As usual, let *q* ≥ 3 be a positive integer. For any positive integer *n* ≥ 2, the classical *n*-th Gauss sums *G*(*m*, *n*; *q*) is defined by

$$\begin{array}{}{\displaystyle G(m,n;q)=\sum _{a=0}^{q-1}e\left(\frac{m{a}^{n}}{q}\right),}\end{array}$$

where *e*(*y*) = *e*^{2π iy}.

Many mathematical scholars have studied the arithmetical properties concerning *G*(*m*, *n*; *q*) and have obtained various interesting results, see references [1, 2, 3, 4, 5, 6, 7, 8, 9] and [11]. For example, Shimeng Shen and Wenpeng Zhang [2] studied the computational problem of the number *M*_{n}(*p*) of solutions of the congruence equation

$$\begin{array}{}{\displaystyle {x}_{1}^{4}+{x}_{2}^{4}+\cdots +{x}_{n}^{4}\equiv 0modp,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0\le {x}_{i}\le p-1,\phantom{\rule{thinmathspace}{0ex}}i=1,\phantom{\rule{thinmathspace}{0ex}}2,\cdots ,n,}\end{array}$$

and proved the following conclusions:

Let *p* be a prime with *p* = 8*k* + 5, *U*_{n}(*p*) = *M*_{n}(*p*) − *p*^{n−1}. Then for any positive integer *n* ≥ 5, one has the fourth-order linear recurrence formula

$$\begin{array}{}{\displaystyle {U}_{n}(p)=-2p{U}_{n-2}(p)+4p\alpha (p){U}_{n-3}(p)-\left(9{p}^{2}-p{\alpha}^{2}(p)\right){U}_{n-4}(p),}\end{array}$$

where the first four terms are *U*_{1}(*p*) = 0, *U*_{2}(*p*) = −(*p* − 1), *U*_{3}(*p*) = 3(*p* − 1)*α*(*p*) and *U*_{4}(*p*) = −7*p*(*p* − 1) + (*p* − 1)*α*^{2}(*p*).

If *p* = 8*k* + 1, then for any positive integer *n* ≥ 5, one has the fourth-order linear recurrence formula

$$\begin{array}{}{\displaystyle {U}_{n}(p)=6p{U}_{n-2}(p)+4p\alpha (p){U}_{n-3}(p)-\left({p}^{2}-p{\alpha}^{2}(p)\right){U}_{n-4}(p),}\end{array}$$

where the first four terms are *U*_{1}(*p*) = 0, *U*_{2}(*p*) = 3(*p* − 1), *U*_{3}(*p*) = 3(*p* − 1)*α*(*p*), *U*_{4}(*p*) = 17*p*(*p* − 1)+ (*p* − 1)*α*^{2}(*p*), and $\begin{array}{}\alpha (p)=\sum _{a=1}^{\frac{p-1}{2}}\left(\frac{a+\overline{a}}{p}\right),\left(\frac{\ast}{p}\right)\end{array}$ denotes the Legendre’s symbol mod *p*, and *a* denotes the multiplicative inverse of *a* mod *p*.

Xiaoxue Li and Jiayuan Hu [3] obtained the identity

$$\begin{array}{}{\displaystyle \phantom{\rule{1em}{0ex}}\sum _{b=1}^{p-1}{\left|\sum _{a=0}^{p-1}e\left(\frac{b{a}^{4}}{p}\right)\right|}^{2}\cdot {\left|\sum _{c=1}^{p-1}e\left(\frac{bc+\overline{c}}{p}\right)\right|}^{2}}\\ =\{\begin{array}{ll}{\displaystyle 3{p}^{3}-3{p}^{2}-3p+p\left({\tau}^{2}\left({\overline{\chi}}_{4}\right)+{\tau}^{2}({\chi}_{4})\right),}& \mathrm{if}p\equiv 5mod8;\\ ,{\displaystyle 3{p}^{3}-3{p}^{2}-3p-p{\tau}^{2}\left({\overline{\chi}}_{4}\right)-p{\tau}^{2}({\chi}_{4})+2{\tau}^{5}\left({\overline{\chi}}_{4}\right)+2{\tau}^{5}({\chi}_{4}),}& \mathrm{if}p\equiv 1mod8,\end{array}\end{array}$$

where *χ*_{4} denotes any fourth-order character mod $\begin{array}{}p,\tau (\chi )=\sum _{a=1}^{p-1}\chi (a)e\left(\frac{a}{p}\right)\end{array}$ denotes the classical Gauss sums.

At the same time, Xiaoxue Li and Jiayuan Hu [3] also pointed out that how to compute the exact value of *τ*^{2}(*χ*_{4}) + *τ*^{2}(*χ*_{4}) and *τ*^{5}(*χ*_{4}) + *τ*^{5}(*χ*_{4}) are two meaningful problems.

Let *A*(*k*, *p*) = *τ*^{k}(*χ*_{4}) + *τ*^{k}(*χ*_{4}). Zhuoyu Chen and Wenpeng Zhang [9] studied the computational problem of *A*(*k*, *p*), and obtained two interesting linear recurrence formulas. That is, let *p* be an odd prime with *p* ≡ 1 mod 4. Then for any positive integer *k*, one has the linear recurrence formulas

$$\begin{array}{}{\displaystyle A(2k+2,p)=2\sqrt{p}\cdot \alpha (p)\cdot A(2k,p)-{p}^{2}\cdot A(2k-2,p)}\end{array}$$

and

$$\begin{array}{}{\displaystyle A(2k+3,p)=2\sqrt{p}\cdot \alpha (p)\cdot A(2k+1,p)-{p}^{2}\cdot A(2k-1,p),}\end{array}$$

where *A*(0, *p*) = 2, *A*(1, *p*) = *G*(1) − $\begin{array}{}\sqrt{p},\end{array}$ *G*(1) = *G*(1, 4; *p*), *A*(2, *p*) = $\begin{array}{}2\sqrt{p}\alpha \end{array}$(*p*), *A*(3, *p*) = $\begin{array}{}\sqrt{p}\cdot \end{array}$ $\begin{array}{}\left(2\alpha -(-1{)}^{\frac{p-1}{4}}\sqrt{p}\right)\cdot \left(G(1)-\sqrt{p}\right).\end{array}$

In this paper, as a note of [2] and [9], we shall consider the computational problem of one kind hybrid power mean of two different Gauss sums

$$\begin{array}{}{\displaystyle \sum _{m=1}^{p-1}{\left|\sum _{a=0}^{p-1}e\left(\frac{m{a}^{3}}{p}\right)\right|}^{2h}\cdot {\left|\sum _{b=0}^{p-1}e\left(\frac{m{b}^{4}}{p}\right)\right|}^{2k},}\end{array}$$(1)

where *p* = 12*r* + 1 is an odd prime, *k* and *h* are two non-negative integers.

What we are interested in is whether there exits an exact computational formula for (1). Through researches mentioned above we found that for some special prime *p* we can give an an efficient method to compute the value of (1). The main purpose of this paper is to illustrate this point. That is, we shall prove the following main results:

#### Theorem 1.1

*Let* *p* *be a prime with* *p* = 24*r* + 13. *Then for any positive integers* *h* *and* *k*, *we have the identity*

$$\begin{array}{}{\displaystyle \sum _{m=1}^{p-1}{\left|\sum _{a=0}^{p-1}e\left(\frac{m{a}^{3}}{p}\right)\right|}^{2h}\cdot {\left|\sum _{b=0}^{p-1}e\left(\frac{m{b}^{4}}{p}\right)\right|}^{2k}=\frac{1}{2}\cdot {p}^{\frac{k}{2}}\cdot \left[{\left(3\sqrt{p}+2\alpha (p)\right)}^{k}+{\left(3\sqrt{p}-2\alpha (p)\right)}^{k}\right]\cdot \sum _{m=1}^{p-1}{\left|\sum _{a=0}^{p-1}e\left(\frac{m{a}^{3}}{p}\right)\right|}^{2h},}\end{array}$$

*where* $\begin{array}{}\alpha (p)=\sum _{a=1}^{\frac{p-1}{2}}\left(\frac{a+\overline{a}}{p}\right),\left(\frac{\ast}{p}\right)\end{array}$ *denotes the Legendre’s symbol* mod *p*, *a* *denotes the multiplicative inverse of* *a* mod *p*.

From Theorem 1.1 we may immediately deduce the corollaries as follows.

#### Corollary 1.2

*If* *p* *is a prime with* *p* = 24*r* + 13, *then for any positive integer* *k*, *we have*

$$\begin{array}{}{\displaystyle \sum _{m=1}^{p-1}{\left|\sum _{b=0}^{p-1}e\left(\frac{m{b}^{4}}{p}\right)\right|}^{2k}=\frac{1}{2}(p-1)\cdot {p}^{\frac{k}{2}}\cdot \left[{\left(3\sqrt{p}+2\alpha (p)\right)}^{k}+{\left(3\sqrt{p}-2\alpha (p)\right)}^{k}\right].}\end{array}$$

#### Corollary 1.3

*If* *p* *is a prime with* *p* = 24*r* + 13, *then for any positive integer* *k*, *we have*

$$\begin{array}{}{\displaystyle \sum _{m=1}^{p-1}{\left|\sum _{a=0}^{p-1}e\left(\frac{m{a}^{3}}{p}\right)\right|}^{2}\cdot {\left|\sum _{b=0}^{p-1}e\left(\frac{m{b}^{4}}{p}\right)\right|}^{2k}=(p-1)\cdot {p}^{\frac{k+2}{2}}\cdot \left[{\left(3\sqrt{p}-2\alpha (p)\right)}^{k}+{\left(3\sqrt{p}+2\alpha (p)\right)}^{k}\right].}\end{array}$$

#### Corollary 1.4

*If* *p* = 24*r* + 13 *is an odd prime*, *then for any positive integer* *k*, *we have*

$$\begin{array}{}{\displaystyle \sum _{m=1}^{p-1}{\left|\sum _{a=0}^{p-1}e\left(\frac{m{a}^{3}}{p}\right)\right|}^{4}\cdot {\left|\sum _{b=0}^{p-1}e\left(\frac{m{b}^{4}}{p}\right)\right|}^{2k}=3(p-1)\cdot {p}^{\frac{k+4}{2}}\cdot \left[{\left(3\sqrt{p}+2\alpha (p)\right)}^{k}+{\left(3\sqrt{p}-2\alpha (p)\right)}^{k}\right].}\end{array}$$

#### Corollary 1.5

*If* *p* = 24*r* + 13 *is an odd prime*, *then for any positive integer* *k*, *we have*

$$\begin{array}{}{\displaystyle \sum _{m=1}^{p-1}{\left|\sum _{a=0}^{p-1}e\left(\frac{m{a}^{3}}{p}\right)\right|}^{6}\cdot {\left|\sum _{b=0}^{p-1}e\left(\frac{m{b}^{4}}{p}\right)\right|}^{2k}=\frac{1}{2}(p-1)(18p+{d}^{2})\cdot {p}^{\frac{k+4}{2}}\cdot \left[{\left(3\sqrt{p}-2\alpha (p)\right)}^{k}+{\left(3\sqrt{p}+2\alpha (p)\right)}^{k}\right],}\end{array}$$

*where* *d* *is uniquely determined by* 4*p* = *d*^{2} + 27*b*^{2} *and* *d* ≡ 1 mod 3.

Let *k* and *h* be two positive integers, *p* is a prime with *p* = 24*r* + 13, and *M*(*h*, *k*; *p*) denotes the number of solutions of the congruence equation

$$\begin{array}{}{\displaystyle {x}_{1}^{3}+\cdots +{x}_{h}^{3}+{y}_{1}^{4}+\cdots +{y}_{k}^{4}\equiv {z}_{1}^{3}+\cdots +{z}_{h}^{3}+{w}_{1}^{4}+\cdots +{w}_{k}^{4}modp,}\end{array}$$

where 0 ≤ *x*_{i}, *z*_{i} *y*_{j}, *w*_{j} ≤ *p* − 1, *i* = 1, 2, ⋯, *h*, *j* = 1, 2, ⋯, *k*.

Then from Theorem 1.1 we can give an exact computational method for *M*(*h*, *k*; *p*). In particular, we have the following:

#### Corollary 1.6

*If* *p* = 24*r* + 13 *is an odd prime*, *then for any positive integer* *k*, *we have*

$$\begin{array}{}{\displaystyle M(2,k;p)={p}^{2k+3}+3(p-1)\cdot {p}^{\frac{k+2}{2}}\cdot \left[{\left(3\sqrt{p}-2\alpha (p)\right)}^{k}+{\left(3\sqrt{p}+2\alpha (p)\right)}^{k}\right].}\end{array}$$

If prime *p* = 24*r* + 1, then the situation is more complex, we can only give an effective calculation method one by one. Theorem 1.7 indicates some examples of it.

#### Theorem 1.7

*If* *p* *is an odd prime with* *p* = 24*r* + 1, *then we have the identities*

$$\begin{array}{}{\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sum _{m=1}^{p-1}{\left|\sum _{a=0}^{p-1}e\left(\frac{m{a}^{3}}{p}\right)\right|}^{2}\cdot {\left|\sum _{a=0}^{p-1}e\left(\frac{m{a}^{4}}{p}\right)\right|}^{2}=6(p-1)\cdot {p}^{2}.}\\ \\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\sum _{m=1}^{p-1}{\left|\sum _{a=0}^{p-1}e\left(\frac{m{a}^{3}}{p}\right)\right|}^{4}\cdot {\left|\sum _{a=0}^{p-1}e\left(\frac{m{a}^{4}}{p}\right)\right|}^{4}=6{p}^{3}(p-1)\left(17p+4{\alpha}^{2}(p)\right).}\\ \\ {\displaystyle \sum _{m=1}^{p-1}{\left|\sum _{a=0}^{p-1}e\left(\frac{m{a}^{3}}{p}\right)\right|}^{6}\cdot {\left|\sum _{a=0}^{p-1}e\left(\frac{m{a}^{4}}{p}\right)\right|}^{6}=3{p}^{4}(p-1)\left(33p+28{\alpha}^{2}(p)\right)\cdot \left(18p+{d}^{2}\right).}\end{array}$$

**Some notes**: If 3 ∤ (*p* − 1), then for any integer *m* with (*m*, *p*) = 1, we have

$$\begin{array}{}{\displaystyle \left|\sum _{a=0}^{p-1}e\left(\frac{m{a}^{3}}{p}\right)\right|=\left|\sum _{a=0}^{p-1}e\left(\frac{ma}{p}\right)\right|=0.}\end{array}$$

If prime *p* = 4*r* + 3, then we have

$$\begin{array}{}{\displaystyle \left|\sum _{a=0}^{p-1}e\left(\frac{m{a}^{4}}{p}\right)\right|=\left|1+\sum _{a=1}^{p-1}e\left(\frac{m{a}^{2}}{p}\right)+\sum _{a=1}^{p-1}\left(\frac{a}{p}\right)e\left(\frac{m{a}^{2}}{p}\right)\right|=\sqrt{p}.}\end{array}$$

So in these cases, the problem we are studying is trivial.

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