已知0<=x<=y<p,p=1e9+7且有
(x+y)=bmodp
(x×y)=cmodp
求解任意一对x,y，不存在输出−1 −1。

由两式变化可得(y−x)2=(b2−4c+p)%p mod p，那么可以应用二次剩余定理解得y−x的值，我们可以知道(x+y)=b或者(x+y)=b+p，那么直接求解即可。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
ll qpow_num(ll a, ll b, ll mod) {
    ll ans = 1;
    while (b) {
        if (b & 1) {
            ans = (ans * a) % mod;
        }
        b >>= 1;
        a = (a * a) % mod;
    }
    return ans  % mod;
}

pll qmul_pll(pll a, pll b, ll w, ll mod) {
    pll ans;
    ans.first = ((a.first * b.first) % mod + ((a.second * b.second) % mod * w) % mod) % mod;
    ans.second = ((a.second * b.first) % mod + (a.first * b.second) % mod) % mod;
    return ans;
}

pll qpow_pll(pll a, ll b, ll w, ll mod) {
    pll ans = make_pair(1, 0);
    while (b) {
        if (b & 1) {
            ans = qmul_pll(ans, a, w, mod);
        }
        a = qmul_pll(a, a, w, mod);
        b >>= 1;
    }
    return ans;
}

ll solve(ll n, ll mod) {
    n %= mod;
    if (n <= 1) return n;
    if (mod == 2) return n;
    if (qpow_num(n, (mod - 1) / 2, mod) == mod - 1) return -1;
    ll a, w;
    while (1) {
        a = rand()% mod;
        w = ((a * a - n) % mod + mod) % mod;
        if (qpow_num(w, (mod - 1) / 2, mod) == mod - 1) break;
    }
    pll ans = make_pair(a, 1);
    return qpow_pll(ans, (mod + 1) / 2, w, mod).first;
}
ll mod = 1000000007;
ll getans(ll a, ll b, ll &x, ll &y) {
    if ((a + b) % 2 == 0) {
        y = (a + b) / 2;
        x = y - a;
        if (x >= 0 && y >= 0 && x < mod && y < mod) return true;
    }
    return false;
}

int main() {
    int T;
    srand(time(0));
    cin >> T;
    while (T--) {
        ll b, c;
        cin >> b >> c;

        ll n = b * b - 4 * c + mod;
        n = (n % mod + mod) % mod;
        ll ans1 = solve(n, mod);    
        if (ans1 == -1) {
            cout << "-1 -1" << endl; continue;
        }
        ll ans2 = mod - ans1;
        ll x, y;
        if (getans(ans1, b, x, y)) {
            cout << x << " " << y << endl; continue;
        }
        if (getans(ans1, b + mod, x, y)) {
            cout << x << " " << y << endl; continue;
        }
        if (getans(ans2, b, x, y)) {
            cout << x << " " << y << endl; continue;
        }
        if (getans(ans2, b + mod, x, y)) {
            cout << x << " " << y << endl; continue;
        }
        cout << "-1 -1" << endl;

    }
    return 0;
}