习题十一

作者: 洛玖言 | 来源:发表于2019-11-04 18:14 被阅读0次

习题十一

2

计算 $\left(\dfrac{17}{23}\right),\,\left(\dfrac{19}{37}\right),\,\left(\dfrac{60}{79}\right),\,\left(\dfrac{92}{101}\right).$

Sol:

$\left(\dfrac{17}{23}\right)=\left(\dfrac{-6}{23}\right)=\left(\dfrac{-1}{23}\right)\left(\dfrac{2}{23}\right)\dfrac{3}{23}=-\left(\dfrac{3}{23}\right)$
又由二次互反律 $\left(\dfrac{23}{3}\right)\left(\dfrac{3}{23}\right)=-1$ 且 $\left(\dfrac{23}{3}\right)=\left(\dfrac{2}{3}\right)=-1$
$\therefore\left(\dfrac{3}{23}\right)=1$
$\therefore\left(\dfrac{17}{23}\right)=-1$

$\left(\dfrac{19}{37}\right)=\left(\dfrac{-18}{37}\right)=\left(\dfrac{-1}{37}\right)\left(\dfrac{2}{37}\right)\left(\dfrac{3}{37}\right)\left(\dfrac{3}{37}\right)=-\left(\dfrac{3}{37}\right)^2$
又由二次互反律 $\left(\dfrac{3}{37}\right)\left(\dfrac{37}{3}\right)=1$ 且 $\left(\dfrac{37}{3}\right)=\left(\dfrac{1}{3}\right)=1$
$\therefore \left(\dfrac{3}{37}\right)=1$
$\therefore \left(\dfrac{19}{37}\right)=-1$

$\left(\dfrac{60}{79}\right)=\left(\dfrac{-19}{79}\right)=\left(\dfrac{-1}{79}\right)\left(\dfrac{19}{79}\right)=-\left(\dfrac{19}{79}\right)$
由二次互反律 $\left(\dfrac{19}{79}\right)\left(\dfrac{79}{19}\right)=-1$ 且 $\left(\dfrac{79}{19}\right)=\left(\dfrac{3}{19}\right)$
由二次互反律 $\left(\dfrac{3}{19}\right)\left(\dfrac{19}{3}\right)=-1$ 且 $\left(\dfrac{19}{3}\right)=\left(\dfrac{2}{3}\right)=-1$
$\therefore\left(\dfrac{3}{19}\right)=1$
$\therefore\left(\dfrac{19}{79}\right)=-1$
$\therefore\left(\dfrac{60}{19}\right)=1$

$\left(\dfrac{92}{101}\right)=\left(\dfrac{-9}{101}\right)=\left(\dfrac{-1}{101}\right)\left(\dfrac{3}{101}\right)^2=\left(\dfrac{3}{101}\right)^2$
由二次互反律 $\left(\dfrac{3}{101}\right)\left(\dfrac{101}{3}\right)=1$ 且 $\left(\dfrac{101}{3}\right)=\left(\dfrac{-1}{3}\right)=-1$
$\therefore\left(\dfrac{3}{101}\right)=-1$
$\therefore\left(\dfrac{92}{101}\right)=1$

3

(i) 确定以 $-3$ 为二次剩余的素数.
(ii) 确定以 $5$ 为二次剩余的素数.
(iii) 确定以 $15$ 为二次剩余的素数.

Sol:
(i)
由题意得
$\left(\dfrac{-3}{p}\right)=\left(\dfrac{-1}{p}\right)\left(\dfrac{3}{p}\right)=1$
$\therefore\left(\dfrac{-1}{p}\right)=\left(\dfrac{3}{p}\right)=1$ 或 $\left(\dfrac{-1}{p}\right)=\left(\dfrac{3}{p}\right)=-1$
$\therefore p\equiv1\pmod{4}$ 且 $p\equiv\pm1\pmod{12}$
或 $p\equiv3\pmod{4}$ 且 $p\equiv\pm5\pmod{12}$
$\therefore p\equiv1\pmod{12}$ 或 $p\equiv7\pmod{12}$

(ii)
由题意得
$\left(\dfrac{5}{p}\right)=1$
又由二次互反律 $\left(\dfrac{5}{p}\right)\bigg(\dfrac{p}{5}\bigg)=1$
$\therefore p^2\equiv1\pmod{5}\Rightarrow p\equiv\pm1\pmod{5}$

(iii)
$\left(\dfrac{15}{p}\right)=\left(\dfrac{3}{p}\right)\left(\dfrac{5}{p}\right)=1$
即 $\left(\dfrac{3}{p}\right)=\left(\dfrac{5}{p}\right)=1$
或 $\left(\dfrac{3}{p}\right)=\left(\dfrac{5}{p}\right)=-1$
$\therefore p\equiv\pm1\pmod{12}$ 且 $p\equiv\pm1\pmod{5}\Rightarrow p\equiv\pm1,\pm11\pmod{60}$
或 $p\equiv\pm5\pmod{12}$ 且 $p\equiv\pm3\pmod{5}\Rightarrow p\equiv\pm7,\pm17\pmod{60}$

5

设 $p=4k+1$ 是素数， $a$ 是 $k$ 的约数. 证明: $\left(\dfrac{a}{p}\right)=1$

Sol:
若 $a$ 为素数，
$\because p\equiv1\pmod{4}$
$\therefore\left(\dfrac{a}{p}\right)\left(\dfrac{p}{a}\right)=1$
$\because \;a$ 是 $k$ 的约数， $p=4k+1$
$\therefore\bigg(\dfrac{p}{a}\bigg)=\left(\dfrac{1}{a}\right)=1$
$\therefore\left(\dfrac{a}{p}\right)=1$
当 $a$ 不为素数时，可分解为素因子乘积
$a=q_1q_2\cdots q_n$
$\therefore\left(\dfrac{a}{p}\right)=\left(\dfrac{q_1}{p}\right)\left(\dfrac{q_2}{p}\right)\cdots\left(\dfrac{q_n}{p}\right)=1\cdot1\cdots1=1$
Q.E.D.

6

设 $p=10n-1$ 是素数，证明: $p|5^{5n-1}-1$

Sol:
$\because p\equiv-1\pmod{5}$
$\therefore\left(\dfrac{5}{p}\right)=1$
$\therefore\left(\dfrac{5}{p}\right)\equiv5^{\frac{p-1}{2}}\equiv5^{5n-1}\equiv1\pmod{p}$
$\therefore p|5^{5n-1}-1$