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 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 integersvoid 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 Codeint main(){ int n = 7; solve(n); return 0;} |
Java
// Java Program to find two numbers// with difference of their// squares equal to Nimport java.util.*;class GFG{ // Function to check and print// the required two positive integersstatic 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 Codepublic 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 Codeif __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 Nusing System;class GFG{ // Function to check and print// the required two positive integersstatic 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 Codepublic 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)
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