Dado un solo entero n 
[1, 1000000000], genere un triplete de Pitágoras que incluya n como uno de sus lados si es posible.
Ejemplos:
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
Explicación:
Definición: Los “tripletes de Pitágoras” son soluciones enteras del Teorema de Pitágoras, es decir, satisfacen la ecuación 
. Nuestra tarea es generar un triplete a partir de un valor integral. Esta puede ser una tarea confusa porque el lado que se nos da puede ser una hipotenusa o un lado sin hipotenusa.
Comenzando a calcular los tripletes poniéndolos en una fórmula, se puede deducir que solo para 1 y 2, no es posible ningún triplete.
Además,
si n es par, nuestros tripletes se calculan mediante fórmula 
si n es impar, nuestros tripletes se calculan mediante fórmula 
Prueba:
el teorema de Pitágoras también se puede escribir como, por 
ejemplo, a*a = (cb)(c+b)
a*ax 1 = a*a, por lo tanto 
y 
, esta solución funciona si n es impar.
Para una solución uniforme, 
, por lo tanto, obtenemos la fórmula anterior cuando n es par.
Código
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>
Producción:
Pythagoras Triplets exist i.e. 22 120 122