Number of engagements in C program?

Here we will see the engagement number. This is a pair of numbers in which the sum of the true factors of one number is 1 more than that of the other number. We have to find these pairs

For example, this pair is like (48, 75). So the divisors of 48 are {1, 2, 3, 4, 6, 8, 12, 16, 24}, and the sum is 76. Likewise, the divisors of 75 are {1, 3, 5, 15, 25}, so the sum is 49.

Algorithm

Engaged Pairs (n) -

begin
   for num in range 1 to n, do
      sum := 1
      for i in range 2 to num, do
         if num is divisible by i, then
            sum := sum + i
            if i * i is not same as num, then
               sum := sum + num / i
            end if
         end if
         if sum > num, then
            num2 := sum – 1
            sum2 := 1
            for j in range 2 to num2, do
               if num2 is divisible by j, then
                  sum2 := sum2 + j
                  if j * j is not same as num2, then
                     sum2 := sum2 + num2 / j
                  end if
               end if
            done
            if sum2 = num + 1, then
               print the pair num and num2
            end if
         end if
      done
   done
end

Example

#include <iostream>
using namespace std;
void BetrothedPairs(int n) {
   for (int num = 1; num < n; num++) {
      int sum = 1;
      for (int i = 2; i * i <= num; i++) { //go through each number to get proper divisor
         if (num % i == 0) {
            sum += i;
            if (i * i != num) //avoid to include same divisor twice
            sum += num / i;
         }
      }
      if (sum > num) {
         int num2 = sum - 1;
         int sum2 = 1;
         for (int j = 2; j * j <= num2; j++){
            if (num2 % j == 0) {
               sum2 += j;
               if (j * j != num2)
               sum2 += num2 / j;
            }
         }
         if (sum2 == num+1)
         cout << "(" << num << ", " << num2 <<")" << endl;
      }
   }
}
int main() {
   int n = 5000;
   BetrothedPairs(n);
}

Output

1

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