# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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 ` `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 `

Output:

```4 3
```

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

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : AnkitRai01, Code_Mech