[二次剩余]求解二次剩余

DescriptionDescription

求解x2≡n(modp)x^2 \equiv n \pmod p。pp是一个奇质数。

SolutionSolution

由费马小定理

np−1≡1(modp)

n^{p-1}\equiv 1 \pmod p所以

np−12≡±1(modp)

n^{{p-1}\over 2}\equiv \pm1 \pmod p由欧拉准则

(np)≡np−12(modp)

\left(n\over p\right)\equiv n^{{p-1}\over 2}\pmod p其中

(np)

\left(n\over p\right)为勒让德符号。因为 x≡n12(modp)x \equiv n^{1\over2}\pmod p，所以有 xp−1≡np−12≡1(modp)x^{p-1}\equiv n^{{p-1}\over 2}\equiv 1 \pmod p。即

(np)=⎧⎩⎨⎪⎪1−10n在模p意义下是二次剩余n在模p意义下不是二次剩余n≡0(modp)

\left(\frac n p\right) =\left\{ \begin{array}{ll}1 & \text{$n$在模$p$意义下是二次剩余} \\-1 & \text{$n$在模$p$意义下不是二次剩余} \\0 & \text{$n\equiv 0\pmod p$} \end{array}\right.设

k=a2−n,ω=k√

k=a^2-n,\omega=\sqrt{k}则该方程的解为

x≡(a+ω)p+12(modp)

x\equiv (a+\omega)^{\frac {p+1} 2} \pmod p证明：若 kk是该模意义下的非二次剩余，则

wp−1=np−12=−1

w^{p-1}=n^{{p-1}\over 2}=-1同时:

(a+b)p≡ap+bp(modp)

(a+b)^{p}\equiv a^p+b^p\pmod p则

x21≡≡≡≡≡≡≡≡(a+ω)p+1(a+ω)p(a+ω)(ap+ωp)(a+ω)(ap−1a+ωp−1ω)(a+ω)(a−ω)(a+ω)a2−ω2a2−(a2−n)n(modp)

\begin{eqnarray} x_1^2 & \equiv & (a+\omega)^{p+1} \\& \equiv & (a+\omega)^p(a+\omega) \\& \equiv & (a^p+\omega^p)(a+\omega) \\& \equiv & (a^{p-1}a+{\omega}^{p-1}\omega)(a+\omega) \\& \equiv & (a-\omega)(a+\omega) \\& \equiv & a^2-\omega^2 \\& \equiv & a^2-(a^2-n) \\& \equiv & n\pmod p \end{eqnarray}另一解为

x2≡−x1(modp)

x_2\equiv -x_1\pmod p
如何找到这个 kk呢。因为非二次剩余的个数大约有一半，随机几次即可。
后面去实现了一下，发现要重新定义一个复数域Fp2\mathbb{F}_{p^2}，然后就好啦。

struct Complex {ll r, i;Complex(ll _r = 0, ll _i = 0):r(_r), i(_i) {}inline Complex operator +(ll x) {return Complex((x + r) % P, i);}inline Complex operator *(Complex a) {ll rr = (a.r * r % P + w * a.i % P * i % P + P * 3) % P,ii = (a.i * r % P + a.r * i % P) % P;return Complex(rr, ii);}
};inline ll Qr(ll x, ll p) {if (Pow(x, (P - 1) / 2) == P - 1) return -1;if (Pow(x, (P - 1) / 2) == 0) return 0;ll a;Complex k(0, 1);while (true) {a = rand();if (Pow((a * a - x + P) % P, (P - 1) / 2) == P - 1) break;}w = (a * a - x + P) % P;return Pow(k + a, (P + 1) / 2);
}