最后更新于4年前
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ppp为素数
\left(\frac{a}{p}\right)=\left\{ \begin{array}{**lr**} 1 &a为模p的平方剩余\\ -1 &a为模p的非平方剩余\\ 0 &p\mid a \end{array} \right.
ppp为奇素数，对整数aaa
(ap)≡ap−12(mod p)\left(\frac{a}{p}\right)\equiv a^{\frac{p-1}{2}}\left(\mod p\right)(pa)≡a2p−1(modp)
(1p)=1\left(\frac{1}{p}\right)=1(p1)=1
(−1p)=(−1)p−12\left(\frac{-1}{p}\right)={\left(-1\right)}^{\frac{p-1}{2}}(p−1)=(−1)2p−1
ppp为奇素数，则
(a+pp)=(ap)\left(\frac{a+p}{p}\right)=\left(\frac{a}{p}\right)(pa+p)=(pa)
(a⋅bp)=(ap)(bp)\left(\frac{a\cdot b}{p}\right)=\left(\frac{a}{p}\right)\left(\frac{b}{p}\right)(pa⋅b)=(pa)(pb)
设(a,p)=1\left(a,p\right)=1(a,p)=1，则(a2p)=1\left(\frac{a^2}{p}\right)=1(pa2)=1
ppp为奇素数，aaa为整数，(a,p)=1\left(a,p\right)=1(a,p)=1，整数a⋅1,a⋅2,⋯ ,a⋅p−12a\cdot1,a\cdot2,\cdots,a\cdot\frac{p-1}{2}a⋅1,a⋅2,⋯,a⋅2p−1中模ppp的最小正剩余大于p2\frac{p}{2}2p的个数是mmm，则(ap)=(−1)m\left(\frac{a}{p}\right)={\left(-1\right)}^m(pa)=(−1)m