问题及解答

设 $f(x)$ 是关于 $x$ 的 $k$ 次多项式,

\[
f(x)=a_k x^k +a_{k-1}x^{k-1}+\cdots+a_2 x^2+a_1 x+a_0,
\]

这里 $a_0=0$.

若 $x=q_1 y+q_2 z$, $q_1,q_2,y,z\in\mathbb{Z}$, 则

\[f(q_1 y+q_2 z)\equiv f(q_1 y)+f(q_2 z)\pmod{q_1 q_2}.\]

若记 $e_q(f(x))=e^{2\pi i\frac{f(x)}{q}}$, 则对于一般的多项式 $f(x)$, 即常数项不一定为 0 ($f(x)=a_k x^k +a_{k-1}x^{k-1}+\cdots+a_2 x^2+a_1 x+a_0$). 有

\[
e_{q_1 q_2}(f(q_1 y+q_2 z))=e_{q_2}(\frac{f(q_1 z)}{q_1})\cdot e_{q_1}(\frac{f(q_2 z)}{q_2}).
\]

\[
f(x)=a_k x^k +a_{k-1}x^{k-1}+\cdots+a_2 x^2+a_1 x+a_0
\]

这里 $a_i\in\mathbb{Z}$. 

\[
f(q_1 y+q_2 z)=a_k(q_1 y+q_2 z)^k +a_{k-1}(q_1 y+q_2 z)^{k-1}+\cdots+a_2(q_1 y+q_2 z)^2+a_1(q_1 y+q_2 z)+a_0.
\]

注意

\[
(q_1 y+q_2 z)^k=(q_1 y)^k+\sum_{i=1}^{k-1}C_k^i (q_1 y)^{k-i}(q_2 z)^i+(q_2 z)^k,\quad\forall\ k\geqslant 2.
\]

因此, 

\[
(q_1 y+q_2 z)^k\equiv (q_1 y)^k+(q_2 z)^k\ \pmod{q_1 q_2}\quad\forall\ k\geqslant 1 .
\]

于是, 

\[
\begin{split}
f(q_1 y+ q_2 z)&\equiv \Bigl(a_k(q_1 y)^k+a_k(q_2 z)^k\Bigr)+\Bigl(a_{k-1}(q_1 y)^{k-1}+a_{k-1}(q_2 z)^{k-1}\Bigr)+\cdots+\Bigl(a_1(q_1 y)+a_1(q_2 z)+a_0\Bigr)\ \pmod{q_1 q_2},\\
&\equiv f(q_1 y)+f(q_2 z)-a_0\ \pmod{q_1 q_2}.
\end{split}
\]

\[
e_{q_1 q_2}(f(q_1 y+q_2 z))=e^{2\pi i\cdot\frac{f(q_1 y+q_2 z)}{q_1 q_2}}=e^{2\pi i\cdot\frac{f(q_1 y)+f(q_2 z)}{q_1 q_2}}=e_{q_2}(\frac{1}{q_1}f(q_1 y))\cdot e_{q_1}(\frac{1}{q_2}f(q_2 z)).
\]