Project Euler 454 - Diophantine Reciprocals

/*
https://projecteuler.net/problem=454
In the following equation x, y, and n are positive integers.

(1/x) + (1/y) = (1/n)
For a limit L we define F(L) as the number of solutions which satisfy x < y ≤ L.

We can verify that F(15) = 4 and F(1000) = 1069.
Find F(10^12).
*/


#include <iostream>
#include <stdlib.h>
#include <math.h>

using namespace std;

int findNumOfSolutions(int limit){
    int counter = 0;
    for(int x=1; x<limit; x++){
        for(int y=x+1; y<limit+1; y++){
            double diairesi = double(x * y) / double(x + y);
            // cout << "x = " << x << ", y = " << y << " ---> " << diairesi << " " << int(diairesi) << endl;
            if(diairesi == int(diairesi)){
                counter++;
            }
        }
    }
    return counter;
}

int main()
{
    cout << "I look for pairs of positive integers x, y and n, that satisfy the following condition:\n";
    cout << " (1/x) + (1/y) = (1/n) <==> (x*y) / (x+y) = n = integer, where: x < y <=L, L = a positive integer limit\n\n";
    cout << "Number of solutions from 1 to 15 = F(15) = " << findNumOfSolutions(15) << endl;
    cout << "Number of solutions from 1 to 1000 = F(1000) = " << findNumOfSolutions(1000) << endl;
    return 0;
}