From d9ef9f5c17660e4d53e03cc92481163f0b9a9bf8 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E3=82=BF=E3=82=AF=E3=83=A4=E3=83=9E?= Date: Sun, 7 Jul 2024 23:48:30 +0800 Subject: [PATCH] =?UTF-8?q?=E5=AE=8C=E6=88=90=E4=BA=86=E7=AD=94=E6=A1=88?= =?UTF-8?q?=E7=9A=84=E7=BC=96=E5=86=99?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 信息安全数学基础（一）/期末试卷/手做答案.md | 17 ++++++++++++++++- 1 file changed, 16 insertions(+), 1 deletion(-) diff --git a/信息安全数学基础（一）/期末试卷/手做答案.md b/信息安全数学基础（一）/期末试卷/手做答案.md index 569c7dd..ef46ca7 100644 --- a/信息安全数学基础（一）/期末试卷/手做答案.md +++ b/信息安全数学基础（一）/期末试卷/手做答案.md @@ -31,9 +31,24 @@ 2. 勒让德符号的计算较为简单，这里不给出解题过程，两问的答案分别是-1，-1 # 四、证明题 -1. 要证121是基3的拟素数，即证 $3^{120}\equiv1\ (mod\ 121)$ + +1.要证121是基3的拟素数，即证 $3^{120}\equiv1\ (mod\ 121)$ 一种常见的思路： 显然121与3互素，由欧拉定理， $\varphi(121)=11^2-11=110,3^{\varphi(121)}=3^{110}\equiv1\ (mod\ 121)$ 所以 $3^{120}\equiv3^{10}\ (mod\ 121)$, $3^{10}$显然可以手动验算，得证 另一种可能性： 尝试逐个检验后发现 $3^{5}=243\equiv1\ (mod\ 121),5|120$，直接得证 +2. 显然p不为2 + $\because p|n^4+1$ + $\therefore n^4+1\equiv 0\ (mod \ p)$ + $\therefore n^4+2n^2+1\equiv 2n^2\ (mod \ p)$ + $\therefore (n^2+1)^2\equiv 2n^2\ (mod \ p)$ + 由二次剩余的定义，知式子右边是模p的二次剩余 + $\therefore(\frac{2n^2}{p})=1$ + 又 $\because (n,p)=1$ + $\therefore(\frac{2}{p})=1$ + $\therefore p\equiv 1,-1\ (mod\ 8)$ + + 类似的，有 $n^4-2n^2+1\equiv -2n^2\ (mod \ p),(\frac{-2}{p})=1$ + 分别检验 $p\equiv 1\ (mod\ 8)$ 与 $p\equiv -1\ (mod\ 8)$，发现只有 $p\equiv 1\ (mod\ 8)$满足条件，得证 +