我们先从最简单的情形理解之. 设 $k=1$, 则恒等式为
\[
\sum_{a_1}\biggl|\sum_{x=1}^{p}e_p(a_1 x)\biggr|^2=\sum_{x_1}\sum_{y_1}\sum_{a_1}e_p\bigl(a_1(x_1-y_1)\bigr).
\]
这很容易验证. 我们只需证明
\[
\biggl|\sum_{x=1}^{p}e^{2\pi i a_{_1}x/p}\biggr|^2=\sum_{x_1}\sum_{y_1}e^{2\pi i a_{_1}(x_{_1}-y_{_1})/p}.
\]
\[
\begin{split}
\text{LHS}&=\biggl(\sum_{x=1}^{p}e^{2\pi i a_{_1}x/p}\biggr)\cdot\biggl(\overline{\sum_{x=1}^{p}e^{2\pi i a_{_1}x/p}}\biggr)\\
&=\Bigl(e^{2\pi ia_{_1}1/p}+e^{2\pi ia_{_1}2/p}+\cdots+e^{2\pi ia_{_1}p/p}\Bigr)\cdot\Bigl(e^{-2\pi ia_{_1}1/p}+e^{-2\pi ia_{_1}2/p}+\cdots+e^{-2\pi ia_{_1}p/p}\Bigr)\\
&=\sum_{x_1=1}^{p}\sum_{y_1=1}^{p}e^{2\pi i a_{_1}(x_{_1}-y_{_1})/p}\\
&=\text{RHS}.
\end{split}
\]
故对于 $k=1$ 是成立的.
当 $k=2$ 时, 恒等式为
\[
\sum_{a_2}\sum_{a_1}\biggl|\sum_{x=1}^{p}e_p(a_2 x^2+a_1 x)\biggr|^4=\sum_{x_1}\sum_{x_2}\sum_{y_1}\sum_{y_2}\sum_{a_2}\sum_{a_1}e_p\bigl(a_2(x_1^2+x_2^2-y_1^2-y_2^2)+a_1(x_1+x_2-y_1-y_2)\bigr).
\]
我们只需证明
\[
\biggl|\sum_{x=1}^{p}e^{2\pi i(a_{_2}x^2+a_{_1}x)/p}\biggr|^4=\sum_{x_1}\sum_{x_2}\sum_{y_1}\sum_{y_2}e^{2\pi i\bigl(a_{_2}(x_1^2+x_2^2-y_1^2-y_2^2)+a_{_1}(x_{_1}+x_{_2}-y_{_1}-y_{_2})\bigr)/p}.
\]
注意
\[
\biggl|\sum_{x=1}^{p}e^{2\pi i(a_{_2}x^2+a_{_1}x)/p}\biggr|^2=\biggl(\sum_{x=1}^{p}e^{2\pi i(a_{_2}x^2+a_{_1}x)/p}\biggr)\cdot\biggl(\sum_{x=1}^{p}e^{-2\pi i(a_{_2}x^2+a_{_1}x)/p}\biggr)=\sum_{x_1=1}^{p}\sum_{y_1=1}^{p}e^{2\pi i(a_{_2}x_1^2-a_{_2}y_1^2+a_{_1}x_{_1}-a_{_1}y_{_1})/p}
\]
然后再平方, 就可以得到上面的式子.
对于一般情形, 只要是有限和, 类似可以证明.