Check if N can be represented as sum of squares of two consecutive integers

Given an integer N, the task is to check whether N can be represented as a sum of squares of two consecutive integers or not.

Examples:

Input: N = 5 
Output: Yes 
Explanation:
The integer 5 = 1² + 2² where 1 and 2 are consecutive numbers.

Input: 13 
Output: Yes 
Explanation:
13 = 2² + 3²

Approach: This equation can be represented as: 

=> 
=> 
=> 

Finally, check the value of computed using this formula is an integer, which means that N can be represented as the sum of squares 2 consecutive integers

Below is the implementation of the above approach:

C++

// C++ implementation to check that
// a number is sum of squares of 2
// consecutive numbers or not

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

// Function to check that the
// a number is sum of squares of 2
// consecutive numbers or not
bool isSumSquare(int N)
{
    float n
        = (2 + sqrt(8 * N - 4))
          / 2;

    // Condition to check if the
    // a number is sum of squares of 2
    // consecutive numbers or not
    return (n - (int)n) == 0;
}

// Driver Code
int main()
{
    int i = 13;

    // Function call
    if (isSumSquare(i)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
    return 0;
}

Java

// Java implementation to check that
// a number is sum of squares of 2
// consecutive numbers or not
import java.lang.Math;

class GFG{

// Function to check that the
// a number is sum of squares of 2
// consecutive numbers or not
public static boolean isSumSquare(int N)
{
    double n = (2 + Math.sqrt(8 * N - 4)) / 2;

    // Condition to check if the
    // a number is sum of squares of 2
    // consecutive numbers or not
    return(n - (int)n) == 0;
}

// Driver code
public static void main(String[] args)
{
    int i = 13;

    // Function call
    if (isSumSquare(i))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 implementation to check that
# a number is sum of squares of 2
# consecutive numbers or not
import math

# Function to check that the a
# number is sum of squares of 2
# consecutive numbers or not
def isSumSquare(N):

    n = (2 + math.sqrt(8 * N - 4)) / 2

    # Condition to check if the a
    # number is sum of squares of
    # 2 consecutive numbers or not
    return (n - int(n)) == 0

# Driver code
if __name__=='__main__':

    i = 13

    # Function call
    if isSumSquare(i):
        print('Yes')
    else :
        print('No')

# This code is contributed by rutvik_56

C#

// C# implementation to check that
// a number is sum of squares of 2
// consecutive numbers or not
using System;
class GFG{

// Function to check that the
// a number is sum of squares of 2
// consecutive numbers or not
public static bool isSumSquare(int N)
{
    double n = (2 + Math.Sqrt(8 * N - 4)) / 2;

    // Condition to check if the
    // a number is sum of squares of 2
    // consecutive numbers or not
    return(n - (int)n) == 0;
}

// Driver code
public static void Main(String[] args)
{
    int i = 13;

    // Function call
    if (isSumSquare(i))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}

// This code is contributed by sapnasingh4991

Javascript

<script>

// Javascript implementation to check that
// a number is sum of squares of 2
// consecutive numbers or not

// Function to check that the
// a number is sum of squares of 2
// consecutive numbers or not
function isSumSquare(N)
{
    var n = (2 + Math.sqrt(8 * N - 4)) / 2;

    // Condition to check if the
    // a number is sum of squares of 2
    // consecutive numbers or not
    return (n - parseInt( n)) == 0;
}

// Driver code
var i = 13;

// Function call
if (isSumSquare(i))
{
    document.write("Yes");
}
else
{
    document.write("No");
}

// This code is contributed by todaysgaurav

</script>

Output:

Yes

Note: In order to print the integers, we can easily solve the above equation to get the roots.

Time Complexity: O(1)

Auxiliary Space: O(1)