Problem 9(Special Pythagorean triplet)

Problem 9(Special Pythagorean triplet)

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

a² + b² = c²

For example, 3² + 4² = 9 + 16 = 25 = 5².

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

In Python:

import math
import time

start_time=time.time()

for i in range(1,334):
    for j in range(i,500):
        k = math.sqrt(pow(i,2)+pow(j,2))
        if k.is_integer() and i+j+k==1000:
            print(i, j, k, i*j*k)
        else:
            pass

print(time.time() - start_time,"seconds")

Run time: 0.23833203315734863 seconds

In Java:

//PE9 Special Pythagorean triplet

package project_euler;

public class Euler9 {

public static void main(String[] args) {

long startTime = System.currentTimeMillis();

double k = 0;

for (int i = 1; i < 334; i++) {

for (int j = i; j < 500; j++) {

k = Math.sqrt(Math.pow(i, 2) + Math.pow(j, 2));

if (k == (int)k && i+j+k == 1000) {

System.out.println(i * j * k);

}

}

}

long endTime = System.currentTimeMillis();

System.out.println((double)(endTime - startTime)/(double)1000 + "seconds");

}

}

Run time: 0.007seconds

solution: 31875000

[from Project Euler: https://projecteuler.net/problem=9]