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

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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:

=> $N = K^{2} + (K - 1)^{2}$
=> $N = 2*K^{2} - 2*K + 1$
=> $K = \frac{2 + \sqrt{8*N - 4}}{2}$

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(logN) because it is using inbuilt sqrt function
Auxiliary Space: O(1)

Approach 2: Iterative Method:

Here’s another approach to check whether a number is the sum of squares of two consecutive numbers or not:

Iterate through all possible pairs of consecutive integers (x, x+1) such that x is less than the square root of N.
For each pair (x, x+1), compute their sum of squares as x^2 + (x+1)^2.
If the sum of squares is equal to N, return true. Otherwise, continue iterating through the pairs of consecutive integers.
If no pair of consecutive integers satisfies the condition, return false.

Here’s the implementation of this approach in C++:

C++

#include <bits/stdc++.h>
using namespace std;
 
bool isSumSquare(int N) {
    for (int x = 0; x * x < N; x++) {
        int sum = x * x + (x + 1) * (x + 1);
        if (sum == N) {
            return true;
        }
    }
    return false;
}
 
int main() {
    int N = 13;
    if (isSumSquare(N)) {
        cout << "Yes";
    } else {
        cout << "No";
    }
    return 0;
}

Java

public class Main {
    // Function to check if a number can be expressed as the 
    //sum of squares of two consecutive numbers
    static boolean isSumSquare(int N) {
        for (int x = 0; x * x < N; x++) {
            int sum = x * x + (x + 1) * (x + 1);
            if (sum == N) {
                return true;
            }
        }
        return false;
    }
//   Driver code
    public static void main(String[] args) {
        int N = 13;
        if (isSumSquare(N)) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}

Python3

def isSumSquare(N):
    for x in range(0, N):
        sum = x * x + (x + 1) * (x + 1)
        if sum == N:
            return True
        elif sum > N:
            return False
    return False
 
N = 13
if isSumSquare(N):
    print("Yes")
else:
    print("No")

C#

using System;
 
public class Program {
  public static bool IsSumSquare(int N)
  {
    for (int x = 0; x * x < N; x++) {
      int sum = x * x + (x + 1) * (x + 1);
      if (sum == N) {
        return true;
      }
    }
    return false;
  }
 
  public static void Main()
  {
    int N = 13;
    if (IsSumSquare(N)) {
      Console.WriteLine("Yes");
    }
    else {
      Console.WriteLine("No");
    }
  }
}
// This code is contributed by user_dtewbxkn77n

Javascript

//Javascript Code
function isSumSquare(N) {
    for (let x = 0; x * x < N; x++) {
        let sum = x * x + (x + 1) * (x + 1);
        if (sum === N) {
            return true;
        }
    }
    return false;
}
 
let N = 13;
if (isSumSquare(N)) {
    console.log("Yes");
} else {
    console.log("No");
}

Output:

Yes

Time Complexity: O(N)
Auxiliary Space: O(1)

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