P5491 【模板】二次剩余

\(\text{Summary}\)

实际上是做法的归纳

一切皆是结论性的，没有证明！

模 \(p\) 意义下的二次剩余有 \(\frac{p-1}2\) 个，二次非剩余也恰有那么多

考虑解关于 \(x\) 的同余方程

\[x^2 \equiv n \pmod p
\]

当 \(n=0\) 时，\(x=0\) 是唯一解

当 \(n \not= 0\) 时，若方程有解，则只有两个互为相反数的解

判断有无解：欧拉准则

考虑 \(n^{\frac{p-1}{2}}\) 模 \(p\) 的结果

由 \((n^{\frac{p-1}{2}})^2 \equiv 1 \pmod p\)

只其结果只为 \(1\) 或 \(-1\)

\(1\) 时有解，\(-1\) 时无解

求解的话，先随机找到一个 \(a\) 满足 \(a^2 - n\) 为二次非剩余，令 \(i^2 = a^2 - n \equiv -1 \pmod p\)

类似实部和虚部，定义这个 \(i\)

有

\[(a+i)^{p+1} \equiv n \pmod p
\]

证明的话，有

\(\text{Lemma 1}\)

\[i^p \equiv -i \pmod p
\]

\(\text{Lemma 2}\)

\[(x+y)^p \equiv x^p+y^p \pmod p
\]

然后

\[\begin{aligned}
(a+i)^{p+1} & \equiv (a+i)^p(a+i) \\
&\equiv (a+i)(a^p+i^p) \\
&\equiv (a+i)(a-i) \\
&\equiv a^2-i^2 \\
&\equiv n \pmod p
\end{aligned}
\]

\(p+1\) 为偶数，开方就很容易了

具体实现弄上“复数”即可

且 \((a+i)^{\frac{p+1}2}\) 的“虚部” 为 \(0\)

\(\text{Code}\)

#include <cstdio>
#include <algorithm>
#include <iostream>
#define IN inline
using namespace std;
typedef long long LL;

int T, n, p;
LL i2;

struct complex {
	LL x, y;
	IN complex(LL _x, LL _y) {x = _x, y = _y;}
	IN bool operator == (complex a) {return (a.x == x && a.y == y);}
	IN complex operator * (complex a) {
		return complex((a.x * x % p + a.y * y % p * i2 % p) % p, (a.x * y % p + a.y * x % p) % p);
	}
};

IN complex power(complex x, int y) {
	complex s = complex(1, 0);
	for(; y; y >>= 1, x = x * x) if (y & 1) s = s * x;
	return s;
}
IN int check(LL a) {return power(complex(a, 0), p - 1 >> 1) == complex(1, 0);}

IN void solve() {
	if (!n) {printf("0\n"); return;}
	if (power(complex(n, 0), p - 1 >> 1) == complex(p - 1, 0)) {printf("Hola!\n"); return;}
	LL a = rand() % p;
	while (!a || check((a * a % p - n + p) % p)) a = rand() % p;
	i2 = (a * a % p - n + p) % p;
	int x0 = power(complex(a, 1), p + 1 >> 1).x, x1 = p - x0;
	if (x0 > x1) swap(x0, x1);
	printf("%d %d\n", x0, x1);
}

int main() {
	scanf("%d", &T);
	for(; T; --T) scanf("%d%d", &n, &p), solve();
}