\[x^2\equiv a(mod\ p)\]
\[b^2-a\equiv w(mod\ p)\]
\[i=\sqrt{w}\]
\[a+bi=a+b\sqrt{w}\]
\[x=(b+i)^{\frac{p+1}{2}}\]
\[x^2=(b+i)^{p}(b+i)=\sum\limits_{j=0}^{p}\binom{p}{j}b^ji^{p-j}(b+i)\]
\[(b+i)^p\equiv b^p+i^p\equiv b^{p-1}b+w^{\frac{p-1}{2}}i\equiv b-i(mod\ p)\]
\[x^2\equiv (b-i)(b+i)\equiv b^2-w\equiv a(mod\ p)\]
\[a^{\frac{p-1}{2}}=-1/0/1\]
\[(x^2)^{\frac{p-1}{2}}\equiv 1\]
\[g^i\equiv a\]
\[a^{\frac{p-1}{2}}\equiv g^{i\frac{p-1}{2}}\equiv 1(mod\ p)\]
\[(p-1)|i\frac{p-1}{2}\]
\[2|i\]
\[x=g^{\frac{i}{2}}\]
\[x^2\equiv a(mod\ p^k)\]
\[a=p^wq,(p,q)=1\]
\[w>=k\]
\[w<k\]
\[2|w\]
\[x=p^es\]
\[x^2=p^{2e}s^2\]
\[2e<k\]
\[(p^{2e}s^2)\% p^k=p^{2e}(s^2\%p^{k-2e})\]
\[p\not| a\]
\[x^2\equiv a(mod p^k)\]
\[r^2-a\equiv 0(mod p)\]
\[(r^2-a)^k\equiv 0(mod\ p^k)\]
原文：https://www.cnblogs.com/Lrefrain/p/12539420.html