Count of pairs of integers whose difference of squares is equal to N

Given a positive integer N, the task is to find the count of pairs of integers (x, y) whose difference of squares is equal to N, i.e., 

Examples:

Input: N = 20 
Output: 4 
Explanation:
The 4 possible pairs are (10, 2), (-10, 2), (-10, -2) and (10, -2).

Input: N = 80 
Output: 12 
Explanation:
The 12 possible pairs are: 
1. (40, 2), (-40, 2), (-40, -2) and (40, -2). 
2. (20, 4), (-20, 4), (-20, -4) and (20, -4). 
3. (10, 8), (-10, 8), (-10, -8) and (10, -8). 

Approach:
The given equation can also be written as:  

=> 
=> 

Now for an integral solution of the given equation:  

(x+y)(x-y) 

is always an integer 
=> (x+y)(x-y) 
 are divisors of N

Let  (x + y) = p1 and (x + y) = p2 
be the two equations where p1 & p2 are the divisors of N
such that p1 * p2 = N.

Solving for the above two equations we have:  

=> 
and 

From the above calculations, for x and y to be integral, then the sum of divisors must be even. Since there are 4 possible values for two values of x and y as (+x, +y), (+x, -y), (-x, +y) and (-x, -y). 
Therefore the total number of possible solutions is given by 4*(count pairs of divisors with even sum).

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 the integral
// solutions of the given equation
void findSolutions(int N)
{

    // Initialise count to 0
    int count = 0;

    // Iterate till sqrt(N)
    for (int i = 1; i <= sqrt(N); i++) {

        if (N % i == 0) {

            // If divisor's pair sum is even
            if ((i + N / i) % 2 == 0) {
                count++;
            }
        }
    }

    // Print the total possible solutions
    cout << 4 * count << endl;
}

// Driver Code
int main()
{
    // Given number N
    int N = 80;

    // Function Call
    findSolutions(N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{

// Function to find the integral
// solutions of the given equation
static void findSolutions(int N)
{

    // Initialise count to 0
    int count = 0;

    // Iterate till sqrt(N)
    for(int i = 1; i <= Math.sqrt(N); i++)
    {
       if (N % i == 0)
       {

           // If divisor's pair sum is even
           if ((i + N / i) % 2 == 0)
           {
               count++;
           }
       }
    }

    // Print the total possible solutions
    System.out.print(4 * count);
}

// Driver code
public static void main(String[] args)
{

    // Given number N
    int N = 80;

    // Function Call
    findSolutions(N);
}
}

// This code is contributed by Shubham Prakash.

Python3

# Python3 program for the above approach
import math;

# Function to find the integral
# solutions of the given equation
def findSolutions(N):

    # Initialise count to 0
    count = 0;

    # Iterate till sqrt(N)
    for i in range(1, int(math.sqrt(N)) + 1):

        if (N % i == 0):

            # If divisor's pair sum is even
            if ((i + N // i) % 2 == 0):
                count += 1;

    # Print the total possible solutions
    print(4 * count);

# Driver Code

# Given number N
N = 80;

# Function Call
findSolutions(N);

# This code is contributed by Code_Mech

C#

// C# program for the above approach
using System;
class GFG{

// Function to find the integral
// solutions of the given equation
static void findSolutions(int N)
{

    // Initialise count to 0
    int count = 0;

    // Iterate till sqrt(N)
    for(int i = 1; i <= Math.Sqrt(N); i++)
    {
        if (N % i == 0)
        {

            // If divisor's pair sum is even
            if ((i + N / i) % 2 == 0)
            {
                count++;
            }
        }
    }

    // Print the total possible solutions
    Console.Write(4 * count);
}

// Driver code
public static void Main(String[] args)
{

    // Given number N
    int N = 80;

    // Function Call
    findSolutions(N);
}
}

// This code is contributed by sapnasingh4991

Javascript

<script>

// Javascript program for the above approach

// Function to find the integral
// solutions of the given equation
function findSolutions(N)
{

    // Initialise count to 0
    let count = 0;

    // Iterate till sqrt(N)
    for(let i = 1; i <= Math.sqrt(N); i++)
    {
        if (N % i == 0)
        {

            // If divisor's pair sum is even
            if ((i + parseInt(N / i)) % 2 == 0)
            {
                count++;
            }
        }
    }

    // Print the total possible solutions
    document.write(4 * count + "<br>");
}

// Driver Code

// Given number N
let N = 80;

// Function Call
findSolutions(N);

// This code is contributed by souravmahato348

</script>

Output:
12

Time Complexity: O(sqrt(N))
Auxiliary Space: O(1)