55 lines
1.9 KiB
Markdown
55 lines
1.9 KiB
Markdown
注：扫描件中模糊处均不影响做题
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# 一、选择题
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## BCBDA CBCCA
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# 二、填空题
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1. 2
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2. 4
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3. $x\equiv7\ (mod\ 11)$
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4. $s=47,t=-18,(a,b)=3$
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5. $x\equiv11\ (mod\ 47), y\equiv1\ (mod\ 47)$
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6. -1
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7. $x\equiv2,5,8\ (mod\ 9)$
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8. 2
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9. 2
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10. 0
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# 三、计算题
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1. $\because\varphi(41)=40,\ (\varphi(41),15)=5$
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$\therefore\ $方程有5个解
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\
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$x^{15}\equiv14\ (mod\ 41)$
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查表得 $14\equiv6^{25}\ (mod\ 41)$
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令 $x\equiv\ 6^a\ (mod\ 41)$
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则有 $6^{a^{15}}\equiv6^{25}\ (mod\ 41)$
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即 $6^{15a}\equiv6^{25}\ (mod\ 41)$
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则 $15a\equiv25\ (mod\ 40)$
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化为 $3a\equiv5\ (mod\ 8)$，该式解为 $a\equiv7\ (mod\ 8)$
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故解为 $a\equiv7,15,23,31,39\ (mod\ 40)$
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查表得原式解为 $x\equiv29,3,30,13,7\ (mod\ 41)$
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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
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# 四、证明题
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1.要证121是基3的拟素数，即证 $3^{120}\equiv1\ (mod\ 121)$
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一种常见的思路：
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显然121与3互素，由欧拉定理， $\varphi(121)=11^2-11=110,3^{\varphi(121)}=3^{110}\equiv1\ (mod\ 121)$
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所以 $3^{120}\equiv3^{10}\ (mod\ 121)$, $3^{10}$显然可以手动验算，得证
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另一种可能性：
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尝试逐个检验后发现 $3^{5}=243\equiv1\ (mod\ 121),5|120$，直接得证
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2. 显然p不为2
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$\because p|n^4+1$
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$\therefore n^4+1\equiv 0\ (mod \ p)$
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$\therefore n^4+2n^2+1\equiv 2n^2\ (mod \ p)$
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$\therefore (n^2+1)^2\equiv 2n^2\ (mod \ p)$
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由二次剩余的定义，知式子右边是模p的二次剩余
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$\therefore(\frac{2n^2}{p})=1$
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又 $\because (n,p)=1$
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$\therefore(\frac{2}{p})=1$
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$\therefore p\equiv 1,-1\ (mod\ 8)$
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类似的，有 $n^4-2n^2+1\equiv -2n^2\ (mod \ p),(\frac{-2}{p})=1$
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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
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