Comprobar si N se puede representar como la suma de los cuadrados de dos enteros consecutivos

Dado un número entero N , la tarea es comprobar si N se puede representar como una suma de cuadrados de dos números enteros consecutivos o no.

Ejemplos: 

Entrada: N = 5 
Salida: Sí 
Explicación: 
El entero 5 = 1 ² + 2 ² donde 1 y 2 son números consecutivos.

Entrada: 13 
Salida: Sí 
Explicación: 
13 = 2 ² + 3 ² 

Enfoque: Esta ecuación se puede representar como: 

=> 
=> 
=> 

Finalmente, verifique que el valor de calculado usando esta fórmula sea un número entero, lo que significa que N se puede representar como la suma de los cuadrados de 2 números enteros consecutivos

A continuación se muestra la implementación del enfoque anterior:

// 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 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 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# 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

<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>

Yes

Nota: para imprimir los números enteros, podemos resolver fácilmente la ecuación anterior para obtener las raíces.
 

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)