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function isPythagoreanTriple(a, b, c) {
  return (a ** 2 + b ** 2 === c ** 2);
}

let max = prompt('Maximale Hypothenusenlänge:');
for (let c = 1; c <= max; c++)
  for (let b = 1; b <= max; b++)
    for (let a = 1; a <= max; a++)
      if (a < b && isPythagoreanTriple(a, b, c))
        console.log(`${a} | ${b} | ${c}`);
                

Lösung von: Lisa Salander (Heidi-Klum-Gymnasium Bottrop)

const int b_max = 30;

for (var a = 1; a <= b_max; a++)
    for (var b = a; b <= b_max; b++) {
        var t = a * a + b * b;
        var c = (int)Math.Sqrt(t);
        if (c * c == t)
            Console.WriteLine($"({a}, {b}, {c})");
    }
                

Lösung von: Jens Kelm (@JKooP)

#include <vector>
#include <iostream>
#include <tuple>

std::vector<std::tuple<int, int, int>> pythagorean_triple(const int b_max) {
	std::vector<std::tuple<int, int, int>> v;
	for (auto a{ 1 }; a <= b_max; a++) {
		for (auto b{ a }; b <= b_max; b++) {
			auto t{ a * a + b * b };
			auto c{ (int)sqrt(t) };
			if (c * c == t)
				v.push_back(std::make_tuple(a, b, c));
		}
	}
	return v;
}

int main() {
	const int b_max = 30;
	for (const auto& pt : pythagorean_triple(b_max)) {
		std::cout << "(" << std::get<0>(pt) << ", " << std::get<1>(pt) << ", " << std::get<2>(pt) << ")\n";
	}
}
                

Lösung von: Jens Kelm (@JKooP)

// Hier als Funktion für große Werte:
const int b_max = 10_000;
var pt = PythagoreanTriple(b_max).ToList();
pt.ForEach(x => Console.WriteLine(x));

static IEnumerable<(int a, int b, int c)> PythagoreanTriple(int b_max = 1) {
    for (var a = 1; a <= b_max; a++)
        for (var b = a; b <= b_max; b++) {
            var t = a * a + b * b;
            var c = (int)Math.Sqrt(t);
            if (c * c == t)
                yield return (a, b, c);
        }
}
                

Lösung von: Jens Kelm (@JKooP)

#m und n seien beliebige, natürliche Zahlen
#für (m > n) and (m != 0 and n != 0) gilt:
#a = m**2 - n**2
#b = 2mn
#c = m**2 + n**2

import random
max = 1000
counter = 1
while counter < max:
    m = random.randint(1,max)
    n = random.randint(1,max)
    if m > n:
        a = m**2 - n**2
        b = 2 * m * n
        c = m**2 + n**2
        #print(f"Counter: {counter}, beliebig m = {m}, beliebig n = {n}, \ta = {a}, \tb = {b}, \tc = {c}")
        print(f"m = {m}, n = {n}, \ta = {a}, \tb = {b}, \tc = {c}")
    counter+=1
                

Lösung von: Py Thon ()

// Achtung: F#
let pt m =
    let mutable lst = []
    for a in 1 .. m do
        for b in a .. m do
            let t = a * a + b * b
            let c = sqrt(t |> double) |> int
            if c * c = t then lst <- [(a, b, c)] |> List.append lst
    lst

printfn "%A" (pt 30)
                

Lösung von: Jens Kelm (@JKooP)

def find_triples(limit):
    import math
    c_to_a_factor = math.sqrt(1 / 2)
    for c in range(1, limit + 1):
        a_limit = int(c * c_to_a_factor)
        for a in range(1, a_limit + 1):
            b_squared = c**2 - a**2
            b = math.isqrt(b_squared)
            if b**2 == b_squared:
                yield a, b, c

for triple in find_triples(30):
    print(triple)
                

Lösung von: Name nicht veröffentlicht

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
max=25
for a in range(1,max):
    for b in range(a,max):
        z=a*a+b*b
        c=int(z**0.5)
        if z==c*c:
            print("{} {} {}".format(a,b,c))
                

Lösung von: Name nicht veröffentlicht

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
max=25
for a in range(1,max):
    for b in range(a,max):
        z=a*a+b*b
        c=int(z**0.5)
        if z==c*c:
            print("{} {} {}".format(a,b,c))
                

Lösung von: Name nicht veröffentlicht