# 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:
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: is always an integer
=> 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 equation 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 solution 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  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

Output:
12

Time Complexity: O(sqrt(N))

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