Some new hybrid power mean formulae of trigonometric sums

Abstract

We apply the analytic method and the properties of the classical Gauss sums to study the computational problem of a certain hybrid power mean of the trigonometric sums and to prove several new mean value formulae for them. At the same time, we also obtain a new recurrence formula involving the Gauss sums and two-term exponential sums.

1 Introduction

For any integer m and odd prime $p\ge 3$, the cubic Gauss sums $A(m, p)=A(m)$ are defined as follows:

$$ A(m) = \sum_{a=0}^{p-1} e \biggl( \frac{ma^{3} }{p} \biggr), $$

where, as usual, $e(y) = e^{2\pi i y}$.

We found that several scholars studied the hybrid mean value problems of various trigonometric sums and obtained many interesting results. For example, Chen and Hu [1] studied the computational problem of the hybrid power mean

$$\begin{aligned} S_{k}(p)=\sum_{m=1}^{p-1} \Biggl( \sum_{a=0}^{p-1} e \biggl( \frac{ma^{3} }{p} \biggr) \Biggr)^{k}\cdot \Biggl\vert \sum _{c=1}^{p-1}e \biggl( \frac{mc+\overline{c}}{p} \biggr) \Biggr\vert ^{2}, \end{aligned}$$

where c̅ denotes the multiplicative inverse of $c\bmod p$, that is, $c\cdot \overline{c}\equiv 1\bmod p$.

For $p\equiv 1\bmod 3$, they proved an interesting third-order linear recurrence formula for $S_{k}(p)$.

Li and Hu [2] studied the computational problem of the hybrid power mean

$$\begin{aligned} \sum_{b=1}^{p-1} \Biggl\vert \sum _{a=0}^{p-1}e \biggl(\frac{ba^{4}}{p} \biggr) \Biggr\vert ^{2}\cdot \Biggl\vert \sum_{c=1}^{p-1}e \biggl( \frac{bc+\overline{c}}{p} \biggr) \Biggr\vert ^{2} \end{aligned}$$

(1)

and proved an exact computational formula for (1).

Zhang and Zhang [3] proved the identity

$$ \sum_{m=1}^{p-1} \Biggl\vert \sum _{a=0}^{p-1}e \biggl(\frac{ma^{3}+na}{p} \biggr) \Biggr\vert ^{4} = \textstyle\begin{cases} 2p^{3}-p^{2} &\text{if } 3\nmid p-1, \\ 2p^{3}-7p^{2} &\text{if } 3\mid p-1. \end{cases} $$

Other related contents can also be found in [4–12], which will not be repeated here.

In this paper, inspired by [1] and [2], we consider the following mean value:

$$\begin{aligned} H_{k}(c,p)=\sum_{m=1}^{p-1} \Biggl(\sum_{a=0}^{p-1}e \biggl( \frac{cma^{3}}{p} \biggr) \Biggr)^{k}\cdot \Biggl(\sum _{a=0}^{p-1}e \biggl(\frac{ma^{3}+a}{p} \biggr) \Biggr)^{3}. \end{aligned}$$

(2)

We do not know whether there exists a precise computational formula for (2), where c is any integer with $(c, p)=1$, and $p\equiv 1\bmod 3$.

Actually, there also exists a third-order linear recurrence formula of $H_{k}(c, p)$ for all integers $k\geq 1$ and c. But for some integers c, the initial value of $H_{k}(c, p)$ is very simple, whereas for other c, the initial value of $H_{k}(c, p)$ is more complex. So a satisfactory recursive formula for $H_{k}(c, p)$ is not available.

The main purpose of this paper is using an analytic method and the properties of classical Gauss sums to give an effective calculation method for $H_{k}(c, p)$ with some special integers c. We will prove the following two theorems.

Theorem 1

Letpbe a prime with$p\equiv 1\bmod 3$. If 3 is not a cubic residue$\bmod~p$, then we have

$$\begin{aligned}& \sum_{m=1}^{p-1} \Biggl(\sum _{a=0}^{p-1}e \biggl(\frac{3ma^{3}}{p} \biggr) \Biggr) \Biggl(\sum_{b=0}^{p-1}e \biggl( \frac{mb^{3}+b}{p} \biggr) \Biggr)^{3} =3p^{2}+dp^{2}, \\& \sum_{m=1}^{p-1} \Biggl(\sum _{a=0}^{p-1}e \biggl(\frac{ma^{3}}{p} \biggr) \Biggr)^{2} \Biggl(\sum_{b=0}^{p-1}e \biggl( \frac{mb^{3}+b}{p} \biggr) \Biggr)^{3} =p^{2} (3p-5d ), \end{aligned}$$

and

$$\begin{aligned} \sum_{m=1}^{p-1} \Biggl(\sum _{a=0}^{p-1}e \biggl(\frac{3ma^{3}}{p} \biggr) \Biggr)^{3} \Biggl(\sum_{b=0}^{p-1}e \biggl( \frac{mb^{3}+b}{p} \biggr) \Biggr)^{3} =p^{2} \bigl(5dp+9p-d^{2} \bigr). \end{aligned}$$

Theorem 2

Letpbe an odd prime with$p\equiv 1\bmod 3$. If 3 is a cubic residue$\bmod~p$, then for any integer$k\geq 3$, we have the third-order linear recurrence formula

$$ H_{k}(1,p)=3pH_{k-2}(1,p)+ dpH_{k-3}(1,p), $$

where the first three terms are$H_{0}(1,p)=2p^{2}-pd$, $H_{1}(1,p)=p^{2} (d-6 )$, and$H_{2}(1,p)=p^{2}(6p-5d)$.

Some notes: First, in Theorem 1, if $(3,p-1)=1$, then the question we are discussing is trivial, because in this case, we have

$$ \sum_{a=0}^{p-1}e \biggl(\frac{ma^{3}}{p} \biggr)=\sum_{a=0}^{p-1}e \biggl( \frac{ma}{p} \biggr)=0. $$

Second, in the first and third formulas of Theorem 1, we take $c=3$ (and $c=1$ in the second formula). These are all for getting the exact value of the mean value. Otherwise, the results will not be pretty.

2 Several lemmas

To complete the proofs of our theorems, several lemmas are essential. Hereafter, we will use related properties of the classical Gauss sums and the third-order character $\bmod~p$, all of which can be found in books concerning elementary number theory or analytic number theory, such as [13] and [14]. First we have the following:

Lemma 1

Letpbe a prime with$p\equiv 1\bmod 3$. Then for any third-order character$\psi \bmod p$, we have the identity

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

Proof

First, applying the trigonometric identity

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

(3)

and noting that $\psi ^{3}=\chi _{0}$, the principal character $\bmod~p$, we have

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

(4)

Noting that $\psi ^{2}=\overline{\psi }$ and $\tau (\psi )\tau (\overline{\psi } )=p$, from the properties of Gauss sums we have

$$\begin{aligned}& \begin{aligned}[b] \sum_{a=1}^{p-1}\overline{ \psi } \bigl(a^{3}+1 \bigr)&=\sum_{a=1}^{p-1} \overline{\psi }(a+1) \bigl(1+\psi (a)+\overline{\psi }(a) \bigr) \\ &=\sum_{a=1}^{p-1}\overline{\psi }(a+1)+\sum _{a=1}^{p-1} \overline{\psi } (1+\overline{a} )+\sum_{a=1}^{p-1} \overline{\psi } \bigl(a^{2}+a \bigr) \\ &=-2+\frac{1}{\tau (\psi )}\sum_{b=1}^{p-1}\psi (b)\sum_{a=1}^{p-1} \overline{\psi }(a)e \biggl(\frac{b(a+1)}{p} \biggr) \\ &=-2+\frac{\tau ^{2} (\overline{\psi } )}{\tau (\psi )}=-2+ \frac{\tau ^{3} (\overline{\psi } )}{p}, \end{aligned} \end{aligned}$$

(5)

$$\begin{aligned}& \begin{aligned}[b] \sum_{a=0}^{p-1}\overline{ \psi } \bigl(a^{3}-(a+1)^{3}+1 \bigr)&= \sum _{a=0}^{p-1}\overline{\psi } \bigl(-3a(a+1) \bigr) \\ &=\overline{\psi }(3)\sum_{a=1}^{p-1} \overline{\psi } \bigl(a(a+1) \bigr)= \frac{\overline{\psi }(3)\tau ^{3} (\overline{\psi } )}{p}. \end{aligned} \end{aligned}$$

(6)

Since ψ is a third-order character $\bmod~p$, for any integer c with $(c,p)=1$, from the properties of the classical Gauss sums we have

$$\begin{aligned} \sum_{a=0}^{p-1}e \biggl( \frac{ca^{3}}{p} \biggr)=1+\sum_{a=1}^{p-1} \bigl(1+\psi (a)+\overline{\psi }(a) \bigr)e \biggl(\frac{ca}{p} \biggr) = \overline{\psi }(c)\tau (\psi )+\psi (c)\tau ( \overline{\psi } ). \end{aligned}$$

(7)

Applying (7), we have

$$\begin{aligned} \sum_{a=0}^{p-1}\sum _{b=0}^{p-1}\overline{\psi } \bigl(a^{3}+b^{3}+1 \bigr) =&\frac{1}{\tau (\psi )} \sum_{c=1}^{p-1} \psi (c)\sum_{a=0}^{p-1} \sum _{b=0}^{p-1}e \biggl(\frac{ca^{3}+cb^{3}+c}{p} \biggr) \\ =&\frac{1}{\tau (\psi )} \sum_{c=1}^{p-1}\psi (c)e \biggl( \frac{c}{p} \biggr) \Biggl(\sum_{a=0}^{p-1}e \biggl(\frac{ca^{3}}{p} \biggr) \Biggr)^{2} \\ =&\frac{1}{\tau (\psi )} \sum_{c=1}^{p-1}\psi (c)e \biggl( \frac{c}{p} \biggr) \bigl(\psi (c)\tau ^{2}(\psi )+2p+ \overline{\psi }(c)\tau ^{2} (\overline{\psi } ) \bigr) \\ =&\tau (\psi )\tau (\overline{\psi } )+2p- \frac{\tau ^{3} (\overline{\psi } )}{p}=3p- \frac{\tau ^{3} (\overline{\psi } )}{p}. \end{aligned}$$

(8)

Combining (4), (5), (6), and (8), we have the identity

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

This proves Lemma 1. □

Lemma 2

Letpbe a prime with$p\equiv 1\bmod 3$, and letψbe any third-order character$\bmod~p$. Then we have

$$ \tau ^{3} (\psi )+ \tau ^{3} (\overline{\psi } )=dp, $$

where$\tau (\psi )$denotes the classical Gauss sums, anddis uniquely determined by$4p=d^{2}+27b^{2}$and$d\equiv 1\bmod 3$.

Proof

See [4] or [9]. □

Lemma 3

Letpbe a prime with$p\equiv 1\bmod 3$. Then we have the identity

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

Proof

Since the congruence equation $x^{3}+1\equiv 0\bmod p$ has three solutions in a reduced residue system $\bmod~p$, from (3) we have

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

(9)

It is clear that the conditions $a^{3}+b^{3}+1\equiv 0\bmod p$ and $a+b+1\equiv 0\bmod p$ ($0\leq a$, $b\leq p-1$) imply $a(a+1)\equiv 0\bmod p$ and $a+b+1\equiv 0\bmod p$, or $(a, b)=(0,p-1)$ and $(a, b)=(p-1, 0)$. So we have

$$\begin{aligned} p^{2}\mathop{\mathop{\sum_{a=0}^{p-1} \sum_{b=0}^{p-1}}_{a^{3}+b^{3}+1 \equiv 0\bmod p}}_{a+b+1\equiv 0\bmod p}1=2p^{2}. \end{aligned}$$

(10)

From (3), (7), Lemma 2, and the properties of Gauss sums we have

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

(11)

Combining (9), (10), and (11), we have the identity

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

This proves Lemma 3. □

3 Proofs of the theorems

We achieve our main results in this part. First, we prove Theorem 1. For any integer m with $(m, p)=1$, from (7) and Lemma 2 we have

$$\begin{aligned} A^{3}(3m) =& \Biggl(\sum_{a=0}^{p-1}e \biggl(\frac{3ma^{3}}{p} \biggr) \Biggr)^{3}= \bigl(\overline{\psi }(3m)\tau (\psi )+\psi (3m)\tau (\overline{\psi } ) \bigr)^{3} \\ =&\tau ^{3}(\psi )+ \tau ^{3} (\overline{\psi } )+3p \bigl(\overline{\psi }(3m)\tau (\psi )+\psi (3m)\tau ( \overline{\psi } ) \bigr)=dp+3pA(3m). \end{aligned}$$

(12)

Applying (7) and Lemmas 1 and 2, we have

$$\begin{aligned} &\sum_{m=1}^{p-1}A(3m) \Biggl(\sum _{a=0}^{p-1}e \biggl( \frac{ma^{3}+a}{p} \biggr) \Biggr)^{3} \\ &\quad =\sum_{m=1}^{p-1} \bigl(\overline{\psi }(3m)\tau (\psi )+\psi (3m) \tau (\overline{\psi } ) \bigr) \Biggl(\sum _{a=0}^{p-1}e \biggl(\frac{ma^{3}+a}{p} \biggr) \Biggr)^{3} \\ &\quad =\overline{\psi }(3)\tau (\psi ) \bigl(\psi (3)p\tau ^{2} ( \psi )-3p \tau (\overline{\psi } ) \bigr)+\psi (3) \tau (\overline{\psi } ) \bigl( \overline{\psi }(3)p\tau ^{2} (\overline{\psi } )-3p \tau (\psi ) \bigr) \\ &\quad =p \bigl(\tau ^{3}(\psi )+\tau ^{3} (\overline{\psi } ) \bigr)-3p^{2} \bigl(\psi (3)+ \overline{\psi }(3) \bigr)=dp^{2}+3p^{2}-3p^{2} \bigl(1+\psi (3)+ \overline{\psi }(3) \bigr) \\ &\quad =p^{2} (d+3 ), \end{aligned}$$

(13)

where we have used the identity $1+\psi (3)+ \overline{\psi }(3)=0$.

Applying Lemmas 1, 2, and 3 and (7), we have

$$\begin{aligned} &\sum_{m=1}^{p-1} \Biggl(\sum _{a=0}^{p-1}e \biggl(\frac{ma^{3}}{p} \biggr) \Biggr)^{2} \Biggl(\sum_{a=0}^{p-1}e \biggl( \frac{ma^{3}+a}{p} \biggr) \Biggr)^{3} \\ &\quad =\sum_{m=1}^{p-1} \bigl(\overline{\psi }(m)\tau (\psi )+\psi (m) \tau (\overline{\psi } ) \bigr)^{2} \Biggl( \sum_{a=0}^{p-1}e \biggl(\frac{ma^{3}+a}{p} \biggr) \Biggr)^{3} \\ &\quad =2p \bigl(2p^{2}-dp \bigr)+ \tau ^{2}(\psi ) \bigl( \overline{\psi }(3)p\tau ^{2} (\overline{\psi } )-3p \tau ( \psi ) \bigr) \\ &\qquad {}+\tau ^{2} (\overline{\psi } ) \bigl(\psi (3)p\tau ^{2} (\psi )-3p \tau (\overline{\psi } ) \bigr) \\ &\quad =2p^{2} (2p-d )+ \bigl(\psi (3)+\overline{\psi }(3) \bigr)p^{3}- 3p \bigl(\tau ^{3} (\psi )+\tau ^{3} ( \overline{\psi } ) \bigr) \\ &\quad =p^{2} (3p-5d ). \end{aligned}$$

(14)

Applying Lemmas 1, 2, and 3 and (12), we have

$$\begin{aligned} &\sum_{m=1}^{p-1} \Biggl(\sum _{a=0}^{p-1}e \biggl(\frac{3ma^{3}}{p} \biggr) \Biggr)^{3} \Biggl(\sum_{a=0}^{p-1}e \biggl( \frac{ma^{3}+a}{p} \biggr) \Biggr)^{3} \\ &\quad =\sum_{m=1}^{p-1} \bigl(\overline{\psi }(3m)\tau (\psi )+\psi (3m) \tau (\overline{\psi } ) \bigr)^{3} \Biggl( \sum_{a=0}^{p-1}e \biggl(\frac{ma^{3}+a}{p} \biggr) \Biggr)^{3} \\ &\quad =dp\sum_{m=1}^{p-1} \Biggl(\sum _{a=0}^{p-1}e \biggl( \frac{ma^{3}+a}{p} \biggr) \Biggr)^{3}+ 3p\sum_{m=1}^{p-1}A(3m) \Biggl(\sum_{a=0}^{p-1}e \biggl( \frac{ma^{3}+a}{p} \biggr) \Biggr)^{3} \\ &\quad =dp \bigl(2p^{2}-pd \bigr)+ 3p \bigl(3p^{2}+dp^{2} \bigr)=p^{2} \bigl(5dp+9p-d^{2} \bigr). \end{aligned}$$

(15)

Now Theorem 1 follows from (13), (14), and (15).

If $p\equiv 1\bmod 3$ and 3 is a cubic residue $\bmod~p$, then $\psi (3)=\overline{\psi }(3)=1$. From Lemma 3 we have

$$\begin{aligned} H_{0}(1,p)=2p^{2}-pd. \end{aligned}$$

(16)

From (7) and Lemmas 1 and 2 we have

$$\begin{aligned} H_{1}(1,p) =&\tau (\psi ) \bigl(p\tau ^{2}(\psi )-3p\tau ( \overline{\psi } ) \bigr)+\tau (\overline{\psi } ) \bigl(p\tau ^{2} ( \overline{\psi } )-3p \tau (\psi ) \bigr) \\ =&p \bigl(\tau ^{3}(\psi )+\tau ^{3} (\overline{\psi } ) \bigr)-6p^{2}=dp^{2}-6p^{2}. \end{aligned}$$

(17)

From (7) and Lemmas 1, 2, and 3 we also have

$$\begin{aligned} H_{2}(1,p) =& 2pH_{0}(1,p)+\tau ^{2} (\overline{ \psi } ) \bigl(p\tau ^{2}(\psi )-3p\tau (\overline{\psi } ) \bigr)+ \tau ^{2} (\psi ) \bigl(p\tau ^{2} ( \overline{\psi } )-3p\tau ( \psi ) \bigr) \\ =&2p^{2}(2p-d)+ 2p^{3}-3p \bigl(\tau ^{3}(\psi )+ \tau ^{3} ( \overline{\psi } ) \bigr) =p^{2} (6p-5d ). \end{aligned}$$

(18)

If $k\geq 3$, then applying (12), we have

$$\begin{aligned} H_{k}(1,p) =&\sum_{m=1}^{p-1}A^{k}(m) \Biggl(\sum_{a=0}^{p-1}e \biggl( \frac{ma^{3}+a}{p} \biggr) \Biggr)^{3} \\ =&\sum_{m=1}^{p-1}A^{k-3}(m) \bigl(dp+3pA(m) \bigr) \Biggl(\sum_{a=0}^{p-1}e \biggl(\frac{ma^{3}+a}{p} \biggr) \Biggr)^{3} \\ =&3p H_{k-2}(1,p)+dpH_{k-3}(1,p). \end{aligned}$$

(19)

Now Theorem 2 follows from (16), (17), (18), and (19).

This completes the proofs of all our results.

4 Conclusion

The main work of this paper includes two theorems. In Theorem 1, we obtained some exact values of (2) when $k=1, 2$, and 3. In Theorem 2, we showed that $H_{k}(1,p)$ satisfies an interesting third-order linear recurrence formula. These works not only profoundly reveal the regularity of a certain hybrid power mean of the trigonometric sums, but also provide some new ideas and methods for further study of such problems.