Problem

A Pythagorean triplet is a set of three natural numbers, $a \lt b \lt c$,
for which

$$a^2 + b^2 = c^2$$

For example, $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.

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

Answer

31875000

---

Python

def compute():
	PERIMETER = 1000
	for a in range(1, PERIMETER + 1):
		for b in range(a + 1, PERIMETER + 1):
			c = PERIMETER - a - b
			if a * a + b * b == c * c:
				# It is now implied that b < c, because we have a > 0
				return str(a * b * c)


if __name__ == "__main__":
	print(compute())

---

JavaScript

for (let a = 1; a < 1000; a++) {
  for (let b = a, lb = 1000 - a; b < lb; b++) {
    const c = 1000 - (a + b)
    if (a * a + b * b === c * c) {
      return console.log(a * b * c)
    }
  }
}

---

Go

package main

import "fmt"

func main() {
	var a, b, c, lb int
	max := 1000
	for a = 1; a < max; a++ {
		lb = max - a
		for b = a; b < lb; b++ {
			c = max - (a + b)
			if a*a+b*b == c*c {
				fmt.Println(a * b * c)
				return
			}
		}
	}
}

---

Ruby

#!/usr/bin/env ruby
for a in (1..500)
  for b in (a..500)
    for c in (b..500)
      if a**2 + b**2 == c**2 and a+b+c == 1000
        puts a*b*c
      end
    end
  end
end

---

Haskell

main ::  IO ()
main = print $ head [a*b*c | a <- [1..500], b <- [a..500], c <- [b..500],
                             a+b+c == 1000, a*a + b*b == c*c]

---

Clojure

#!/usr/bin/env clojure
(defn square [n]
  (* n n))

(defn triple? [a b c]
  (and
    (and (> a 0) (> b 0) (> c 0))
    (and (< a b) (< b c))
    (= (+ (square a) (square b)) (square c))))

(defn candidates [limit]
  (for [a (range 1 (inc limit)) 
        b (range a (inc limit)) 
        c (range b (inc limit))
        :when (and 
                (= (+ a b c) 1000)
                (triple? a b c))] 
    (list a b c)))

(println (reduce * (first (candidates 500))))

---

Mathematica

For[a = 1, a <= 1000, a++,
  For[b = a + 1, b <= 1000, b++,
    Block[{c = 1000 - a - b},
      If[a < b < c && a^2 + b^2 == c^2,
        Print[a, " ", b, " ", c, " ", a * b * c]]]]]

---

Java

public final class p009 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p009().run());
	}
	
	
	/* 
	 * Computers are fast, so we can implement a brute-force search to directly solve the problem.
	 * Note that a^2 + b^2 is bounded above by 2*(1000^2), which fits in a Java int type.
	 */
	private static final int PERIMETER = 1000;
	
	public String run() {
		for (int a = 1; a < PERIMETER; a++) {
			for (int b = a + 1; b < PERIMETER; b++) {
				int c = PERIMETER - a - b;
				if (a * a + b * b == c * c) {
					// It is now implied that b < c, because we have a > 0
					return Integer.toString(a * b * c);
				}
			}
		}
		throw new AssertionError("Not found");
	}
	
}