本帖最后由 tommywong 于 2018-5-14 18:44 编辑
$\displaystyle \sigma_m(A)=\sum_{j_1<j_2<\dots<j_m\atop j_1,j_2,\dots,j_m\in A} x_{j_1}\dots x_{j_m}$
证明$d^m|\sigma_m(A),m=1,2,\dots,|A| \Leftrightarrow d|x_j,\forall j\in A$
n=1:
$d|\sigma_1(\{1\})=x_1 \Leftrightarrow d|x_1$
n=k+1:
$d^m|\sigma_m(\{1,2,\dots,k+1\}),m=1,2,\dots,k+1$
$d^{k+1}|\sigma_{k+1}(\{1,2,\dots,k+1\})=x_1 x_2\dots x_{k+1}$
$\displaystyle d=\prod_i p_i^{k_i}$
$\forall i,p^{k_i(k+1)}|x_1 x_2\dots x_{k+1}\Rightarrow \forall i,\exists r_i\in \{1,2,\dots,k+1\},s.t. p^{k_i}|x_{r_i}$
$\sigma_1(\{1,2,\dots,k+1\})=\sigma_1(\{1,2,\dots,k+1\}-\{r_i\})+x_{r_i}$
$d|\sigma_1(\{1,2,\dots,k+1\})\Rightarrow p^{k_i}|\sigma_1(\{1,2,\dots,k+1\})\Rightarrow p^{k_i}|\sigma_1(\{1,2,\dots,k+1\}-\{r_i\})=\sigma_1(\{1,2,\dots,k+1\})-x_{r_i}$
$\sigma_{m+1}(\{1,2,\dots,k+1\})=\sigma_{m+1}(\{1,2,\dots,k+1\}-\{r_i\})+x_{r_i}\sigma_m(\{1,2,\dots,k+1\}-\{r_i\})$
$d^{m+1}|\sigma_{m+1}(\{1,2,\dots,k+1\})\Rightarrow p^{k_i(m+1)}|\sigma_{m+1}(\{1,2,\dots,k+1\})$
$p^{k_i(m+1)}|\sigma_{m+1}(\{1,2,\dots,k+1\}),p^{k_i m}|\sigma_m(\{1,2,\dots,k+1\}-\{r_i\}),p^{k_i}|x_{r_i}\Rightarrow p^{k_i(m+1)}|\sigma_{m+1}(\{1,2,\dots,k+1\}-\{r_i\})=\sigma_{m+1}(\{1,2,\dots,k+1\})-x_{r_i}\sigma_m(\{1,2,\dots,k+1\}-\{r_i\})$
$\Rightarrow \forall i,p^{k_i m}|\sigma_m(\{1,2,\dots,k+1\}-\{r_i\}),m=1,2,\dots,k\Rightarrow \forall i,p^{k_i}|x_j,j\in \{1,2,\dots,k+1\}-\{r_i\}$
$\Rightarrow \forall i,p^{k_i}|x_j,j\in \{1,2,\dots,k+1\}\Rightarrow d|x_j,j\in \{1,2,\dots,k+1\}$ |