The use of quadratic remainder is
a bit like an inverse element. When solving the problem, there should be no decimals in
the modulus. All that is needed is this x. The following is the case where p is a prime number.
Theorem
1. There are a total of (p-1)/2+1 kinds of n to make the equation have a solution.
2
3. 
Solving code
#include<iostream>
#include<cstring>
#include<iomanip>
#include<cstdio>
#include<cstdlib>
#include<queue>
#include<map>
#include<stack>
#include<vector>
#include<iomanip>
#include<algorithm>
#include<unordered_map>
#include<string>
#include<cmath>
#include<cstdio>
#include<fstream>
#include<time.h>
#define lson rt<<1
#define rson rt<<1|1
#define fmid (tree[rt].l + tree[rt].r) / 2
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 10;
const ll mod = 1e9 + 9;
const double eps = 1e-6;
const double PI = acos(-1);
#define MOD(a, b) a >= b ? a % b + b : a
#define random(a,b) (rand()%(b-a+1)+a)
void acc_ios()
{
ios::sync_with_stdio(false);
cin.tie(0);
}
ll w;
bool ok;
struct QuadraticField
{
ll x, y;
QuadraticField operator*(QuadraticField T)
{
QuadraticField ans;
ans.x = (this->x * T.x % mod + this->y * T.y % mod * w % mod) % mod;
ans.y = (this->x * T.y % mod + this->y * T.x % mod) % mod;
return ans;
}
QuadraticField operator^(ll b)
{
QuadraticField ans;
QuadraticField a = *this;
ans.x = 1;
ans.y = 0;
while(b)
{
if(b & 1)
{
ans = ans * a;
b--;
}
b /= 2;
a = a * a;
}
return ans;
}
};
ll fast_pow(ll a, ll b)
{
ll res = 1;
while(b)
{
if(b & 1) res = (res * a) % mod;
a = (a * a) % mod;
b >>= 1;
}
return res;
}
ll legender(ll a)
{
ll ans = fast_pow(a, (mod - 1) / 2);
if(ans + 1 == mod)
return -1;
else
return ans;
}
ll getW(ll n, ll a)
{
return ((a * a - n) % mod + mod) % mod;
}
ll solve(ll n)
{
ll a;
if(mod == 2)
return n;
if(legender(n) == -1)
ok = false;
srand((unsigned)time(NULL));
while(1)
{
a = random(0, mod - 1);
w = getW(n, a);
if(legender(w) == -1)
break;
}
QuadraticField ans, res;
res.x = a;
res.y = 1;
ans = res^((mod + 1) / 2);
return ans.x;
}
ll comp(ll a, ll b)
{
if(a < b) return 0;
if(a == b) return 1;
if(b > a - b) b = a - b;
ll ans = 1, ca = 1, cb = 1;
for(int i = 0; i < b; i++)
{
ca = ca * (a - i) % mod;
cb = cb * (b - i) % mod;
}
ans = ca * fast_pow(cb, mod - 2) % mod;
return ans;
}
int main()
{
acc_ios();
ll n, ans1, ans2;
cin>>n;
ok = true;
n %= mod;
ans1 = solve(n);
ans2 = mod - ans1;
if(!ok)
{
cout<<"no root"<<endl;
return 0;
}
if(ans1 == ans2)
cout<<ans1<<endl;
else
cout<<ans1<<" "<<ans2<<endl;
return 0;
}