Generate a pythagoras triplet from a single integer

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Given a single integer n $\in$ [1, 1000000000], generate a Pythagoras triplet that includes n as one of its sides if possible.

Examples :

Input : 22
Output : Pythagoras Triplets exist i.e. 22 120 122

Input : 4
Output : Pythagoras Triplets exist i.e.  4 3 5

Input : 2
Output : No Pythagoras Triplet exists

Explanation:
Definition: “Pythagorean triplets” are integer solutions to the Pythagorean Theorem, i.e. they satisfy the equation $a^2 + b^2 = c^2$
Our task is to generate a triplet from an integral value. This can be a confusing task because, the side given to us can be a hypotenuse or a non-hypotenuse side.
Starting to calculate triplets by putting them in a formula, it can be deduced that only for 1 and 2, no triplets are possible.
Further,
if n is even, our triplets are calculated by formula $(n^2/4 - 1)^2 + n^2 = (n^2/4 + 1)^2$
if n is odd, our triplets are calculated by formula $(n^2-1)^2/2 + n^2 = (n^2+1)^2/2$
Proof:
Pythagoras’s Theorem can also be written as $c^2 - b^2 = a^2$
i.e a*a = (c-b)(c+b)
a*a x 1 = a*a, thus $c = (a^2+1)/2$ and $b = (a^2 - 1)/2$ , this solution works if n is odd.
For even solution, $c+b = n^2/2, c-b=2$ , thus, we get the above formula when n is even.

Below is the implementation:

C++

// CPP program to find Pythagoras triplet 
// with one side as given number. 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function, to evaluate the Pythagoras triplet 
// with includes 'n' if possible 
void evaluate(long long int n) 
{ 
  
    if (n == 1 || n == 2) 
        printf("No Pythagoras Triplet exists"); 
  
    else if (n % 2 == 0) { 
  
        // Calculating for even case 
        long long int var = 1LL * n * n / 4; 
        printf("Pythagoras Triplets exist i.e. "); 
        printf("%lld %lld %lld", n, var - 1, var + 1); 
    } 
  
    else if (n % 2 != 0) { 
  
        // Calculating for odd case 
        long long int var = 1LL * n * n + 1; 
        printf("Pythagoras Triplets exist i.e. "); 
        printf("%lld %lld %lld", n, var / 2 - 1, var / 2); 
    } 
} 
  
// Driver function 
int main() 
{ 
    long long int n = 22; 
    evaluate(n); 
    return 0; 
} 

Java

// Java program to find  
// Pythagoras triplet  
// with one side as  
// given number. 
import java.io.*; 
  
class GFG 
{ 
      
// Function, to evaluate  
// the Pythagoras triplet 
// with includes 'n' if  
// possible 
static void evaluate( int n) 
{ 
    if (n == 1 || n == 2) 
        System.out.println("No Pythagoras " +  
                           "Triplet exists"); 
  
    else if (n % 2 == 0)  
    { 
  
        // Calculating for even case 
        int var = 1 * n * n / 4; 
        System.out.print("Pythagoras Triplets " + 
                                  "exist i.e. "); 
        System.out.print(n + " "); 
        System.out.print(var - 1+ " "); 
        System.out.println(var + 1 +" "); 
    } 
  
    else if (n % 2 != 0)  
    { 
  
        int var = 1 * n * n + 1; 
        System.out.print("Pythagoras Triplets " +  
                                  "exist i.e. "); 
        System.out.print(n + " "); 
        System.out.print(var / 2 - 1 + " "); 
        System.out.println(var / 2 + " "); 
    } 
} 
  
// Driver Code 
public static void main(String[] args)  
{ 
    int n = 22; 
    evaluate(n); 
} 
} 
  
// This code is contributed 
// by ajit 

Python3

# Python3 program to find  
# Pythagoras triplet with  
# one side as given number. 
  
# Function, to evaluate the 
# Pythagoras triplet with  
# includes 'n' if possible 
def evaluate(n): 
    if (n == 1 or n == 2): 
        print("No Pythagoras" +
            " Triplet exists"); 
    elif (n % 2 == 0): 
          
        # Calculating for 
        # even case 
        var = n * n / 4; 
        print("Pythagoras Triplets" +
             " exist i.e. ", end = ""); 
        print(int(n), " ", int(var - 1), 
                      " ", int(var + 1)); 
    elif (n % 2 != 0): 
          
        # Calculating for odd case 
        var = n * n + 1; 
        print("Pythagoras Triplets " + 
             "exist i.e. ", end = ""); 
        print(int(n), " ", int(var / 2 - 1), 
                         " ", int(var / 2)); 
  
# Driver Code 
n = 22; 
evaluate(n); 
  
# This code is contributed by mits 

C#

// C# program to find  
// Pythagoras triplet  
// with one side as  
// given number. 
using System; 
  
class GFG 
{ 
      
// Function, to evaluate  
// the Pythagoras triplet 
// with includes 'n' if  
// possible 
static void evaluate(int n) 
{ 
    if (n == 1 || n == 2) 
        Console.WriteLine("No Pythagoras " +  
                          "Triplet exists"); 
      
    else if (n % 2 == 0)  
    { 
  
        // Calculating for even case 
        int var = 1 * n * n / 4; 
        Console.Write("Pythagoras Triplets " + 
                               "exist i.e. "); 
        Console.Write(n + " "); 
        Console.Write(var - 1+ " "); 
        Console.WriteLine(var + 1 +" "); 
    } 
  
    else if (n % 2 != 0)  
    { 
        int var = 1 * n * n + 1; 
        Console.Write("Pythagoras Triplets " +  
                               "exist i.e. "); 
        Console.Write(n + " "); 
        Console.Write(var / 2 - 1 + " "); 
        Console.WriteLine(var / 2 + " "); 
    } 
} 
  
// Driver Code 
static public void Main () 
{ 
    int n = 22; 
    evaluate(n); 
} 
} 
  
// This code is contributed 
// by ajit 

PHP

<?php 
// PHP program to find Pythagoras triplet 
// with one side as given number. 
  
// Function, to evaluate the 
// Pythagoras triplet with  
// includes 'n' if possible 
function evaluate($n) 
{ 
  
    if ($n == 1 || $n == 2) 
        echo "No Pythagoras Triplet exists"; 
  
    else if ($n % 2 == 0) { 
  
        // Calculating for even case 
        $var = $n * $n / 4; 
        echo "Pythagoras Triplets exist i.e. "; 
        echo $n, " ", $var - 1, " ", $var + 1; 
    } 
  
    else if ($n % 2 != 0) { 
  
        // Calculating for odd case 
        $var = $n * $n + 1; 
        echo "Pythagoras Triplets exist i.e. "; 
        echo $n, " ", $var / 2 - 1, " ", $var / 2; 
    } 
} 
  
    // Driver Code 
    $n = 22; 
    evaluate($n); 
  
// This code is contributed by ajit 
?> 

Javascript

<script> 
    // Javascript program to find  
    // Pythagoras triplet  
    // with one side as  
    // given number. 
      
    // Function, to evaluate  
    // the Pythagoras triplet 
    // with includes 'n' if  
    // possible 
    function evaluate(n) 
    { 
        if (n == 1 || n == 2) 
            document.write("No Pythagoras Triplet exists"); 
  
        else if (n % 2 == 0)  
        { 
  
            // Calculating for even case 
            let Var = 1 * n * n / 4; 
            document.write("Pythagoras Triplets " + 
                                   "exist i.e. "); 
            document.write(n + " "); 
            document.write(Var - 1+ " "); 
            document.write(Var + 1 +" "); 
        } 
  
        else if (n % 2 != 0)  
        { 
            let Var = 1 * n * n + 1; 
            document.write("Pythagoras Triplets " +  
                                   "exist i.e. "); 
            document.write(n + " "); 
            document.write(parseInt(Var / 2, 10) - 1 + " "); 
            document.write(parseInt(Var / 2, 10) + " "); 
        } 
    } 
      
    let n = 22; 
    evaluate(n); 
          
</script>

Output

Pythagoras Triplets exist i.e. 22 120 122

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

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