Skip to content
Related Articles

Related Articles

Table of Contents

Improve Article
Save Article
Like Article

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

  • Difficulty Level : Basic
  • Last Updated : 07 May, 2021

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: 

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

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)
 



My Personal Notes
arrow_drop_up
Previous
Floor value Kth root of a number using Recursive Binary Search
Next

How to create a custom String class in C++ with basic functionalities
Recommended Articles
Page :
Find K numbers with sum equal to N and sum of their squares maximized
07, May 20
Split squares of first N natural numbers into two sets with minimum absolute difference of their sums
02, Dec 20
Maximum sum of a subsequence having difference between their indices equal to the difference between their values
17, Jun 21
Possible values of Q such that, for any value of R, their product is equal to X times their sum
09, Aug 20
Find two numbers made up of a given digit such that their difference is divisible by N
07, May 20
Subsets of size K with product equal to difference of two perfect squares
04, Jun 20
Find two distinct numbers such that their LCM lies in given range
15, Aug 20
Count of pairs in a given range with sum of their product and sum equal to their concatenated number
14, Jun 20
Count of pairs in given range having their ratio equal to ratio of product of their digits
08, Sep 21
Count of pairs in given Array with product of their values equal to sum of their indices (arr[i]*arr[j] = i+j)
18, Nov 21
Count pairs from 1 to N such that their Sum is divisible by their XOR
17, Mar 20
Count the pairs in an array such that the difference between them and their indices is equal
09, Oct 19
Sort perfect squares in an array at their relative positions
26, Nov 18
Maximize sum of squares of array elements possible by replacing pairs with their Bitwise AND and Bitwise OR
12, Oct 20
Modify array by sorting nearest perfect squares of array elements having their digits sorted in decreasing order
05, Mar 21
Modify Binary Tree by replacing all nodes at even and odd levels by their nearest even or odd perfect squares respectively
22, Feb 21
Sort a linked list after converting elements to their squares
11, Nov 21
Make two numbers equal by multiplying with their prime factors minimum number of times
17, Jun 21
Find three integers less than or equal to N such that their LCM is maximum
29, Nov 18
Find the first N integers such that the sum of their digits is equal to 10
06, Sep 19
Find any K distinct odd integers such that their sum is equal to N
26, Mar 20
Find the pair (a, b) with minimum LCM such that their sum is equal to N
10, Aug 20
Maximize count of equal numbers in Array of numbers upto N by replacing pairs with their sum
28, Aug 20
Count ways to split array into two subsets having difference between their sum equal to K
09, Jun 21
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
shobhitgupta907
@shobhitgupta907
Vote for difficulty
Current difficulty :
Basic





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox

Start Your Coding Journey Now!