2026.5.3 闲话

波谱分析 / 解密：波谱分析

折叠

Wolstenholme：对于素数 \(p>3\)，

\[\sum_{i=1}^{p-1}\dfrac1i\equiv 0\pmod{p^2} \]

图

\(\newcommand{\inv}{\operatorname{inv}}\inv_p(n)\) 表示 \(n\) 模 \(p\) 意义下的逆元，早已知道 \(\inv_{p^2}(n)\equiv \inv_p(n)(2-n\inv_p(n))\pmod{p^2}\) .

后面纯属乱写，别看了、、（详见评论区）
upd. 应该改成对的了

\[\begin{aligned}\sum_{i=1}^{p-1}\inv_{p^2}(i)&\equiv\sum_{i=1}^{p-1}\inv_p(i)(2-i\inv_p(i))&\pmod{p^2}\\&\equiv2\sum_{i=1}^{p-1}\inv_p(i)-\sum_{i=1}^{p-1}i\inv_p(i)^2&\pmod{p^2}\\&\equiv-\sum_{i=1}^{p-1}i\inv_p(i)^2-p&\pmod{p^2}\\&\equiv-\dfrac12\left(\sum_{i=1}^{p-1}i\inv_p(i)^2+\sum_{i=1}^{p-1}i^2\inv_p(i)\right)-p&\pmod{p^2}\\&\equiv-\dfrac16\sum_{i=1}^{p-1}((i+\inv_p(i))^3-i^3-\inv_p(i)^3)-p&\pmod{p^2}\\&\equiv-\dfrac16\sum_{i=1}^{p-1}(i+\inv_p(i))^3-p&\pmod{p^2}\\&\equiv-\dfrac1{12}\left(\sum_{i=1}^{p-1}(2p-i-\inv_p(i))^3+\sum_{i=1}^{p-1}(i+\inv_p(i))^3\right)-p&\pmod{p^2}\\&\equiv-\dfrac12\sum_{i=1}^{p-1}p(i+\inv_p(i))^2-p&\pmod{p^2}\\&\equiv0&\pmod{p^2}\end{aligned} \]

第七个等号：亦可以在模 \(p^3\) 意义下考虑 \(\displaystyle\sum_{i=1}^{p-1}(i+\inv_p(i))^4=\sum_{i=1}^{p-1}(2p-i+\inv_p(i))^4\) 得到 .