Find two numbers such that difference of their squares equal to N

Given an integer N, the task is to find two non-negative integers A and B, such that A² – B² = N. If no such integers exist, then print -1.

Examples:

Input: N = 7 
Output: 4 3 
Explanation:
The two integers 4 and 3 can be represented as 4² – 3² = 7.
Input: N = 6 
Output: -1 
Explanation:
No pair of (A, B) exists that satisfies the required condition.  

Approach:

• A² – B² can be represented as (A – B) * (A + B).
  

A² – B² = (A – B) * (A + B)

• Thus, for A² – B² to be equal to N, both (A + B) and (A – B) should be divisors of N.
• Considering A + B and A – B to be equal to C and D respectively, then C and D must be divisors of N such that C ≤ D and C and D should be of the same parity.
• Hence, in order to solve this problem, we just need to find any pair C and D satisfying the above condition. If no such C & D exists, then the print -1.

Below is the implementation of the above approach:

C++

// C++ Program to find two numbers
// with difference of their
// squares equal to N

#include <bits/stdc++.h>
using namespace std;

// Function to check and print
// the required two positive integers
void solve(int n)
{

    // Iterate till sqrt(n) to find
    // factors of N
    for (int x = 1; x <= sqrt(n); x++) {

        // Check if x is one
        // of the factors of N
        if (n % x == 0) {

            // Store the factor
            int small = x;

            // Compute the other factor
            int big = n / x;

            // Check if the two factors
            // are of the same parity
            if (small % 2 == big % 2) {

                // Compute a and b
                int a = (small + big) / 2;
                int b = (big - small) / 2;

                cout << a << " "
                     << b << endl;
                return;
            }
        }
    }

    // If no pair exists
    cout << -1 << endl;
}

// Driver Code
int main()
{
    int n = 7;

    solve(n);

    return 0;
}

Java

// Java Program to find two numbers
// with difference of their
// squares equal to N
import java.util.*;
class GFG{

// Function to check and print
// the required two positive integers
static void solve(int n)
{

    // Iterate till sqrt(n) to find
    // factors of N
    for (int x = 1; x <= Math.sqrt(n); x++)
    {

        // Check if x is one
        // of the factors of N
        if (n % x == 0)
        {

            // Store the factor
            int small = x;

            // Compute the other factor
            int big = n / x;

            // Check if the two factors
            // are of the same parity
            if (small % 2 == big % 2)
            {

                // Compute a and b
                int a = (small + big) / 2;
                int b = (big - small) / 2;

                System.out.print(a + " " + b);
                return;
            }
        }
    }

    // If no pair exists
    System.out.print(-1);
}

// Driver Code
public static void main(String args[])
{
    int n = 7;

    solve(n);
}
}

// This code is contributed by Code_Mech

Python3

# Python3 Program to find two numbers
# with difference of their
# squares equal to N
from math import sqrt

# Function to check and print
# the required two positive integers
def solve(n) :

    # Iterate till sqrt(n) to find
    # factors of N
    for x in range(1, int(sqrt(n)) + 1) :

        # Check if x is one
        # of the factors of N
        if (n % x == 0) :

            # Store the factor
            small = x;

            # Compute the other factor
            big = n // x;

            # Check if the two factors
            # are of the same parity
            if (small % 2 == big % 2) :

                # Compute a and b
                a = (small + big) // 2;
                b = (big - small) // 2;
                print(a, b) ;
                return;

    # If no pair exists
    print(-1);

# Driver Code
if __name__ == "__main__" :
    n = 7;
    solve(n);

# This code is contributed by AnkitRai01

C#

// C# Program to find two numbers
// with difference of their
// squares equal to N
using System;
class GFG{

// Function to check and print
// the required two positive integers
static void solve(int n)
{

    // Iterate till sqrt(n) to find
    // factors of N
    for (int x = 1; x <= Math.Sqrt(n); x++)
    {

        // Check if x is one
        // of the factors of N
        if (n % x == 0)
        {

            // Store the factor
            int small = x;

            // Compute the other factor
            int big = n / x;

            // Check if the two factors
            // are of the same parity
            if (small % 2 == big % 2)
            {

                // Compute a and b
                int a = (small + big) / 2;
                int b = (big - small) / 2;

                Console.WriteLine(a + " " + b);
                return;
            }
        }
    }

    // If no pair exists
    Console.WriteLine(-1);
}

// Driver Code
public static void Main()
{
    int n = 7;

    solve(n);
}
}

// This code is contributed by Code_Mech

Javascript

<script>
// javascript Program to find two numbers
// with difference of their
// squares equal to N

    // Function to check and print
    // the required two positive integers
    function solve(n) {

        // Iterate till sqrt(n) to find
        // factors of N
        for (var x = 1; x <= Math.sqrt(n); x++) {

            // Check if x is one
            // of the factors of N
            if (n % x == 0) {

                // Store the factor
                var small = x;

                // Compute the other factor
                var big = n / x;

                // Check if the two factors
                // are of the same parity
                if (small % 2 == big % 2) {

                    // Compute a and b
                    var a = (small + big) / 2;
                    var b = (big - small) / 2;

                    document.write(a + " " + b);
                    return;
                }
            }
        }

        // If no pair exists
        document.write(-1);
    }

    // Driver Code

        var n = 7;

        solve(n);

// This code contributed by aashish1995
</script>

Output:

4 3

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