Find smallest perfect square number A such that N + A is also a perfect square number


Given a positive number N. The task is to find out the smallest perfect square number A such that N + A is also a perfect square number or return -1.

Examples:

Input: N = 3
Output: 1
Explanation: 
As 1 + 3 = 4 = 22

Input: N=1
Output: -1

Naive Approach:
Traverse M from {1, 2, 3, 4, 5…} and check whether (N + M * M) is a perfect square number or not.

Efficient Approach:

  • On observing, we have an equation like:



    • N + (X * X) = (M * M) where N is given and M and X are unknown.
    • We can rearrange it and get:
      • N = (M * M) – (X * X)
      • N = (M + X) * (M – X)
  • Now we can see that for obtaining N, we need to find the factor of N.The factor of N can be obtained in O(N) time. But it can be optimized to O(N^1/2) by this method.
  • Let the factor of N be a and b = (N / a). So, from the above equation a = (M – X) and b = (M + X), and after solving this we can obtain the value of X = (b – a)/2.

Below is the implementation of the above approach:

C++

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// C++ code to find out the smallest 
// perfect square X which when added to N 
// yields another perfect square number. 
#include<bits/stdc++.h>
using namespace std;
  
long SmallestPerfectSquare(long N){
      
    // X is the smallest perfect 
    // square number 
    long X = (long)1e9; 
    long ans; 
      
    // Loop from 1 to square root of N 
    for(int i = 1; i < sqrt(N); i++)
    
      
        // Condition to check whether i 
        // is factor of N or not 
        if (N % i == 0)
        
            long a = i; 
            long b = N / i; 
      
            // Condition to check whether 
            // factors satisfies the 
            // equation or not 
            if((b - a != 0) && ((b - a) % 2 == 0))
            
                          
                // Stores minimum value 
                X = min(X, (b - a) / 2); 
            
        
    
      
    // Return if X * X if X is not equal 
    // to 1e9 else return -1 
    if (X != 1e9) 
        ans = X * X; 
    else
        ans = -1; 
              
    return ans; 
      
// Driver code 
int main() 
{
    long N = 3; 
    cout << SmallestPerfectSquare(N); 
    return 0;
  
// This code is contributed by AnkitRai01 

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Java

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// Java code to find out the smallest 
// perfect square X which when added to N 
// yields another perfect square number. 
  
public class GFG {
      
  
    static long SmallestPerfectSquare(long N)
    {
      
        // X is the smallest perfect 
        // square number 
        long X = (long)1e9;
        long ans;
      
        // Loop from 1 to square root of N 
        for(int i = 1; i < Math.sqrt(N); i++){ 
      
            // Condition to check whether i 
            // is factor of N or not 
            if (N % i == 0){
                long a = i ;
                long b = N / i; 
      
                // Condition to check whether 
                // factors satisfies the 
                // equation or not 
                if ((b - a != 0) && ((b - a) % 2 == 0)){
                          
                    // Stores minimum value 
                    X = Math.min(X, (b - a) / 2) ;
                }
            }
        }
      
        // Return if X * X if X is not equal 
        // to 1e9 else return -1 
        if (X != 1e9)
            ans = X * X;
        else
            ans = -1;
              
        return ans; 
    }
      
    // Driver code 
    public static void main (String[] args){
        long N = 3;
          
        System.out.println(SmallestPerfectSquare(N)) ;
      
        }
}
// This code is contributed by AnkitRai01

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Python3

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# Python3 code to find out the smallest
# perfect square X which when added to N
# yields another perfect square number.
import math
def SmallestPerfectSquare(N):
  
    # X is the smallest perfect 
    # square number 
    X = 1e9
  
    # Loop from 1 to square root of N
    for i in range(1, int(math.sqrt(N)) + 1):
  
        # Condition to check whether i 
        # is factor of N or not
        if N % i == 0:
            a = i
            b = N // i  
  
            # Condition to check whether  
            # factors satisfies the 
            # equation or not 
            if b - a != 0 and (b - a) % 2 == 0:
                     
                # Stores minimum value
                 X = min(X, (b - a) // 2)
  
    # Return if X * X if X is not equal
    # to 1e9 else return -1
    return(X * X if X != 1e9 else -1)
  
# Driver code 
if __name__ == "__main__" :  
    
    N = 3
    
    print(SmallestPerfectSquare(N))

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C#

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// C# code to find out the smallest 
// perfect square X which when added to N 
// yields another perfect square number. 
using System;
  
class GFG { 
      
    static long SmallestPerfectSquare(long N) 
    
        // X is the smallest perfect 
        // square number 
        long X = (long)1e9; 
        long ans; 
      
        // Loop from 1 to square root of N 
        for(int i = 1; i < Math.Sqrt(N); i++){ 
      
            // Condition to check whether i 
            // is factor of N or not 
            if (N % i == 0)
            
                long a = i; 
                long b = N / i; 
      
                // Condition to check whether 
                // factors satisfies the 
                // equation or not 
                if ((b - a != 0) && ((b - a) % 2 == 0))
                
                    // Stores minimum value 
                    X = Math.Min(X, (b - a) / 2); 
                
            
        
      
        // Return if X*X if X is not equal 
        // to 1e9 else return -1 
        if (X != 1e9) 
            ans = X * X; 
        else
            ans = -1; 
              
        return ans; 
    
      
    // Driver code 
    public static void Main (string[] args)
    
        long N = 3; 
        Console.WriteLine(SmallestPerfectSquare(N)); 
    
  
// This code is contributed by AnkitRai01 
  

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Output:

1

Time complexity: sqrt(N)

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