Can I solve this equation without looping for each value of x and y?

Question

Given 2 positive integers c and n, find any 2 positive integer answers for x and y that satisfy the following equation: √x + √y = √z where z = c²n.

Input

The first line of the input contains a single integer t (1≤t≤10⁵), the number of test cases. The first and only line of each test case consists of 2 space-separated integers c and n (4≤c²n≤10¹⁸, 2≤c≤10⁹, 1≤n≤10⁹)

Output

For each test case output 2 space-separated positive integers x and y, the answer to the equation.

If there are multiple answers, print any.

I tried to make y = n, and then calculate x by quadratic analysis:

√x + √y = c*√n
=> x + 2√y√x + y = c²n
=> x + 2√y√x + y - c²n =0

Then, I calculated x by quadratic analysis:

=> x = (-b + √(b² - 4ac))/2a

Note: the time limit for this problem is 1.3 seconds, which means in the worst case I shouldn't use more than 10¹⁰ or 10¹¹ steps.

#include <iostream>
#include <cmath>

using namespace std;

int main()
{
    int t, n , c;
    double x, num1, num2;

    cin >> t;

    for(int i=0; i<t; i++)
    {
        cin >> c >> n;

        num1 = (double)n - (pow(c, 2)*n);
        num2 = (double) ((-2*sqrt(n))+sqrt((4*n)-(4*num1)))/2;

        cout << (double)pow(num2 , 2) << ' ' << n;
    }
}

I get the wrong answer but I don't know what's wrong.

Answer 1

#include <iostream>

void solve_test_case() {
    long long c, n;
    std::cin >> c >> n;
    
    // Strategy: Let's make x = (c-1)*(c-1)*n and y = n
    // This ensures √x + √y = c√n
    long long x = (c-1)*(c-1)*n;
    long long y = n;
    
    std::cout << x << " " << y << "\n";
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    int t;
    std::cin >> t;
    while (t--) {
        solve_test_case();
    }
    return 0;
}