指数和与著名的华林问题有着密切的联系, 它在华林问题的主项研究中起着重要的作用.许多著名的学者如华罗庚、Weil、高斯等对指数和的上界估计做出了重要的贡献^[1-6].近些年来, 指数和高次均值的计算成了这一领域的热点, 相关研究成果丰硕^{[7-11, 13-17]}.

设$q, m, s, n, k, t$为整数, 且$q\ge 3$, 定义二项指数和与三项指数和如下:

$\begin{eqnarray*} &&\sum\limits_{a = 1}^q {'e(\frac{{m{a^k} + na}}{q})}, \\ &&\sum\limits_{a = 1}^q {'e(\frac{{m{a^k} + s{a^t} + na}}{q})}, \end{eqnarray*}$

其中$e(x)=e^{2\pi ix}$, $ \sum\limits_{a = 1}^q {'}$表示对所有满足$(a, q)=1$的整数a求和.

国内外众多学者对于二项指数和的各种性质做了深入细致的研究, 取得了众多的研究成果^{[7-11, 16, 17]}.但是关于三项指数和的各类性质, 国内外的相关研究尚不多见.

1972年, Mordel^[12]利用三项指数和成功的研究$\mod p$剩余类的有理函数表示.

2012年, 陈华等人^[13]研究了带Dirichlet特征的三项指数和四次均值的计算公式, 并给出了$\sum\limits_{t = 1}^q{'\sum\limits_{m = 1}^q} {\sum\limits_{\chi {\rm mod} q} {|\sum\limits_{a = 1}^q {'\chi (a)e(\frac{{m{a^k} + t{a^2} + na}}{q})} {|^4}} } $的精确计算公式.

2014年, 论文^[14]给出了三项指数和的四次均值

$\sum\limits_{s = 1}^p {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{m{a^k} + s{a^t} + na}}{p})} } \right|} } ^4$

在$k = 2t, (t, p -1) = 1, 2$时的精确计算公式, 主要结果如下:

命题1.1  设$p$为素数, 则对任意固定的满足条件$(n, p)=1$的正整数$n$, 有

$\begin{eqnarray*} &&\sum\limits_{s = 1}^p {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{m{a^{2t}} + s{a^t} + na}}{p})} } \right|} } ^4 = {p^2}(2{p^2} - 5p + 3), (t, p - 1) = 1;\\ &&\sum\limits_{s = 1}^p {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{m{a^{2t}} + s{a^t} + na}}{p})} } \right|} } ^4 = {p^2}(2{p^2} - 11p + 16), (t, p - 1) = 2. \end{eqnarray*}$

本文将进一步深入讨论三项指数和四次均值的计算问题, 在命题1.1的条件下, 给出在$(t, p-1)=3, 4$时的精确计算公式, 本文主要结果见定理1.1.此外, 本文还将给出命题1.1和定理1.1的一个简化证明方法.

定理1.1  设$p$为奇素数, 则对任意固定的满足条件$(n, p)=1$的正整数$n$, 有

$\begin{align} & \sum\limits_{s=1}^{p}{{{\sum\limits_{m=1}^{p}{\left| \sum\limits_{a=1}^{p-1}{e(\frac{m{{a}^{2t}}+s{{a}^{t}}+na}{p})} \right|}}^{4}}}={{p}^{2}}(2{{p}^{2}}-17p+45),(t,p-1)=3; \\ & \sum\limits_{s=1}^{p}{{{\sum\limits_{m=1}^{p}{\left| \sum\limits_{a=1}^{p-1}{e(\frac{m{{a}^{2t}}+s{{a}^{t}}+na}{p})} \right|}}^{4}}}=\left\{ \begin{array}{*{35}{l}} {{p}^{2}}(2{{p}^{2}}-25p+96),p\not{|}{{(-4)}^{t/4}}-1, \\ {{p}^{2}}(2{{p}^{2}}-29p+96),p\left| {{(-4)}^{t/4}} \right.-1, \\ \end{array} \right.(t,p-1)=4. \\ \end{align}$

首先给出定理证明中需要用到的几个引理.

引理2.1  设$p \ge 3, t$是正整数, $(n, p)=1, $则有

$\begin{equation*} {\sum\limits_{s = 1}^p {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{m{a^{2t}} + s{a^t} + na}}{p})} } \right|} } ^4} = {p^4} - {p^3} + [M + {(l- 1)^2}]{p^3} - {l^2}{p^2}(2p - 2 - l), \end{equation*}$

其中

$l = \left( {t,p - 1} \right),M = \mathop {\sum\limits_{a = 2}^{p - 1} {\sum\limits_{c = 2}^{p - 1} 1 } }\limits_{\begin{array}{*{20}{c}} {{{\left( {a\bar c} \right)}^t} \equiv 1\left( {\,{\rm mod} \,p} \right)}\\ {\begin{array}{*{20}{c}} {{{[(\left( {a - 1} \right))\overline {(\left( {c - 1} \right))} ]}^t} \equiv 1\,\left( {{\rm mod} \,p} \right)}\\ {\;\;\;\,{c^t}\not \equiv 1\,\left( {{\rm mod} \,p} \right)} \end{array}} \end{array}} .$

证  参见文献[14]中的引理3.

引理2.2  $\mathop {\sum\limits_{a = 2}^{p - 1} {\sum\limits_{c = 2}^{p - 1} 1 } }\limits_{\begin{array}{*{20}{c}} {{{\left( {a\bar c} \right)}^t} \equiv 1\left( {{\rm mod} \,p} \right)}\\ {\begin{array}{*{20}{c}} {{{\left[ {\left( {a - 1} \right)\overline {\left( {c - 1} \right)} } \right]}^t} \equiv 1\left( {\,{\rm mod} \,p} \right)}\\ {\;\;\;\,{c^t}\not \equiv 1\,\left( {{\rm mod} \,p} \right)} \end{array}} \end{array}} = \left\{ {\begin{array}{*{20}{c}} {p - 2,\quad \quad \quad \quad \quad \quad \left( {t,p - 1} \right) = 3;}\\ {\begin{array}{*{20}{l}} {p - 1,p\not |{{\left( { - 4} \right)}^{t/4}} - 1,\left( {t,p - 1} \right) = 4;}\\ {p - 5,p\left| {{{\left( { - 4} \right)}^{t/4}} - 1} \right.,\;{\mkern 1mu} \left( {t,p - 1} \right) = 4.} \end{array}} \end{array}} \right.$

证  易见$\mathop {\sum\limits_{a = 2}^{p -1} {\sum\limits_{c = 2}^{p -1} 1 } }\limits_{\begin{array}{*{20}{c}} {{{(a\overline c)}^t} \equiv 1~({\rm mod} p)}\\ {\scriptstyle{[(a-1)\overline {(c-1)}]^t} \equiv 1~({\rm mod} p)\atop \scriptstyle\quad {c^t}\not \equiv 1~({\rm mod} p)} \end{array}}$的值为如下同余方程组解的个数,

$\begin{equation} \left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ {(a\overline c )^t} \equiv 1~ ({\rm mod} p), \end{array}\\ {{{[(a-1)\overline {(c-1)}]}^t} \equiv 1~~({\rm mod} p), } \end{array}} \right. \end{equation}$

(2.1)

其中$2 \le a, c \le p -1.$

情形1  若$(t, p -1) = 3, $则${x^t} \equiv 1~({\rm mod} p)$有3个解, 分别记为$1, A, \overline A, $其中$ \overline A \equiv {A^2}({\rm mod} p), {A^3} \equiv 1~({\rm mod} p), $则(2.1) 等价于如下方程组:

$ \begin{equation*} \left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv c, Ac, \overline A c~({\rm mod} p), \end{array}\\ {a - 1 \equiv (c - 1), A(c - 1), \overline A (c - 1)~~({\rm mod} p).} \end{array}} \right. \end{equation*}$

(1) 若$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv c~({\rm mod} p), \end{array}\\ {a -1 \equiv c -1~({\rm mod} p), } \end{array}} \right.$则$\left\{ {\begin{array}{*{20}{c}} {{c^t}\not \equiv 1~({\rm mod} p), }\\ {a \equiv c~~({\rm mod} p), } \end{array}} \right.$满足${c^t} \equiv 1~({\rm mod} p)$的$c$有$(t, p -1) = 3$个, 此种情形下共有$p -1 -3 = p -4$个解;

(2) 若$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~ ({\rm mod} p), \\ a \equiv c~ ({\rm mod} p), \end{array}\\ {a -1 \equiv A(c -1)({\rm mod} p)} \end{array}} \right.$或$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~ ({\rm mod} p), \\ a \equiv c~ ({\rm mod} p), \end{array}\\ {a -1 \equiv \overline A(c -1)({\rm mod} p), } \end{array}} \right.$则$A, \overline A \equiv 1~({\rm mod} p), $与$A, \overline A \not \equiv 1~({\rm mod} p)$矛盾, 无解;

(3) 若$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv Ac~({\rm mod} p), \end{array}\\ {a -1 \equiv (c -1)({\rm mod} p)} \end{array}} \right.$或$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv \overline A c~({\rm mod} p), \end{array}\\ {a -1 \equiv (c -1)({\rm mod} p), } \end{array}} \right.$则$A, \overline A \equiv 1~({\rm mod} p), $与$A, \overline A \not \equiv 1~({\rm mod} p)$矛盾, 无解;

(4) 若$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv Ac~({\rm mod} p), \end{array}\\ {a -1 \equiv A(c -1)({\rm mod} p)} \end{array}} \right.$或$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv \overline A c~({\rm mod} p), \end{array}\\ {a -1 \equiv \overline A (c -1)({\rm mod} p), } \end{array}} \right.$则$A, \overline A \equiv 1~({\rm mod} p), $与$A, \overline A \not \equiv 1~({\rm mod} p)$矛盾, 无解;

(5) 若$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv \overline A c~({\rm mod} p), \end{array}\\ {a -1 \equiv A(c -1)~({\rm mod} p), } \end{array}} \right.$则$\left\{ {\begin{array}{*{20}{c}} {{c^t}\not \equiv 1~({\rm mod} p), }\\ {c \equiv 1 + A~({\rm mod} p), } \end{array}} \right.$由${A^3} \equiv 1~({\rm mod} p)$可得$1 + A + {A^2} \equiv 0({\rm mod} p), $则$1 + A \equiv -{A^2}({\rm mod} p), {c^t} \equiv {(-{A^2})^t} \equiv {(-1)^t}{A^2}^t \equiv -1\not \equiv 1~({\rm mod} p), $保留此解;

(6) 若$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^{2t}}\not \equiv 1~({\rm mod} p), \\ a \equiv Ac~({\rm mod} p), \end{array}\\ {a -1 \equiv \overline A (c -1)~({\rm mod} p), } \end{array}} \right.$则$\left\{ {\begin{array}{*{20}{c}} {{c^t}\not \equiv 1~({\rm mod} p), }\\ {c \equiv -A~({\rm mod} p), } \end{array}} \right.$同(5) 的情形, 保留此解.所以

$\begin{equation*} M = p - 2. \end{equation*}$

情形2  若$(t, p -1) = 4, $则${x^t} \equiv 1~({\rm mod} p)$有4个解, 分别记为$ \pm 1, \pm A, $其中${A^2} \equiv -1~({\rm mod} p)$所以(2.1) 式等价于如下方程组:

$\begin{equation*} \left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1\quad \quad \quad \quad\quad \quad \quad \quad({\rm mod} p), \\ a \equiv \pm c; \pm cA\quad \quad \quad \quad \quad \quad ({\rm mod} p), \end{array}\\ {a - 1 \equiv \pm (c - 1); \pm A(c - 1)({\rm mod} p), } \end{array}} \right. \end{equation*}$

且易见$-2\overline {(A -1)} \equiv A + 1~({\rm mod} p), {A^2} \equiv -1~({\rm mod} p), \overline A \equiv -A~({\rm mod} p), {A^4} \equiv 1~({\rm mod} p).$

(1) 若$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv c~({\rm mod} p), \end{array}\\ {a -1 \equiv c -1~({\rm mod} p), } \end{array}} \right.$则$\left\{ {\begin{array}{*{20}{c}} {{c^t}\not \equiv 1~({\rm mod} p), }\\ {a \equiv c({\rm mod} p), } \end{array}} \right.$满足${c^t} \equiv 1~({\rm mod} p)$的$c$有$(t, p -1) = 4$个, 则此种情形下共有$p -1 -(t, p -1) = p -5$个解;

(2) 若$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv -c~({\rm mod} p), \end{array}\\ {a -1 \equiv c -1~({\rm mod} p), } \end{array}} \right.$此种情形无解;

(3) 若$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv c~ ({\rm mod} p), \end{array}\\ {a -1 \equiv -c + 1~({\rm mod} p), } \end{array}} \right.$则$\left\{ {\begin{array}{*{20}{c}} {{c^t}\not \equiv 1~({\rm mod} p), }\\ {a \equiv c \equiv 1~({\rm mod} p), } \end{array}} \right.$此种情形无解;

(4) 若$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv -c~({\rm mod} p), \end{array}\\ {a -1 \equiv -c + 1~({\rm mod} p), } \end{array}} \right.$此种情形无解;

(5) 若$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv c~({\rm mod} p), \end{array}\\ {a -1 \equiv \pm A(c -1)({\rm mod} p), } \end{array}} \right.$ $ \left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv \pm Ac~({\rm mod} p), \end{array}\\ {a -1 \equiv c -1~({\rm mod} p), } \end{array}} \right.$或$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv \pm Ac~ ({\rm mod} p), \end{array}\\ {a -1 \equiv \pm A(c -1)({\rm mod} p), } \end{array}} \right.$则$A \equiv \pm 1~({\rm mod} p), $与$A\not \equiv \pm 1~({\rm mod} p)$矛盾, 无解;

(6) 若$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv -c~({\rm mod} p), \end{array}\\ {a -1 \equiv \pm A(c -1)({\rm mod} p), } \end{array}} \right.$则$\left\{ {\begin{array}{*{20}{c}} {{c^t}\not \equiv 1\; \; ({\rm mod} p)}, \\ \begin{array}{l} a \equiv \mp A~({\rm mod} p), \\ c \equiv \pm A~({\rm mod} p), \end{array} \end{array}} \right.$此时${c^t} \equiv {A^t} \equiv 1~({\rm mod} p)$与条件${c^t}\not \equiv 1~({\rm mod} p)$矛盾, 无解;

(7) 若$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv Ac~({\rm mod} p), \end{array}\\ {a -1 \equiv -c + 1~({\rm mod} p), } \end{array}} \right.$则$\left\{ {\begin{array}{*{20}{c}} {{c^t}\not \equiv 1\; \; ({\rm mod} p), }\\ \begin{array}{l} a \equiv A + 1~({\rm mod} p), \\ c \equiv 1 -A~({\rm mod} p), \end{array} \end{array}} \right.$此时

${c^t} \equiv {({c^4})^{t/4}} \equiv {[{(1-A)^4}]^{t/4}} \equiv {[{(1-2A + {A^2})^2}]^{t/4}} \equiv {[{(-2A)^2}]^{t/4}} \equiv {[4{A^2}]^{t/4}} \equiv {( - 4)^{t/4}}({\rm mod} p).$

若$p\not |{( - 4)^{t/4}} - 1,$则${c^t} \equiv {(1 -A)^t} \equiv {(-4)^{t/4}}\not \equiv 1~({\rm mod} p), $此时$(a, c) = (A + 1, 1 -A)$是解.

若$p\left. \right|{(-4)^{t/4}} -1, $则${c^t} \equiv {(1 -A)^t} \equiv {(-4)^{t/4}} \equiv 1~({\rm mod} p), $此时$(a, c) = (A + 1, 1 -A)$不是解;

(8) 若$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv -Ac~({\rm mod} p), \end{array}\\ {a -1 \equiv -c + 1~({\rm mod} p)}, \end{array}} \right.$ $ \left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv -Ac~({\rm mod} p), \end{array}\\ {a -1 \equiv A(c -1)({\rm mod} p), } \end{array}} \right.$或$\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} {c^t}\not \equiv 1~({\rm mod} p), \\ a \equiv Ac~ ({\rm mod} p), \end{array}\\ {a -1 \equiv -A(c -1)({\rm mod} p), } \end{array}} \right.$这三种情形类同(7).

所以

$\begin{equation*} M = \left\{ \begin{array}{l} p - 1, p\not | {( - 4)^{t/4}} - 1, \\ p - 5, p\left. \right|{( - 4)^{t/4}} - 1. \end{array} \right. \end{equation*}$

下面证明定理1.1.

在引理2.1中令$(t, p -1) = 3, $结合引理2.2可得

$\begin{eqnarray*} {\sum\limits_{s = 1}^p {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{m{a^{2t}} + s{a^t} + na}}{p})} } \right|} } ^4} &=& {p^4} - {p^3} + [M + {(l-1)^2}]{p^3} - {l^2}{p^2}(2p - 2 - l)\\ & = &{p^4} - {p^3} + [p-2 + {(3-1)^2}]{p^3} - {3^2}{p^2}(2p - 2 - 3), \\ & = &= {p^2}(2{p^2} - 17p + 45). \end{eqnarray*}$

类似的, 在引理2.1中令$(t, p -1) = 4, $结合引理2.2可得

$\begin{equation*} {\sum\limits_{s = 1}^p {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{m{a^{2t}} + s{a^t} + na}}{p})} } \right|} } ^4} = \left\{ \begin{array}{l} {p^2}(2{p^2} - 25p + 96), p\not | {( - 4)^{t/4}} - 1, \\ {p^2}(2{p^2} - 29p + 96), p\left. \right|{( - 4)^{t/4}} - 1 \end{array} \right.(t, p - 1) = 4. \end{equation*}$

在命题1.1及定理1.1中令$t = 1, 2, 3, 4, $可得如下结论.

推论2.1  设$p$为素数, 则对任意固定的满足条件$(n, p)=1$的正整数$n$, 有

$\begin{eqnarray*} &&{\sum\limits_{s = 1}^p {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{m{a^4} + s{a^2} + na}}{p})} } \right|} } ^4} = {p^2}(2{p^2} - 11p + 16), p \ge 5.\\ &&{\sum\limits_{s = 1}^p {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{m{a^6} + s{a^3} + na}}{p})} } \right|} } ^4} = \left\{ {\begin{array}{*{20}{c}} {{p^2}(2{p^2} - 17p + 45), p \equiv 1({\rm mod} 3), }\\ {{p^2}(2{p^2} - 5p + 3), \quad p \equiv 2({\rm mod} 3), } \end{array}} \right.p \ge 5, \\ &&{\sum\limits_{s = 1}^p {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{m{a^8} + s{a^4} + na}}{p})} } \right|} } ^4} = \left\{ {\begin{array}{*{20}{c}} {{p^2}(2{p^2} - 25p + 96), \quad p \equiv 1({\rm mod} 4), }\\ {{p^2}(2{p^2} - 11p + 16), \quad p \equiv 3({\rm mod} 4), } \end{array}} \right.p \ge 7. \end{eqnarray*}$

这一部分将给出三项指数和的四次均值

${\sum\limits_{s = 1}^p {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^p {'e(\frac{{m{a^k} + s{a^t} + na}}{p})} } \right|} } ^4}$

在$k=2t$时的简化证明方法.

当$k = 2t$时, 有

$\begin{eqnarray*} &&{\sum\limits_{s = 1}^p {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^p {'e(\frac{{m{a^k} + s{a^t} + na}}{p})} } \right|} } ^4} = {\sum\limits_{s = 1}^p {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^p {'e(\frac{{m{a^{2t}} + s{a^t} + na}}{p})} } \right|} } ^4}\\ &= &\sum\limits_{m = 1}^p {\sum\limits_{s = 1}^p {\sum\limits_{a = 1}^{p - 1} {\sum\limits_{b = 1}^{p - 1} {\sum\limits_{c = 1}^{p - 1} {\sum\limits_{d = 1}^{p - 1} {e(\frac{{m({a^{2t}} + {b^{2t}} - {c^{2t}} - {d^{2t}}) + s({a^t} + {b^t} - {c^t} - {d^t}) + n(a + b - c - d)}}{p})} } } } } } \\ &= &{p^2}\mathop {\sum\limits_{a = {\rm{1}}}^{p - 1} {\sum\limits_{b = 1}^{p - 1} {\sum\limits_{c = {\rm{1}}}^{p - 1} {\sum\limits_{d = 1}^{p - 1} \, } } } }\limits_{\begin{array}{*{20}{c}} {{a^{2t}} + {b^{2t}} \equiv {c^{2t}} + {d^{2t}}({\rm mod} p)}\\ {{a^t} + {b^t} \equiv {c^t} + {d^t}({\rm mod} p)} \end{array}} e(\frac{{n(a + b - c - d)}}{p})\\ &=& {p^2}\mathop {\sum\limits_{a = {\rm{1}}}^{p - 1} {\sum\limits_{b = 1}^{p - 1} {\sum\limits_{c = {\rm{1}}}^{p - 1} {\sum\limits_{d = 1}^{p - 1} \, } } } }\limits_{\begin{array}{*{20}{c}} {{a^{2t}} + {b^{2t}} \equiv {c^{2t}} + {d^{2t}}({\rm mod} p)}\\ {{a^t} + {b^t} \equiv {c^t} + {d^t}({\rm mod} p)} \end{array}} e(\frac{{n(a + b - c - d)}}{p})\\ &=& {p^2}\mathop {\sum\limits_{a = {\rm{1}}}^{p - 1} {\sum\limits_{b = 1}^{p - 1} {\sum\limits_{c = {\rm{1}}}^{p - 1} {} } } }\limits_{\begin{array}{*{20}{c}} {{a^{2t}} + {b^{2t}} \equiv {c^{2t}} + 1~({\rm mod} p)}\\ {{a^t} + {b^t} \equiv {c^t} + 1~({\rm mod} p)} \end{array}} \sum\limits_{d = 1}^{p - 1} {\, e(\frac{{nd(a + b - c - 1)}}{p})} \\ &=& {p^2}\mathop {\sum\limits_{a = {\rm{1}}}^{p - 1} {\sum\limits_{b = 1}^{p - 1} {\sum\limits_{c = {\rm{1}}}^{p - 1} {} } } }\limits_{\begin{array}{*{20}{c}} {{a^{2t}} + {b^{2t}} \equiv {c^{2t}} + 1~({\rm mod} p)}\\ {\scriptstyle{a^t} + {b^t} \equiv {c^t} + 1~({\rm mod} p)\atop \scriptstyle a + b \equiv c + 1~({\rm mod} p)} \end{array}} e(\frac{{nd(a + b - c - 1)}}{p}) \\ &&- {p^2}\mathop {\sum\limits_{a = {\rm{1}}}^{p - 1} {\sum\limits_{b = 1}^{p - 1} {\sum\limits_{c = {\rm{1}}}^{p - 1} {} } } }\limits_{\begin{array}{*{20}{c}} {{a^{2t}} + {b^{2t}} \equiv {c^{2t}} + 1~({\rm mod} p)}\\ {{a^t} + {b^t} \equiv {c^t} + 1~({\rm mod} p)} \end{array}} e(\frac{{nd(a + b - c - 1)}}{p})\\ &=& {p^2}\mathop {\sum\limits_{a = {\rm{1}}}^{p - 1} {\sum\limits_{b = 1}^{p - 1} {\sum\limits_{c = {\rm{1}}}^{p - 1} {} } } }\limits_{\begin{array}{*{20}{c}} {({a^t} - 1)({b^t} - 1) \equiv 0({\rm mod} p)}\\ {\scriptstyle{a^t} + {b^t} \equiv {c^t} + 1~({\rm mod} p)\atop \scriptstyle a + b \equiv c + 1~({\rm mod} p)} \end{array}} e(\frac{{nd(a + b - c - 1)}}{p})\\ &&-{p^2}\mathop {\sum\limits_{a = {\rm{1}}}^{p - 1} {\sum\limits_{b = 1}^{p - 1} {\sum\limits_{c = {\rm{1}}}^{p - 1} {} } } }\limits_{\begin{array}{*{20}{c}} {({a^t} - 1)({b^t} - 1) \equiv 0({\rm mod} p)}\\ {{a^t} + {b^t} \equiv {c^t} + 1~({\rm mod} p)} \end{array}} e(\frac{{nd(a + b - c - 1)}}{p}), \end{eqnarray*}$

上式的计算可归结为求解同余方程组的问题, 经分析和计算亦可获得命题1.1及定理1.1中的结果, 此处不再展开讨论.

本文进一步深入研究了三项指数和四次均值

${\sum\limits_{s = 1}^p {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{p - 1} {e(\frac{{m{a^k} + s{a^t} + na}}{p})} } \right|} } ^4}$

的计算问题, 获得了$k = 2t, (k, p -1) = 3, 4$时的精确计算公式, 推广了论文^[14]中的结果.下一步将用本文提出的方法研究

${\sum\limits_{s = 1}^p {\sum\limits_{m = 1}^p {\left| {\sum\limits_{a = 1}^{{p^\alpha }} {'e(\frac{{m{a^{2t}} + s{a^t} + na}}{{{p^\alpha }}})} } \right|} } ^4}, \alpha \ge 2$

的计算问题.本文所使用的思想和方法对于四项指数和与更多项指数和也有一定的借鉴作用, 但其算法复杂度大大增加.