Let \(q\ge3\) be an integer. For any positive integer \(k\geq2\), the kth Gauss sums \(G(m, k; q)\) are defined as
$$G(m,k;q) = \sum_{a=0}^{q-1} e \biggl( \frac{ma^{k} }{q} \biggr), $$
where, as usual, \(e(y) = e^{2\pi i y}\).
Recently, some scholars have studied the properties of \(G(m,4;p)\) and obtained many interesting results, where p is an odd prime with \(p\equiv1\bmod4\). For example, Shimeng Shen and Wenpeng Zhang [1] proved a recurrence formula related to \(G(m,4;p)\). The author and Jiayuan Hu [2] studied the computational problem of the hybrid power mean
$$ \sum_{b=1}^{p-1} \Biggl\vert \sum_{a=0}^{p-1}e \biggl(\frac{ba^{4}}{p} \biggr) \Biggr\vert ^{2}\cdot \Biggl\vert \sum _{c=1}^{p-1}e \biggl(\frac{bc+\overline{c}}{p} \biggr) \Biggr\vert ^{2}. $$
(1)
We proved the identity
$$ \begin{gathered} \sum_{b=1}^{p-1} \Biggl\vert \sum_{a=0}^{p-1}e \biggl( \frac{ba^{4}}{p} \biggr) \Biggr\vert ^{2}\cdot \Biggl\vert \sum _{c=1}^{p-1}e \biggl(\frac{bc+\overline{c}}{p} \biggr) \Biggr\vert ^{2} \\ \quad=\left \{ \textstyle\begin{array}{l@{\quad}l} 3p^{3}-3p^{2}-3p+p (\tau^{2} (\overline{\chi }_{4} )+\tau^{2}(\chi_{4}) ), &\text{if } p\equiv5\bmod8; \\ 3p^{3}-3p^{2}-3p -p\tau^{2} (\overline{\chi}_{4} )-p\tau^{2}(\chi_{4})+2\tau^{5} (\overline{\chi}_{4} )+2\tau^{5}(\chi_{4}), &\textrm{if }p\equiv1\bmod8, \end{array}\displaystyle \right . \end{gathered} $$
where \(\chi_{4}\) denotes any fourth-order character modp, \(\tau(\chi)= \sum_{a=1}^{p-1} \chi(a)e (\frac{a}{p} )\) denotes the classical Gauss sums, and c̅ denotes the multiplicative inverse of \(c \bmod p\).
At the same time, the author and Jiayuan Hu [2] also pointed out how to compute the exact value of \(\tau^{2} (\overline{\chi}_{4} )+\tau^{2}(\chi_{4})\) and \(\tau^{5} (\overline{\chi}_{4} )+\tau^{5}(\chi_{4})\), these are two meaningful problems.
Zhuoyu Chen and Wenpeng Zhang [3] studied the properties of the Gauss sums
$$G(k,p)= \tau^{k} (\psi )+ \tau^{k} (\overline{\psi} ). $$
By using the analytic method and the properties of classical Gauss sums, they obtained an exact computational formula for \(G(k,p)\), which completely solved the problem proposed by the author and Jiayuan Hu in [2]. Some related works can also be found in references [4–11].
Inspired by reference [3], we will consider the following hybrid power mean:
$$ M_{k}(p)=\sum_{m=1}^{p-1} \Biggl( \sum_{a=0}^{p-1}e \biggl(\frac{ma^{4}}{p} \biggr) \Biggr)^{k}\cdot \Biggl\vert \sum _{a=0}^{p-1}e \biggl(\frac{ma^{4}+a}{p} \biggr) \Biggr\vert ^{2}. $$
For convenience, hereinafter, we always assume that p is a prime with \(p\equiv1\bmod4\), \((\frac{*}{p} )=\chi_{2}\) denotes the Legendre symbol modp, and
$$\alpha=\alpha(p)=\sum_{a=1}^{\frac{p-1}{2}} \biggl( \frac{a+\overline {a}}{p} \biggr), $$
a̅ denotes the solution of the equation \(ax\equiv 1\bmod p\). The number α is closely related to prime p. In fact, we have a very important formula
$$p= \Biggl(\sum_{a=1}^{\frac{p-1}{2}} \biggl( \frac{a+\overline{a}}{p} \biggr) \Biggr)^{2}+ \Biggl(\sum _{a=1}^{\frac{p-1}{2}} \biggl(\frac{ra+\overline {a}}{p} \biggr) \Biggr)^{2}\equiv\alpha^{2}+\beta^{2}, $$
where r is any integer with \((\frac{r}{p} )=-1\) (see Theorems 4–11 in [12]).
In this paper, by using the analytic method, the properties of the classical Gauss sums, and trigonometric sums, we will study the computational problem of \(M_{k}(p)\), and give an interesting fourth-order linear recurrence formula for it. That is, we will prove the following two results.
Theorem 1
If
p
is a prime with
\(p\equiv5\bmod8\), then for any integer
\(k\geq4\), we have the linear recurrence formula
$$M_{k}(p)=-2pM_{k-2}(p)+8p\alpha M_{k-3}(p)-p \bigl(9p-4\alpha^{2} \bigr)M_{k-4}(p), $$
where the first four items in the sequence
\(\{M_{k}(p)\}\)
are: \(M_{0}(p)=p(p-3)\); \(M_{1}(p)=2p\alpha\); \(M_{2}(p)=-p (p^{2}-3p-4\alpha ^{2} )\), and
\(M_{3}(p)=2p^{2}\alpha (3p-14 )\).
Theorem 2
If
p
is a prime with
\(p\equiv1\bmod8\), then for any integer
\(k\geq4\), we have the linear recurrence formula
$$M_{k}(p)=6pM_{k-2}(p)+8p\alpha M_{k-3}(p)-p \bigl(p-4\alpha^{2} \bigr)M_{k-4}(p), $$
where the first four items in the sequence
\(\{M_{k}(p)\}\)
are: \(M_{0}(p)=p(p-3)\); \(M_{1}(p)=-6p\alpha\); \(M_{2}(p)=p (3p^{2}-17p-4\alpha ^{2} )\), and
\(M_{3}(p)=6p^{2}\alpha(p-8)\).
For some special integers \(k=2\) or \(k=4\), from our theorems we may immediately deduce the following three corollaries.
Corollary 1
If
p
is an odd prime with
\(p\equiv5\bmod 8\), then we have
$$ \sum_{m=1}^{p-1} \Biggl\vert \sum _{a=0}^{p-1}e \biggl(\frac{ma^{4}}{p} \biggr) \Biggr\vert ^{2}\cdot \Biggl\vert \sum_{c=0}^{p-1}e \biggl(\frac{mc^{4}+c}{p} \biggr) \Biggr\vert ^{2} =p \bigl(3p^{2}-9p-4\alpha^{2} \bigr). $$
Corollary 2
If
p
is an odd prime with
\(p\equiv 1\bmod8\), then we have the identity
$$ \sum_{m=1}^{p-1} \Biggl\vert \sum _{a=0}^{p-1}e \biggl(\frac{ma^{4}}{p} \biggr) \Biggr\vert ^{2}\cdot \Biggl\vert \sum_{c=0}^{p-1}e \biggl(\frac{mc^{4}+c}{p} \biggr) \Biggr\vert ^{2} =p \bigl(3p^{2}-17p-4\alpha^{2} \bigr). $$
Corollary 3
If
p
is an odd prime with
\(p\equiv 1\bmod8\), then we have
$$ \sum_{m=1}^{p-1} \Biggl\vert \sum _{a=0}^{p-1}e \biggl(\frac{ma^{4}}{p} \biggr) \Biggr\vert ^{4}\cdot \Biggl\vert \sum_{c=0}^{p-1}e \biggl(\frac{mc^{4}+c}{p} \biggr) \Biggr\vert ^{2} =p^{2} \bigl(17p^{2}+4p\alpha^{2}-99p-84\alpha^{2} \bigr). $$
Notes
If \(p=4k+3\), then \((\frac{-1}{p} )=-1\). Then, in this case, for any integer m with \((m,p)=1\), we have
$$ \begin{aligned}\sum_{a=0}^{p-1}e \biggl(\frac{ma^{4}}{p} \biggr)&=1+\sum_{a=1}^{p-1} \biggl(1+ \biggl(\frac{a}{p} \biggr) \biggr)e \biggl(\frac{ma^{2}}{p} \biggr) \\ &=\sum_{a=0}^{p-1}e \biggl(\frac{ma^{2}}{p} \biggr)+\sum_{a=1}^{p-1} \biggl( \frac{a}{p} \biggr)e \biggl(\frac{ma^{2}}{p} \biggr) =\sum _{a=0}^{p-1}e \biggl(\frac{ma^{2}}{p} \biggr)= \biggl( \frac{m}{p} \biggr)i\sqrt{p}, \end{aligned} $$
where \(i^{2}=-1\). Therefore, the hybrid power mean \(M_{k}(p)\) can be easily obtained.