Posted by haifeng on 2015-12-19 00:29:54 last update 2015-12-19 00:43:58 | Answers (0) | 收藏
求解同余式 $x^2\equiv 5(\mod 11)$.
[Hint]
\[(0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100) \mod 11= (0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1)\]
因此 $x\equiv 7(\mod 11)$, 于是解为 $x=11k+7$.
