最后更新于3年前
METHOD1: 定理设ppp为奇素数(2p)=(−1)p2−18\left(\frac{2}{p}\right)={\left(-1\right)}^{\frac{p^2-1}{8}}(p2)=(−1)8p2−1若(a,2p)=1\left(a,2p\right)=1(a,2p)=1，则(ap)=(−1)T(a,p)\left(\frac{a}{p}\right)={\left(-1\right)}^{T_{\left(a,p\right)}}(pa)=(−1)T(a,p)，其中T_{\left(a,p\right)}=\sum_\limits{k=1}^{\frac{p-1}{2}}{\left[\frac{a\cdot k}{p}\right]}METHOD2: 二次互反律若p,qp,qp,q为互素奇素数，则(pq)=(−1)p−12⋅q−12(pq)\left(\frac{p}{q}\right)={\left(-1\right)}^{\frac{p-1}{2}\cdot\frac{q-1}{2}}\left(\frac{p}{q}\right)(qp)=(−1)2p−1⋅2q−1(qp)
METHOD1: 定理
设ppp为奇素数
(2p)=(−1)p2−18\left(\frac{2}{p}\right)={\left(-1\right)}^{\frac{p^2-1}{8}}(p2)=(−1)8p2−1
若(a,2p)=1\left(a,2p\right)=1(a,2p)=1，则(ap)=(−1)T(a,p)\left(\frac{a}{p}\right)={\left(-1\right)}^{T_{\left(a,p\right)}}(pa)=(−1)T(a,p)，其中T_{\left(a,p\right)}=\sum_\limits{k=1}^{\frac{p-1}{2}}{\left[\frac{a\cdot k}{p}\right]}
METHOD2: 二次互反律
若p,qp,qp,q为互素奇素数，则(pq)=(−1)p−12⋅q−12(pq)\left(\frac{p}{q}\right)={\left(-1\right)}^{\frac{p-1}{2}\cdot\frac{q-1}{2}}\left(\frac{p}{q}\right)(qp)=(−1)2p−1⋅2q−1(qp)