推广威尔逊定理
\((p-1)! \equiv -1 \pmod{p}\)\(\displaystyle \implies (p-r)! \equiv -\prod_{k=1}^{r-1} \frac{1}{p-k} \equiv -\prod_{k=1}^{r-1} \frac{1}{-k} \equiv \frac{(-1)^r}{(r-1)!}\pmod{p}\)
\(\displaystyle \implies -1 \equiv (p-1)! \equiv \frac{(p-1)!(-1)^r}{(p-r)!(r-1)!} \equiv (-1)^r C_{p-1}^{r-1} \pmod{p}\)
\(\implies C_{p-1}^r \equiv (-1)^r \pmod{p},\;C_{p-1}^n C_n^r = C_{p-1}^r C_{p-1-r}^{p-1-n}\)
\(\displaystyle \implies C_n^r \equiv (-1)^{n-r} C_{p-1-r}^{p-1-n} \pmod{p} ~ (p-1\ge n)\)
1.对于\(p-1< n\)还有什么关系式？
2.能否证明\((mp-1)! \equiv (-1)^m (m-1)!p^{m-1} \pmod{p^m}\)？ \(\displaystyle (x^{p-1}-1)^m - \prod_{k=1 \atop p\nmid k}^{mp-1} (x-k) \equiv 0 \pmod{p^m} \\
\implies \displaystyle (-1)^m - (-1)^{m(p-1)}\prod_{k=1 \atop p\nmid k}^{mp-1} k \equiv 0 \pmod{p^m} \\
\implies \displaystyle\prod_{k=1 \atop p\nmid k}^{mp-1} k \equiv (-1)^m \pmod{p^m} \\
\implies \displaystyle(mp-1)! \equiv (-1)^m (m-1)!p^{m-1} \pmod{p^m} \) \(\displaystyle (mp-1)! \equiv (-1)^m (m-1)!p^{m-1} \pmod{p^m}\)这个实验值没有问题
我以为2#的\(\displaystyle \prod_{k=1 \atop p\nmid k}^{mp-1} k \equiv (-1)^m \pmod{p^m}\)也没有问题
可是p=2,m=2时\(3! \equiv 2\pmod{2^2}\)，而\(3 \neq 1\pmod{2^2}\)
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