Count of Ordered Pairs (X, Y) satisfying the Equation 1/X + 1/Y = 1/N

Given a positive integer N, the task is to find the number of ordered pairs (X, Y) where both X and Y are positive integers, such that they satisfy the equation 1/X + 1/Y = 1/N.

Examples:

Input: N = 5 
Output: 3 
Explanation: Only 3 pairs {(30,6), (10,10), (6,30)} satisfy the given equation.

Input: N = 360 
Output: 105 

Approach:
Follow the steps to solve the problem: 

• Solve for X using the given equation. 
  

1/X + 1/Y = 1/N 
=> 1/X = 1/N – 1/Y 
=> 1/X = (Y – N) / NY 
=> X = NY / (Y – N) 

X = [ NY / (Y – N) ] * (1)

=> X = [ NY / (Y – N) ] * [1 – N/Y + N/Y]

=> X = [ NY / (Y – N) ] * [(Y- N)/Y + N/Y]

=> X = N + N² / (Y – N)

• Therefore, it can be observed that, to have a positive integer X, the remainder when N² is divided by (Y – N) needs to be 0.
• It can be observed that the minimum value of Y can be N + 1 (so that denominator Y – N > 0) and the maximum value of Y can be N² + N so that N²/(Y – N) remains a positive integer ≥ 1.
• Then iterate over the maximum and minimum possible values of Y, and for each value of Y for which N² % (Y – N) == 0, increment count.
• Finally, return count as the number of ordered pairs.

Below is the implementation of the above approach:

C++

// C++ Program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find number of ordered
// positive integer pairs (x,y) such
// that  they satisfy the equation
void solve(int n)
{
    // Initialize answer variable
    int ans = 0;

// Iterate over all possible values of y
    for (int y = n + 1; y <= n * n + n; y++) {

        // For valid x and y,
        // (n*n)%(y - n) has to be 0
        if ((n * n) % (y - n) == 0) {

            // Increment count of ordered pairs
            ans += 1;
        }
    }

    // Print the answer
    cout << ans;
}

// Driver Code
int main()
{
    int n = 5;
    // Function call
    solve(n);
    return 0;
}

Java

// Java program for the above approach
class GFG{

// Function to find number of ordered
// positive integer pairs (x,y) such
// that they satisfy the equation
static void solve(int n)
{

    // Initialize answer variable
    int ans = 0;

    // Iterate over all possible values of y
    for(int y = n + 1; y <= n * n + n; y++)
    {

        // For valid x and y,
        // (n*n)%(y - n) has to be 0
        if ((n * n) % (y - n) == 0)
        {

            // Increment count of ordered pairs
            ans += 1;
        }
    }

    // Print the answer
    System.out.print(ans);
}

// Driver Code
public static void main(String[] args)
{
    int n = 5;

    // Function call
    solve(n);
}
}

// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach

# Function to find number of ordered
# positive integer pairs (x,y) such
# that they satisfy the equation
def solve(n):

    # Initialize answer variable
    ans = 0

    # Iterate over all possible values of y
    y = n + 1
    while(y <= n * n + n):

        # For valid x and y,
        # (n*n)%(y - n) has to be 0
        if ((n * n) % (y - n) == 0):

            # Increment count of ordered pairs
            ans += 1

        y += 1

    # Print the answer
    print(ans)

# Driver Code
n = 5

# Function call
solve(n)

# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;

class GFG{

// Function to find number of ordered
// positive integer pairs (x,y) such
// that they satisfy the equation
static void solve(int n)
{

    // Initialize answer variable
    int ans = 0;

    // Iterate over all possible values of y
    for(int y = n + 1; y <= n * n + n; y++)
    {

        // For valid x and y,
        // (n*n)%(y - n) has to be 0
        if ((n * n) % (y - n) == 0)
        {

            // Increment count of ordered pairs
            ans += 1;
        }
    }

    // Print the answer
    Console.Write(ans);
}

// Driver Code
public static void Main(String[] args)
{
    int n = 5;

    // Function call
    solve(n);
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>
// javascript program for the above approach
// Function to find number of ordered
    // positive integer pairs (x,y) such
    // that they satisfy the equation
    function solve(n) {

        // Initialize answer variable
        var ans = 0;

        // Iterate over all possible values of y
        for (y = n + 1; y <= n * n + n; y++) {

            // For valid x and y,
            // (n*n)%(y - n) has to be 0
            if ((n * n) % (y - n) == 0) {

                // Increment count of ordered pairs
                ans += 1;
            }
        }

        // Print the answer
        document.write(ans);
    }

    // Driver Code

        var n = 5;

        // Function call
        solve(n);

// This code contributed by umadevi9616
</script>

Output:

3

Time Complexity: O(N²)
Auxiliary Space: O(1)