Filioi Arithmoi

#include <iostream>
#include <stdlib.h>
#include <math.h>
#include <vector>
#include <algorithm>
#define N 10000

using namespace std;

// Auxiliary Functions
vector <int> findDivisors(unsigned int n){
    vector <int> divisors = vector <int> ();
    divisors.push_back(1);
    for(int i=2; i<=n/2; i++){
        if(n % i == 0){
            divisors.push_back(i);
        }
    }
    return divisors;
}

bool isPrime(unsigned int n){
    // Prime numbers have only 2 divisors: 1 and n
    // If I use the above function, the returned vector will contain only 1 divisor, specifically only the number 1
    // Because I don't push back n as a divisor of number n
    vector <int> divisors = findDivisors(n);
    if(divisors.size() == 1){
        return true;
    }
    else{
        return false;
    }
}

int main()
{
    /*
    // Check the function
    for(int i=2; i<20; i++){
        vector <int> v = vector <int> ();
        v = findDivisors(i);
        cout << i << " ----> ";
        for(unsigned int j=0; j<v.size(); j++){
            cout << v[j] << " ";
        }
        cout << endl;
    }
    */
    vector <int> filioi = vector <int> ();
    for(int i=2; i<=N; i++){
        // First, I have to find the divisors of teach number i that I scan with the for-loop
        vector <int> divisors1 = findDivisors(i);
        int sum1 = 0;
        for(unsigned int j=0; j<divisors1.size(); j++){
            sum1 += divisors1[j];
        }
        // I will find the divisors of the sum1
        // But first I have to check if it is prime
        // Because if it is prime, I will have only 1 divisor (number 1)
        if(isPrime(sum1)){
            continue;
        }
        vector <int> divisors2 = findDivisors(sum1);
        int sum2 = 0;
        for(unsigned int j=0; j<divisors2.size(); j++){
            sum2 += divisors2[j];
        }
        // Check if sum2 (sum of divisors of sum1) is equal to the number i
        bool flag1 = (i == sum2);
        bool flag2 = (i == sum1);
        bool pairHasBeenInserted = false;
        if(filioi.size() != 0){
            for(unsigned int j=0; j<filioi.size(); j++){
                if(i == filioi[j]){
                    pairHasBeenInserted = true;
                    break;
                }
            }
        }
        //cout << i << " ----> " << "flag1 = " << flag1 << ", flag2 = " << flag2 << endl;
        // Now, I have to check if I have a perfect number
        // In this case, i = sum2 but also i = sum1, because I have a perfect number and I don't want to
        // include this case in my problem
        if(flag1 == true && flag2 == false && pairHasBeenInserted == false){
            filioi.push_back(i);
            filioi.push_back(sum1);
        }
    }

    // The process for finding the pairs of filioi numbers has finished
    cout << "Filioi Arithmoi: " << endl;
    int counter = 0;
    for(unsigned int i=0; i<filioi.size(); i+=2){
        counter++;
        cout << "Pair " << counter << ": " << filioi[i] << ", " << filioi[i+1] << endl;
    }

    return 0;
}