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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)



Last Updated : 17 Feb, 2023
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