For integers $m, n, q, k$ with $q\geq3, k\geq2$, we define a two-term exponential sums

$ C(m, n, k;q)= \sum\limits_{a = 1}^q {'e(\frac{{m{a^k} + na}}{q})}, $

(1.1)

where $e(y)=e^{2\pi iy}$ and $ \sum\limits_{a = 1}^q {'}$ denotes the summation over all $a$ with $(a, q)=1$. The two-term exponential sums $C(m, n, k;q)$ originally arose in connection with Waring's problem and the aim is to find optimal bounds. As a pioneer work, Davenport and Heilbronn [2] proved that

$ C(m, n, k;p^\alpha)\ll {}_kp^{\alpha\theta}(m, p^{\alpha}) $

(1.2)

for $(p, m)=1$, where $\theta=2/3$ for $k=3$ and $\theta=3/4$ for $k>3$. Afterwards, Hua [9] showed that $\theta=1/2$ for all $k\geq2$ by using Weil's estimate for exponential sums over finite fields. Till now, many improvements for (1.2) were made by Loxton, Vaughan and Smith [5, 6, 11]. Carlitz [7, 8] studied the computation problem of the two-term exponential $C(m, n, k;p)$ over finite fields and obtained the computational formulas for $k=3$ and $k=p+1$. As to the two-term exponential sums with Dirichlet character $C(m, n, k, \chi, q)=\sum\limits_{a = 1}^q {'\chi (a)e(\frac{{m{a^k}+na}}{q})}$, Xu [13], Liu [3], Chen [14, 15], Ai [16] and Calderon [1] also acquired a lot of research results. More, about the three-term exponential sums, there were also some interesting results [17-19].

Though the single value of $C(m, n, k;q)$ is irregular, the high power means that value of $C(m, n, k;q)$ owns graceful arithmetical properties and it in turn becomes an interesting focus for many attentions. In 2010, Liu [4] acquired the computational formula of the fourth mean value, i.e., when $p$ is an odd prime with $(n, p)=1$, then

$ {\sum\limits_{m = 1}^{p}{'\left|{C(m, n, k;p)}\right|^4}}=\left\{\begin{array}{l} {(p-1)^4+p-2}, \quad \quad \mbox{if} \quad k=1;\\ {p^3-p^2-7p-1-(-1)^{(p-1)/2}\cdot2p}, \quad \mbox{if} \quad k=2;\\ {2p^3-3p^2-3p-1}, \quad \mbox{if} \quad k>0 \quad \mbox{and}\quad k \equiv -1(\bmod p-1). \end{array} \right. $

In 2011, Wang, Zhang [12] studied the computation problem of the fourth moment of two-term mixed exponential sums with elementary algebraic method. They proved that when $p$ is a prime and $(n, p)=1$, then

$ {\sum\limits_{m = 1}^{p}{\left|{C(m, n, 2;p)}\right|^4}}=\left\{\begin{array}{l} {p(p^2-p-9)}, \quad \mbox{if} \quad p \equiv 1(\bmod 4);\\ {p(p^2-p-5)}, \quad \mbox{if} \quad p \equiv 3(\bmod 4). \end{array} \right. $

When $p$ is a prime, $(n, p)=1$ and $(3, p-1)=1$, then

$ {\sum\limits_{m = 1}^{p}{\left|{C(m, n, 3;p)}\right|^4}}= {2p^3-3p^2-3p}. $

Unfortunately, though Liu, Wang got the explicit formulas of $\sum\limits_{m = 1}^{p}{\left|{C(m, n, k;p)}\right|^4}$ with $k\geq1, k=-1, 1, 2, 3 (\bmod(p-1))$, the result under the condition $k\geq1, k\equiv5 (\bmod(p-1))$ was not solved. In this paper, this computation problem will be solved and the explicit formulas will be given. Moreover we shall give a transform formula and a lower bound formula for the fourth moment of two-term exponential sums. The main results are the following two theorems.

Theorem 1.1   Let $p$ be a prime with $p\geq5, (5, p-1)=1$, $n$ be an integer with $(n, p)=1$, then for $k\geq1, k\equiv5 (\bmod(p-1))$, we have

$ {\sum\limits_{m = 1}^{p}{\left|{C(m, n, k;p)}\right|^4}}=\left\{\begin{array}{l} {3p^3-8p^2-3p}, \quad p \equiv 5 (\bmod 12);\\ {3p^3-16p^2-3p}, \quad p \equiv 1 (\bmod 12);\\ {3p^3-10p^2-3p}, \quad p \equiv -5 (\bmod 12);\\ {3p^3-2p^2-3p}, \quad p \equiv -1 (\bmod 12). \end{array} \right. $

Theorem 1.2   Let $p$ be a prime with $p\geq3, (k, p-1)=1$, $n$ be an integer with $(n, p)=1$, then we have

$ {\sum\limits_{m = 1}^{p}{\left|{C(m, n, k;p)}\right|^4}}\geq p(2p^2-3p-3). $

To prove the main results, necessary lemmas are listed and proved as below.

Lemma 2.1   For arbitrary integers $a, b, c$, let $p$ be an odd prime with $(a, p)=1$ and denote $N_1$ as the number of the solutions of the congruence equation $ax^2+bx+c\equiv 0 (\bmod p), $ then

${N_1=1+\left(\frac{b^2-4ac}{p}\right)}.$

Proof   From Theorem 3.5.1 in ref. [10], we immediately get the result.

Lemma 2.2   Let $p$ be an odd prime, $N_2$ denote the number of the solutions of the congruence equation $c^2-c+1\equiv 0 (\bmod p), $ then

$ {N_2}=\left\{\begin{array}{l} 2, \quad p \equiv 1, -5 (\bmod 12);\\ 0, \quad p \equiv5, -1 (\bmod 12). \end{array} \right. $

And if $ p \equiv 1, -5 (\bmod 12), 1, p$ are not solutions.

Proof   Since $(1, p)=1$, by Lemma 2.1, we have

$ {N_2=1+\left(\frac{-3}{p}\right)}=1+\left(\frac{p}{3}\right). $

If $p\equiv 1, -5(\bmod 12), $ then $ \left(\frac{p}{3}\right)=1$;

If $p\equiv -1, 5(\bmod 12), $ then $ \left(\frac{p}{3}\right)=-1$.

In conclusion, we have

$ {N_2=1+\left(\frac{p}{3}\right)=\left\{\begin{array}{l} 2, \quad p \equiv 1, -5(\bmod 12);\\ 0, \quad p \equiv5, -1 (\bmod 12). \end{array} \right.} $

And straight forward calculation shows that $1, p$ are not solutions.

Lemma 2.3   Let $p$ be an odd prime, $a^2-4b\not \equiv0(\bmod p), $ then $\sum\limits_{x = 1}^p {\left(\frac{x^2+ax+b}{p}\right)=-1}$, substituting $0 $ for the term in the formula with $ p\left. \right|{x^2+ax+b}.$

Proof   See Theorem 7.8.2 in ref. [10].

Lemma 2.4   Let $p$ be an odd prime, $k$ be an odd positive integer and $N_{k, p}$ denote the number of the solutions of the congruence equation

$ {(a^k-1)(c-1)^k\equiv(c^k-1)(a-1)^k(\bmod p)}, $

(2.1)

where $a, c$ are integers with $2\leq a, c \leq p-1$, then we have $N_{k, p}\geq2p-5$.

Proof   It is obviously to show that $a\equiv c(\bmod p)$ is fit for equation (2.1), now we consider the case $c\equiv \overline a(\bmod p)$.

After substituting $c\equiv \overline a(\bmod p)$ into the left part of formula (2.1), we have

$ {(a^k-1)(\overline a-1)^k\equiv(a^k-1){\overline a}^k(1-a)^k(\bmod p)}. $

Again, $c\equiv \overline a(\bmod p)$ is substituted into the right part of (2.1). Since $k$ is an odd integer, then

$ {(\overline a^k-1)(a-1)^k\equiv\overline a^k(1-a^k)(a-1)^k\equiv (a^k-1){\overline a}^k(1-a)^k(\bmod p)}. $

Therefore

$ {(a^k-1)(\overline a-1)^k\equiv(\overline a^k-1)(a-1)^k(\bmod p)}. $

So $c\equiv \overline a(\bmod p)$ is also fit for equation (2.1).

Moreover $a\equiv c(\bmod p)$ and $a\equiv \overline c(\bmod p)$ have the same solution $(a, c)=(p-1, p-1)$. Hence $N_{k, p}\geq2p-5.$

Lemma 2.5   Let $p$ be a prime with $p>3$ and $N_3$ denote the number of the solutions of the congruence equation

$ (c^2-c+1)a^2-(c^2+1)a+(c^2-c+1)\equiv 0 (\bmod p), $

(2.2)

where $a, c$ are integers with $2\leq a, c \leq p-1$, then we have

$ {N_3}=\left\{\begin{array}{l} {p-1}, \quad p \equiv 5 (\bmod 12);\\ {p-9}, \quad p \equiv 1 (\bmod 12);\\ {p-7}, \quad p \equiv -5 (\bmod 12);\\ {p+1}, \quad p \equiv -1 (\bmod 12). \end{array} \right. $

Proof   Case 1   For a fixed $c, 2\leq c \leq p-1$, if $c^2-c+1\not\equiv 0(\bmod p), $ from Lemma 2.1, the number of the solutions of equation (2.2) is

$ {1+\left(\frac{(c^2+1)^2-4(c^2-c+1)^2}{p}\right)=1+\left(\frac{-c^2+2c-1}{p}\right)\left(\frac{3c^2-2c+3}{p}\right)}\\ =1+\left(\frac{-(c-1)^2}{p}\right)\left(\frac{3}{p}\right)\left(\frac{c^2-2\cdot\overline3+1}{p}\right) =1+\left(\frac{-3}{p}\right)\left(\frac{c^2-2\cdot\overline3+1}{p}\right), $

where $\overline 3$ satisfies $3\cdot\overline 3\equiv1 (\bmod p)$. If $a\equiv 1 (\bmod p)$ satisfies equation (2.2), then $c\equiv 1(\bmod p)$; If $a\equiv 0 (\bmod p)$ satisfies equation (2.2), then $c^2-c+1\equiv 0 (\bmod p), $ that contradicts.

Case 2  For a fixed $c, 2\leq c \leq p-1$, if $c^2-c+1\equiv 0(\bmod p), $ then equation (2.2) is $(c^2+1)a\equiv 0(\bmod p), $ namely $ca\equiv 0(\bmod p), $ therefore congruence equation (2.2) has no solution. So we have

$ N_3=\mathop {\sum\limits_{c = {\rm{2}}}^{p - 1} {} } \limits_{\begin{array}{*{20}{c}} {{c^2-c+1}\not\equiv 0(\bmod p)} \end{array}} {\left[1+\left(\frac{(c^2+1)^2-4(c^2-c+1)^2}{p}\right)\right]}\\ =\mathop {\sum\limits_{c = {\rm{2}}}^{p - 1} {} } {\left[1+\left(\frac{(c^2+1)^2-4(c^2-c+1)^2}{p}\right)\right]}\\ -\mathop {\sum\limits_{c = {\rm{2}}}^{p - 1} } \limits_{\begin{array}{*{20}{c}} {{c^2-c+1}\equiv 0(\bmod p)} \end{array}} {\left[1+\left(\frac{(c^2+1)^2-4(c^2-c+1)^2}{p}\right)\right]}\\ =A-B.\\ A={\sum\limits_{c = {\rm{2}}}^{p - 1} {}} \left[1+\left(\frac{-c^2+2c-1}{p}\right)\left(\frac{3c^2-2c+3}{p}\right)\right]\\ ={\sum\limits_{c = {\rm{2}}}^{p - 1} {}} \left[1+\left(\frac{-3}{p}\right)\left(\frac{c^2-2\cdot \overline 3 c+1}{p}\right)\right]\\ =p-2+\left(\frac{-3}{p}\right)\cdot \sum\limits_{c = {\rm{2}}}^{p - 1} {}\left(\frac{c^2-2\cdot \overline 3 c+1}{p}\right)\\ =p-2+\left(\frac{-3}{p}\right)\cdot \sum\limits_{c = {\rm{1}}}^{p} {}\left[\left(\frac{c^2-2\cdot \overline 3 c+1}{p}\right)-\left(\frac{3}{p}\right)-1\right]. $

By using Lemma 2.3, we have

$ A=p-2+\left(\frac{-3}{p}\right)\cdot\left[-2-\left(\frac{3}{p}\right)\right]= p-2+\left(\frac{-1}{p}\right)-2\left(\frac{-3}{p}\right). $

Now we compute $B$, from Lemma 2.2, we have

$ B=\mathop {\sum\limits_{c = {\rm{2}}}^{p - 1} {} } \limits_{\begin{array}{*{20}{c}} {{c^2-c+1}\equiv 0 (\bmod p)} \end{array}} {\left[1+\left(\frac{(c^2+1)^2}{p}\right)\right]}\\ =\mathop {\sum\limits_{c = {\rm{2}}}^{p - 1} {} } \limits_{\begin{array}{*{20}{c}} {{c^2-c+1}\equiv 0 (\bmod p)} \end{array}} {\left[1+\left(\frac{c^2}{p}\right)\right]} =2\cdot\mathop {\sum\limits_{c = {\rm{2}}}^{p - 1} {} } \limits_{\begin{array}{*{20}{c}} {{c^2-c+1}\equiv 0 (\bmod p)} \end{array}} {1}\\ =2\cdot N_2, $

where ${N_2}=\left\{\begin{array}{l} 2, \quad p \equiv 1, -5 (\bmod 12);\\ 0, \quad p \equiv5, -1 (\bmod 12). \end{array} \right.$ Therefore

$ N_3=p-2-\left(\frac{-1}{p}\right)-2\left(\frac{-3}{p}\right)-2\cdot N_2 . $

If $p\equiv 5(\bmod 12)$, then $\left(\frac{-1}{p}\right)=1$, $\left(\frac{-3}{p}\right)=-1$, therefore $N_3=p-1$.

If $p\equiv 1(\bmod 12)$, then $\left(\frac{-1}{p}\right)=1$, $\left(\frac{-3}{p}\right)=1$, therefore $N_3=p-9$.

If $p\equiv -5(\bmod 12)$, then $\left(\frac{-1}{p}\right)=-1$, $\left(\frac{-3}{p}\right)=1$, therefore $N_3=p-7$.

If $p\equiv -1(\bmod 12)$, then $\left(\frac{-1}{p}\right)=-1$, $\left(\frac{-3}{p}\right)=-1$, therefore $N_3=p+1$.

In conclusion, we have

$ {N_3}=\left\{\begin{array}{l} {p-1}, \quad p \equiv 5 (\bmod 12);\\ {p-9}, \quad p \equiv 1 (\bmod 12);\\ {p-7}, \quad p \equiv -5 (\bmod 12);\\ {p+1}, \quad p \equiv -1 (\bmod 12). \end{array} \right. $

Lemma 2.6   Let $p$ be a prime, $p>5$ and $N_{5, p}$ denote the number of the solutions of the congruence equation

$ {(a^5-1)(c-1)^5\equiv(c^5-1)(a-1)^5 (\bmod p)}, $

(2.3)

where $a, c$ are integers with $2\leq a, c \leq p-1$, then we have

$ {N_{5, p}}=\left\{\begin{array}{l} {3p-10}, \quad p \equiv 5 (\bmod 12);\\ {3p-18}, \quad p \equiv 1 (\bmod 12);\\ {3p-12}, \quad p \equiv -5 (\bmod 12);\\ {3p-4}, \quad p \equiv -1 (\bmod 12). \end{array} \right. $

Proof   By using factorization method, we know that equation (2.3) equivalents to

$ 5(c-1)(a-1)(a-c)(ac-1)[(c^2-c+1)a^2-(c^2+1)a+(c^2-c+1)]\equiv 0 (\bmod p). $

Noting that $p$ is a prime with $p>5$ and $2\leq a, c \leq p-1$, we have

$ (a-c)(ac-1)[(c^2-c+1)a^2-(c^2+1)a+(c^2-c+1)]\equiv 0 (\bmod p). $

Let

$ S_1=\{(a, c)|a-c\equiv 0 (\bmod p)\}, \\ S_2=\{(a, c)|ac\equiv 1 (\bmod p)\}, \\ S_3={\{(a, c)|(c^2-c+1)a^2-(c^2+1)a+(c^2-c+1)\equiv 0 (\bmod p)\}}, $

then

$ N_{5, p}=\left|S_1\cup S_2\cup S_3\right|=\left|S_1\right|+\left|S_2\right|+\left|S_3\right|-\left|S_1\cap S_2\right|-\left|S_1\cap S_3\right|-\left|S_2\cap S_3\right|+\left|S_1\cap S_2\cap S_3\right|, $

where $\left|\quad\right|$ denotes the number of the elements of the set.

(a) It is obviously that $S_1\cap S_2=\{(p-1, p-1)\}$ and thus $\left|S_1\cap S_2\right|=1.$

(b) If $(a, c)\in S_1\cap S_3, $ then \begin{equation*} (c^2+1)(c-1)^2\equiv 0 (\bmod p). \end{equation*} Since $c\not\equiv1 (\bmod p)$, we have

$ c^2\equiv -1(\bmod p). $

(2.4)

If $p\equiv 1 (\bmod 4), $ then $\left(\frac{-1}{p}\right)=1$, so equation (2.4) has two solutions and obviously $c\equiv0, 1, p-1 (\bmod p)$ are not solutions. Therefore

$ \left|S_1\cap S_3\right|=2, \left|S_1\cap S_2\cap S_3\right|=0. $

If $p\equiv -1 (\bmod 4), $ then $\left(\frac{-1}{p}\right)=-1$, so equation (2.4) has no solution. Therefore

$ \left|S_1\cap S_3\right|=0, \left|S_1\cap S_2\cap S_3\right|=0. $

(c) If $(a, c)\in S_2\cap S_3, $ then we substitute $a\equiv\overline c (\bmod p)$ into the equation

$ (c^2-c+1)a^2-(c^2+1)a+(c^2-c+1)\equiv 0 (\bmod p), $

then

$ (c^2-c+1)(\overline c)^2-(c^2+1)\overline c+(c^2-c+1)\equiv 0 (\bmod p). $

Thus

$ (c^2-c+1)c^2-(c^2+1)c+(c^2-c+1)\equiv 0 (\bmod p), $

namely,

$ (c^2+1)(c-1)^2\equiv 0 (\bmod p). $

Now we can see that the case is similar to case (b). Therefore we have if $p\equiv 1 (\bmod 4)$, then $\left|S_2\cap S_3\right|=2; $ if $p\equiv -1 (\bmod 4)$, then $\left|S_2\cap S_3\right|=0.$ So \begin{equation*} N_{5, p}=2p-5+\left|S_3\right|-\left|S_1\cap S_3\right|-\left|S_2\cap S_3\right|. \end{equation*}

From Lemma 2.5, we have

$ {S_3=\left\{\begin{array}{l} {p-1}, \quad p \equiv 5 (\bmod 12);\\ {p-9}, \quad p \equiv 1 (\bmod 12);\\ {p-7}, \quad p \equiv -5 (\bmod 12);\\ {p+1}, \quad p \equiv -1 (\bmod 12). \end{array} \right.} $

Then

$ {{N_{5, p}}=\left\{\begin{array}{l} {3p-10}, \quad p \equiv 5 (\bmod 12);\\ {3p-18}, \quad p \equiv 1 (\bmod 12);\\ {3p-12}, \quad p \equiv -5 (\bmod 12);\\ {3p-4}, \quad p \equiv -1 (\bmod 12). \end{array} \right.} $

Lemma 2.7   Let $p$ be an odd prime with $(n, p)=1$ and $(k, p-1)=1$, then we have

$ {\sum\limits_{m = 1}^{p}{\left|{C(m, n, k;p)}\right|^4}}=2p^2-3p+p^2\cdot N_{k, p} . $

Proof   For integer $r$ satisfying $(r, p)=1$, we have $(\overline r, p)=1$, where $r\overline r\equiv1 (\bmod p)$. Thus we have

$ {\sum\limits_{m = 1}^{p}{\left|{C(m, n, k;p)}\right|^4}}={\sum\limits_{m = 1}^p{\left| {\sum\limits_{a = 1}^{p} {'e(\frac{{m{a^k} + na}}{p})} } \right|} } ^4={\sum\limits_{m = 1}^p{\left| {\sum\limits_{a = 1}^{p-1} {e(\frac{{m{{\overline r}^ka^k} + na}}{p})} } \right|} } ^4\\ = \frac{1}{{p - 1}}\sum\limits_{r = 1}^{p-1} {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{m{{\overline r}^k(ra)^k} + n(ra)}}{p})} } \right|} } ^4 \\ = \frac{1}{{p - 1}}\sum\limits_{r = 1}^{p-1} {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{ma^k} + n(ra)}{p})}} \right|} } ^4 \\ = \frac{1}{{p - 1}}\sum\limits_{r = 1}^{p} {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{ma^k} + n(ra)}{p})}} \right| } }^4 -\frac{1}{{p - 1}}{\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1}} {e(\frac{{ma^k}}{p})} \right| }} ^4 \\ = T_1-T_2. $

Noting that $(k, p-1)=1$, then

$ T_2=\frac{1}{{p - 1}}{\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{ma^k}}{p})} } \right| } ^4}=\frac{1}{{p - 1}}{\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{ma}}{p})} }\right| }} ^4 \nonumber\\ = \frac{1}{{p - 1}}{\left[(p-1)^4+{\sum\limits_{m = 1}^{p-1} {\left| {\sum\limits_{a = 1}^{p-1} {e(\frac{{ma}}{p})}} \right| }} ^4\right]}=(p-1)^3+1.\\ T_1= \frac{1}{{p - 1}}\sum\limits_{r = 1}^{p} {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1}\sum\limits_{b = 1}^{p - 1} {e(\frac{{m(a^k-b^k)} + nr(a-b)}{p})}}\right| }} ^2\\ = \frac{1}{{p - 1}}\sum\limits_{r = 1}^{p} {\sum\limits_{m = 1}^p {\left| p-1+{\sum\limits_{a = 2}^{p - 1}\sum\limits_{b = 1}^{p - 1} {e(\frac{{mb^k(a^k-1)} + nrb(a-1)}{p})}}\right| }} ^2\\ =\frac{1}{{p - 1}}\sum\limits_{r = 1}^{p} {\sum\limits_{m = 1}^p {\left| p-1+{\sum\limits_{a = 2}^{p - 1}\sum\limits_{b = 1}^{p - 1} {e(\frac{{mb^k(\overline{a-1})^k(a^k-1)} + nrb}{p})}}\right| } }^2\\ =\frac{1}{{p - 1}}\sum\limits_{r = 1}^{p} {\sum\limits_{m = 1}^p {\left[p-1+{\sum\limits_{a = 2}^{p-1}\sum\limits_{b = 1}^{p-1} {e(\frac{{mb^k(\overline{a-1})^k(a^k-1)} + nrb}{p})}}\right] }}\\ \cdot {\left[p-1+{\sum\limits_{c = 2}^{p-1}\sum\limits_{d = 1}^{p-1} {e(\frac{{-md^k(\overline{c-1})^k(c^k-1)}-nrd}{p})}}\right] } $

$ =(p-1)p^2+\sum\limits_{r = 1}^{p} {\sum\limits_{m = 1}^p {\sum\limits_{a = 2}^{p - 1}\sum\limits_{b = 1}^{p - 1} {e(\frac{{m(b(\overline{a-1}))^k(a^k-1)} + nrb}{p})} }}\\ +\sum\limits_{r = 1}^{p} {\sum\limits_{m = 1}^p {\sum\limits_{c = 2}^{p - 1}\sum\limits_{d= 1}^{p - 1} {e(\frac{{-m(d(\overline{c-1}))^k(c^k-1)} - nrd}{p})} }}\\ +\frac{1}{{p - 1}}\sum\limits_{r = 1}^{p} \sum\limits_{m = 1}^p {\sum\limits_{a = 2}^{p - 1}\sum\limits_{b = 1}^{p - 1}{{\sum\limits_{c = 2}^{p - 1}\sum\limits_{d = 1}^{p - 1}} }} {e(\frac{{m \left[(b(\overline{a-1}))^k(a^k-1)-(d(\overline{c-1}))^k(c^k-1)\right]} + nr(b-d)}{p})}\\ =(p-1)p^2+T_{11}+T_{12}+T_{13}.\\ T_{11}=\sum\limits_{r = 1}^{p} {\sum\limits_{m = 1}^p {{\sum\limits_{a = 2}^{p - 1}\sum\limits_{b = 1}^{p - 1} {e(\frac{{m(b(\overline{a-1}))^k(a^k-1)} + nrb}{p})}}} }\\ =\sum\limits_{m = 1}^p {\sum\limits_{a = 2}^{p - 1}\sum\limits_{b = 1}^{p - 1} {e(\frac{m(b(\overline{a-1}))^k(a^k-1)}{p})} }\sum\limits_{r = 1}^{p} {e(\frac{nrb}{p})}. $

With the condition $(n, p)=1$ and from the trigonometric identity,

$ \sum\limits_{a = 1}^{m} {e(\frac{na}{m})}= \left\{ \begin{array}{l} m, \quad m\left. \right|n, \\ 0, \quad m\nmid n. \end{array} \right. $

We have $\sum\limits_{r = 1}^{p} {e(\frac{nrb}{p})}=0, $ therefore $T_{11}=0$. Similarly, $T_{12}=0$,

$ T_{13}=\frac{1}{{p - 1}}\sum\limits_{r = 1}^{p} \sum\limits_{m = 1}^p \sum\limits_{a = 2}^{p - 1}\sum\limits_{b = 1}^{p - 1}\sum\limits_{c = 2}^{p - 1}\sum\limits_{d = 1}^{p - 1} {e(\frac{{m\left[(b(\overline{a-1}))^k(a^k-1)-(d(\overline{c-1}))^k(c^k-1)\right]} + nr(b-d)}{p})}\\ =\frac{1}{{p - 1}} \sum\limits_{m = 1}^p \sum\limits_{a = 2}^{p - 1} \sum\limits_{b = 1}^{p - 1}\sum\limits_{c = 2}^{p - 1}\sum\limits_{d = 1}^{p - 1} {e(\frac{{m\left[(b(\overline{a-1}))^k(a^k-1)-(d(\overline{c-1}))^k(c^k-1)\right]} }{p})}\sum\limits_{r = 1}^{p}{e(\frac{nr(b-d)}{p})}\\ =\frac{p}{{p - 1}} \mathop{\sum\limits_{m = 1}^p \sum\limits_{a = 2}^{p - 1} \sum\limits_{b = 1}^{p - 1}\sum\limits_{c = 2}^{p - 1}\sum\limits_{d = 1}^{p - 1}}\limits_{b\equiv d (\bmod p)} {e(\frac{{m\left[(b(\overline{a-1}))^k(a^k-1)-(d(\overline{c-1}))^k(c^k-1)\right]} }{p})}\\ =\frac{p}{{p - 1}} \sum\limits_{m = 1}^p {\sum\limits_{a = 2}^{p - 1}\sum\limits_{b = 1}^{p - 1}{{\sum\limits_{c = 2}^{p - 1} {e(\frac{{m b\left[(\overline{a-1})^k(a^k-1)-(\overline{c-1})^k(c^k-1)\right]} }{p})} }}}\\ =\frac{p}{{p - 1}} \sum\limits_{c = 2}^{p-1} {\sum\limits_{a = 2}^{p - 1}\sum\limits_{b = 1}^{p - 1}{{\sum\limits_{m = 1}^{p} {e(\frac{{mb\left[(\overline{a-1})^k(a^k-1)-(\overline{c-1})^k(c^k-1)\right]} }{p})} }}}\\ =p^2 \cdot\mathop {\sum\limits_{a = 2}^{p - 1} {\sum\limits_{c = 2}^{p - 1} 1 } }\limits_{ (\overline{a-1})^k(a^k-1)-(\overline{c-1})^k(c^k-1)\equiv 0 (\bmod p)}\\ =p^2\cdot\mathop {\sum\limits_{a = 2}^{p - 1} {\sum\limits_{c = 2}^{p - 1} 1 } }\limits_{ (a^k-1)(c-1)^k\equiv(c^k-1)(a-1)^k (\bmod p)}\\ =p^2\cdot N_{k, p}. $

So

$ {\sum\limits_{m = 1}^{p}{\left|{C(m, n, k;p)}\right|^4}}=2p^2-3p+p^2\cdot N_{k, p}. $

Thus all of the lemmas are shown. Besides, the result of Lemma 2.7 shows that the difficulty of calculating $\sum\limits_{m = 1}^{p}{\left|{C(m, n, k;p)}\right|^4}$ is mainly stemmed from computing exactly number of the solutions of high power congruence equation.

First we prove Theorem 1.1.

Proof   By Lemma 2.7, we have

$ {\sum\limits_{m = 1}^{p}{\left|{C(m, n, k;p)}\right|^4}}={\sum\limits_{m = 1}^{p}{\left|{C(m, n, 5;p)}\right|^4}}=2p^2-3p+p^2\cdot N_{5, p}. $

From Lemma 2.6, we have

$ {{N_{5, p}}=\left\{\begin{array}{l} {3p-10}, \quad p \equiv 5 (\bmod 12);\\ {3p-18}, \quad p \equiv 1 (\bmod 12);\\ {3p-12}, \quad p \equiv -5 (\bmod 12);\\ {3p-4}, \quad p \equiv -1 (\bmod 12). \end{array} \right.} $

Therefore

$ {\sum\limits_{m = 1}^{p}{\left|{C(m, n, k;p)}\right|^4}}=\left\{\begin{array}{l} {3p^3-8p^2-3p}, \quad p \equiv 5 (\bmod 12);\\ {3p^3-16p^2-3p}, \quad p \equiv 1 (\bmod 12);\\ {3p^3-10p^2-3p}, \quad p \equiv -5 (\bmod 12);\\ {3p^3-2p^2-3p}, \quad p \equiv -1 (\bmod 12). \end{array} \right. $

This proves Theorem 1.1.

Finally we complete the proof of Theorem 1.2.

Proof   By Lemma 2.7 and Lemma 2.4, we have

$ {\sum\limits_{m = 1}^{p}{\left|{C(m, n, k;p)}\right|^4}}=2p^2-3p+p^2\cdot N_{k, p}\\ \geq 2p^2-3p+p^2\cdot (2p-5)=p(2p^2-3p-3). $