits_{a = 1}^{\frac{p-1}{2}}\left(\frac{a+\overline{a}}{p}\right)$ is an integer, $a\cdot \overline{a}\equiv 1\bmod p$, $\alpha$ satisfies the following identity (see Theorems 4–11 in ^[3]):

and $r$ is any quadratic non-residue modulo $p$. That is, $\left(\frac{r}{p}\right) = -1$.

Recently, W. P. Zhang and Y. Y. Meng ^[4] studied the sixth power mean of $S(m, n, 3, 1, \chi_0; p)$, and proved that for any odd prime $p$ and integer $n$ with $(n, p) = 1$, we have the identities

where $4p = d^2+27\cdot b^2$, $b$ is an integer and $d$ is uniquely determined by $d\equiv 1\bmod 3$.

X. Y. Liu and W. P. Zhang ^[5] proved the following conclusion: For any odd prime $p$ with $3\nmid(p-1)$, one has the identity

On the other hand, if $p\mid n$, then $S(m, n, k, h, \chi; p)$ becomes the generalized $k$-th Gauss sums. W. P. Zhang and H. N. Liu ^[6] proved that for any prime $p$ with $3\mid (p-1)$, one has the identity

where $U$ is the cubic Gauss sums. That is, $U = \sum\limits_{a = 0}^{p-1}e\left(\frac{a^3}{p}\right)$.

For more works towards this direction, see references ^{[7,8,9,10,11,12]}, to save space, we will not list them all here.

In this paper, we use the elementary methods and the properties of the recurrence sequence to study the calculating problem of the $k$-th power mean

and give an exact calculating formula for (1.1) with the restrictive condition $p\equiv 1\bmod 3$ and $k\nmid (p-1)$. In fact, when $k$ is relatively large, the problem of calculating (1.1) is very difficult. However, we find that as long as $(k, p-1) = 1$, we can transform the studied problem into an interesting recursion relation. Thus, we have completely solved the problem of calculating the $k$-th power mean (1.1). That is, we will prove the following two conclusions:

Theorem 1.1. Let $p$ be a prime with $p\equiv 1\bmod 3$, and

Then for any positive integer $k\geq 3$, we have the recurrence formula

where the initial values $S_0(p) = -1$, $S_1(p) = 2p+1$ and $S_2(p) = dp-6p-1$, $A(m) = \sum\limits_{a = 0}^{p-1}e\left(\frac{ma^3}{p}\right)$ is the cubic Gauss sums, $4p = d^2+27\cdot b^2$, and $d$ is uniquely determined by $d\equiv 1\bmod 3$.

As an application of Theorem 1.1, we immediately give an efficient method for calculating (1.1). That is, we have the following:

Theorem 1.2. Let $p$ be a prime with $p\equiv 1\bmod 3$. Then for any positive integer $k$ with $(k, p-1) = 1$, we have the identity

It is clear that combining Theorem 1.1 and Theorem 1.2 we immediately solved the problem of calculating $k$-th power mean (1.1). Especially take $k = 5$, $7$ and $11$, we can deduce the following three corollaries:

Corollary 1.1. Let $p$ be a prime with $p\equiv 1\bmod 3$. If $5\nmid (p-1)$, then we have the identity

Corollary 1.2. Let $p$ be a prime with $p\equiv 1\bmod 3$. If $7\nmid (p-1)$, then we have the identity

Corollary 1.3. Let $p$ be a prime with $p\equiv 1\bmod 3$. If $11\nmid (p-1)$, then we have the identity

Remarks: Firstly in Theorem 1.2, we only considered the case $p\equiv 1\bmod 3$. In fact, if $3\nmid (p-1)$, then for any non-principal character $\chi$ modulo $p$ and any integer $m$ with $(m, p) = 1$, we have

So in this time, for any real number $k\geq 0$, we have the identity

Secondly, there is no restriction for positive integer $k$ in Theorem 1.1. So for any positive integer $k$ with $(k, p-1) = 1$, combining Theorems 1.1 and 1.2 we can get the exact values of the $k$-th power mean (1.1).

Thirdly, with the mathematical software such as Matlab (See Appendix program A), we can use computers to calculate the exact value of Theorem 1.2. Thus, computers can be truly realized in solving theoretical mathematical problems.

Finally, if $k\geq 5$ and $(k, p-1) > 1$, whether there exists an exact calculating formula for (1.1) is an open problem. This will be further studied by us.

2.   The direct proofs of the theorems

To complete the proof of Theorem 1.1, we need following simple lemma. Of course, the proof of this lemma need some knowledge of elementary or analytic number theory, all them can be found in references ^[13,14] or ^[3], we will not repeat them here. First we have the following:

Lemma 2.1. Let $p$ be a prime with $p\equiv 1\bmod 3$. Then for any third-order primitive character $\lambda$ modulo $p$, we have the identity

where $\tau(\chi) = \sum\limits_{a = 1}^{p-1}\chi(a)e\left(\frac{a}{p}\right)$ denotes the classical Gauss sums, $4p = d^2+27\cdot b^2$, $b$ is an integer and $d$ is uniquely determined by $d\equiv 1\bmod 3$.

Proof. See ^[15] or ^[16].

Now we using this Lemma 2.1 to complete the proof of our Theorem 1.1. Let $\lambda$ be any third-order primitive character modulo $p$. For any integer $m$, let

Then for integer $m$ with $(m, p) = 1$, from Lemma 2.1, the properties of the third-order character modulo $p$ and Gauss sums we have

Note that $\tau(\lambda)\cdot \tau\left(\overline{\lambda}\right) = p$, $\lambda^2 = \overline{\lambda}$ and $\lambda^3(m) = 1$ we also have

It is clear that

From (2.1) we also have

If $k\geq 3$, then from (2.5) we have the recursive formula

Now Theorem 1.1 follows from (2.2)–(2.4) and (2.6).

Now we prove Theorem 1.2. From the properties of the reduced residue system, the orthogonality of the characters modulo $p$ and the trigonometrical identity

we have

Since $(k, p-1) = 1$, when $a_k$ passes through a reduced residue system modulo $p$, then $a_k^k$ also passes through a reduced residue system modulo $p$. From (2.7) we have

This completes the proof of Theorem 1.2.

3.   Conclusions

The main results of this paper is to give two theorems. Theorem 1.1 proved that one kind exponential sums related to cubic Gauss sums satisfies a third-order linear recursion property. Theorem 1.2 gave the transformation form of $k$-th power mean of the generalized cubic Gauss sums, which is closely related to Theorem 1.1. Finally, the calculation of Theorem 1.2 can be achieved by Theorem 1.1 and computer programs such as Matlab.

Of course, our results also provides some new and effective methods for the calculating problem of the $k$-th power mean of the generalized cubic Gauss sums. We have reason to believe that these works will play a positive role in promoting the study of relevant problems.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper.

This work is supported by the N. S. F. (12126357) of China and Shaanxi Fundamental Science Research Project for Mathematics and Physics (Grant No.22JSY007).

Conflict of interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Appendix program A

Program 1:

$\text{function} [ s_n ] = \text{RECURRENCE}(d, p, s_{n1}, s_{n2}, s_{n3})\\ s_n = -3*s_{n1}+3*(p-1)*s_{n2}+(d*p+3*p-1)*s_{n3};\\ \text{Program 2}:\\ \text{clc}\\ \text{clear all}\\ \text{syms}\ p; \\ \text{syms}\ d; \\ \text{syms}\ s0;\\ \text{syms}\ s1;\\ \text{syms}\ s2;\\ \text{syms}\ s3;\\ \text{syms}\ s_{n3};\\ \text{syms}\ s_{n2};\\ \text{syms}\ s_{n1};\\ \text{syms}\ s_n; \\ s0 = -1\\ s1 = 2*p+1\\ s2 = d*p-6*p-1\\ s_{n1} = s2;\\ s_{n2} = s1;\\ s_{n3} = s0;\\ \text{for}\ n = 1:8\\ n+2\\ s_n = \text{RECURRENCE}(d, p, s_{n1}, s_{n2}, s_{n3}); \\ \text{expand}(s_n)\\ s_{n3} = s_{n2};\\ s_{n2} = s_{n1};\\ s_{n1} = s_n; \\ \text{end}$