Hybrid mean value involving some two-term exponential sums and fourth Gauss sums

1.   Introduction

Let $q\ge3$ be a positive integer. For any integers $m$, $n$, $k$, $h$, the two-term exponential sums $C(m, n, k, h; q)$ is defined as

where $k > h$. When $n = 0$, this sum reduces to the $k$-th Gauss sum, defined by

These summations are very important in additive number theory because there are close relations between the summation and the Waring's problem. About its arithmetical properties, many authors had studied it. For example, Davenport and Heilbronn ^[1] proved that

where $\theta = \frac{2}{3}$ if $k = 3$, and $\theta = \frac{3}{4}$ if $k \ge 3.$ Applying Weil's estimate for exponential sums over finite fields, Hua ^[2] proved that $\theta = \frac{1}{2}$ for all $k\ge 2.$ Loxton and Smith ^[3], Smith ^[4], Loxton and Vaughan ^[5], Dabrowski and Fisher ^[6], among others, have made improvements to (1.1). Recently, Li ^[7], utilizing the recursive formula for the fourth Gauss sum and relevant results on two-term exponential sums, further extended the study to derive a fourth-order linear recurrence formula for the fourth Gauss sum and the two-term exponential sum

in the case where $p\equiv1\bmod 4$.

We have been exploring the properties of two-term exponential sums, specifically $\left|\sum\limits_{a = 1}^{p-1}e\left(\frac{ma^4+a^2}{p}\right)\right|^t$. The main challenge arises when the lower-order terms of the two-term exponential sum are 2. If $a$ passes through a residue system modulo $p$ while $a^2$ does not pass through the same residue system, direct application of properties of Gauss sums or trigonometric identity becomes complex. Yuan and Wang ^[8], utilizing the orthogonality and parity properties of characters modulo $p$, ingeniously transformed the problem and ultimately provided a concise formula. Specifically, for $p\equiv3\bmod 4$, the following identity holds:

however, when $p\equiv1\bmod 4$, the research on the problem becomes relatively complex, and they did not provide relevant results.

For the sake of convenience, we have introduced some symbols

The main purpose of this paper is to study the asymptotic properties of the $t$-th power mean

and give some exact formulas for (1.3) with $t = 2$; for convenience, we will denote it as $T_k(p)$. That is, we shall prove the following:

Theorem 1. Let $p$ be a prime with $p\equiv5\bmod 8$. Then we have the identity

Corollary 1. Let $p$ be an odd prime with $p\equiv5\bmod 8$. Then we have

Theorem 2. Let $p$ be a prime with $p\equiv1\bmod 8$. Then we have the identity

where $\tau(\psi) = \sum\limits_{a = 0}^{p-1}\psi(a)e\left(\frac{a}{p}\right),$ which usually denotes the classical Gauss sums.

Corollary 2. Let $p$ be an odd prime with $p\equiv1\bmod 8$. Then we have

Theorem 3. Let $p$ be a prime with $p\equiv3\bmod 4$. Then for any integer $k\geq4$, we have

Some notes:

(1) When $p\equiv3\bmod 4$, we have $\left(\frac{-1}{p}\right) = -1$. In this case, for any integer $m$ with $(m, p) = 1$, according to Theorem 7.5.4 in reference ^[9], we have

where $i^2 = -1$. Combining with the properties of Gauss sums, we obtain

then

when $p\equiv3\bmod 4$, the hybrid means of $T_k(p)$ are easily obtained. Hence, in this section, we only consider the case when $p\equiv1\bmod 4$.

(2) From the recursive formula for $G^4(m, 4;p) = \left(\sum\limits_{a = 0}^{p-1}e\left(\frac{ma^4}{p}\right)\right)^4$ in ^[10], we can derive a fourth-order linear recurrence formula for $T_k(p)$.

When $p = 8k+5$, the recursive formula for $G^4(m, 4;p)$ is as follows:

Through simple calculations, we obtain the recursive formula for $T_k(p)$ as follows:

Similarly, when $p = 8k+1$, according to ^[10], we know

then we have

(3) When $p\equiv1\bmod{4}$ and $k$ is any positive integer, for the hybrid mean given by

as $k$ increases, the computation becomes increasingly complex. Whether precise computational formulas or corresponding recurrence relations can be obtained remains a question worth investigating.

2.   Preliminaries

In this section, we give some lemmas, which will be applied to prove the main results.

Lemma 1. Let $p$ be an odd prime with $p\equiv1\bmod 4$. For any fourth-order character $\psi$ modulo $p$, we have

where $\psi^2 = \chi_2$ and $\chi_2 = \left(\frac{\ast}{p}\right)$ denotes the Legendre symbol.

Proof. In fact, this is Lemma 2.1 of ^[11], so its proof is omitted.

Lemma 2. Let $p$ be an odd prime with $p\equiv1\bmod 4$. For any fourth-order character $\psi$ modulo $p$, we have

Proof. See Lemma 2.2 in ^[11].

Lemma 3. Let $p$ be an odd prime with $p\equiv1\bmod 4$, and $a$ be any integer with $(a, p) = 1$. For any fourth-order character $\psi$ modulo $p$, we have

Proof. First, using the trigonometric identity

Then, utilizing the properties of Dirichlet characters, $\psi^4 = \overline{\psi}^4 = \chi_0$ and $\overline{\psi}\chi_2 = \psi$, we can obtain

When $p\equiv1\bmod 4$, $\tau(\chi_2) = \sqrt{p}$, combined with (2.3) in Lemma 1, we obtain

When $p\equiv1\bmod 8$, $\psi(-1) = \overline{\psi}(-1) = 1$; when $p\equiv5\bmod 8$, $\psi(-1) = \overline{\psi}(-1) = -1$. Therefore, we have

This proves Lemma 3.

Lemma 4. Let $p$ be an odd prime with $p\equiv1\bmod 4$, then

Proof. From the proof of Lemma 3, we obtain

Utilizing the properties of the Legendre symbol, then

Combining (2.5) with (2.2) from Lemma 1, we obtain

from Lemma 2, then

This proves Lemma 4.

Lemma 5. Let $p$ be an odd prime with $p\equiv1\bmod 4$, then

Proof. From the properties of quadratic Gauss sums, we obtain

Further calculations yield

using the properties of the Legendre symbol, then

According to (2.7), we can obtain

This proves Lemma 5.

Lemma 6. Let $p$ be an odd prime with $p\equiv1\bmod 4$, then

Proof. According to the properties of the character modulo $p$, we have

Noting that $p\nmid(a^2+1),$ $p\mid(a^2-1)$, and using (2.8), we obtain

This proves Lemma 6.

3.   Proofs of the Theorems

For any integer $m$ satisfying $(m, p) = 1$, based on the classical properties of Gauss sums and the properties of the fourth-order character $\psi$, combined with (2.4), we have

Let us start by proving Theorem 1.

When $p\equiv5\bmod 8$, we have $\psi(-1) = -1$. Utilizing the property $\overline{\tau(\psi)} = \psi(-1)\tau(\overline{\psi}) = -\tau(\overline{\psi})$, it follows that in this scenario, $\overline{\overline{\psi}(m)\tau(\psi)+\psi(m)\tau(\overline{\psi})} = -(\overline{\psi}(m)\tau(\psi)+\psi(m)\tau(\overline{\psi}))$. Additionally, since $\chi_2(m)\sqrt{p}$ is a real number; based on (3.1), we obtain

In this particular case, according to (3.1) and (3.2), we derive

Now, considering (3.3) and incorporating Lemmas 2, 4, and 5, we obtain

This proves Theorem 1.

When $p \equiv 1 \bmod 8$, it holds that $\psi(-1) = 1$. For any integer $m$ with $(m, p) = 1$, as stated in (3.1), we establish that $\overline{G(m, 4;p)} = G(m, 4;p)$, implying that $G(m, 4;p)$ is a real number, then

Since $\tau(\psi)\tau(\overline{\psi}) = \psi(-1)p = p$, combining with Lemmas 2, 3, 4, and 5, we obtain

This proves Theorem 2.

When $p \equiv 3 \bmod 4$, as stated in (1.2) and (1.4), we obtain

This completes the proof of Theorem 3.

For $p \equiv 5 \bmod 8$, considering $|\alpha|\leq \sqrt{p}$ and Theorem 1, we obtain

This completes the proof of Corollary 1.

In the case of $p \equiv 1 \bmod 8$, where $\left|\tau^2(\psi)\right| = p$, combining with Lemma 6 and Theorem 2, we have

This completes the proof of all our theorems and corollaries.

Use of AI tools declaration

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

Acknowledgments

This work is supported by the N.S.F. (12471006, 12371007) of China and Shaanxi Fundamental Science Research Project for Mathematics and Physics (Grant No. 22JSY007).

Conflict of interest

The authors declare there are no conflicts of interest.