# Pythagorean Triples

A Pythagorean triplet is a set of three numbers a, b, and c where `a^2 + b^2 = c^2`. In this Kata, you will be tasked with finding the Pythagorean triplets whose product is equal to `n`, the given argument to the function `pythagorean_triplet`.

## Your task

In this Kata, you will be tasked with finding the Pythagorean triplets whose product is equal to `n`, the given argument to the function, where `0 < n < 10000000`


## Examples

One such triple is `3, 4, 5`. For this challenge, you would be given the value `60` as the argument to your function, and then it would return the Pythagorean triplet in an array `[3, 4, 5]` which is returned in increasing order. `3^2 + 4^2 = 5^2` since `9 + 16 = 25` and then their product (`3 * 4 * 5`) is `60`.


More examples:

| **argument** | **returns** |
| ---------|---------|
| 60       | [3, 4, 5] |
| 780      | [5, 12, 13] |
| 2040     | [8, 15, 17] |

Source: codewars | Difficulty: EASY
Tags: Mathematics

Starter code: def pythagorean_triplet(n):
	

Solutions:
def pythagorean_triplet(n):
	for a in range(3, n):
		for b in range(a + 1, n):
			c = (a * a + b * b) ** 0.5
			if a * b * c > n:
				break
			if c == int(c) and a * b * c == n:
				return [a, b, c]


def getTriplets(n):
	divs = {y for x in range(1, int(n ** 0.5) + 1) for y in (x, n // x) if not n % x}
	for a in divs:
		for b in divs:
			if n / a / b in divs:
				yield sorted((a, b, n / (a * b)))

def pythagorean_triplet(n):
	return next(([a, b, c] for (a, b, c) in getTriplets(n) if a ** 2 + b ** 2 == c ** 2))


triples = [(3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29), (12, 35, 37), (9, 40, 41), (28, 45, 53), (11, 60, 61), (16, 63, 65), (33, 56, 65), (48, 55, 73), (13, 84, 85), (36, 77, 85), (39, 80, 89), (65, 72, 97), (20, 99, 101), (60, 91, 109), (15, 112, 113), (44, 117, 125), (88, 105, 137), (17, 144, 145), (24, 143, 145), (51, 140, 149), (85, 132, 157), (119, 120, 169), (52, 165, 173), (19, 180, 181), (57, 176, 185), (104, 153, 185), (95, 168, 193), (28, 195, 197), (84, 187, 205), (133, 156, 205), (21, 220, 221), (140, 171, 221), (60, 221, 229), (105, 208, 233), (120, 209, 241), (32, 255, 257), (23, 264, 265), (96, 247, 265), (69, 260, 269), (115, 252, 277), (160, 231, 281), (161, 240, 289), (68, 285, 293)]

def pythagorean_triplet(n):
	for (a, b, c) in triples:
		(a0, b0, c0) = (a, b, c)
		while a * b * c < n:
			a += a0
			b += b0
			c += c0
		if a * b * c > n:
			continue
		else:
			return [a, b, c]


from itertools import combinations
from math import hypot
limit = 10000000
candidates = ((a, b, hypot(a, b)) for (a, b) in combinations(range(1, 1000), 2))
D = {a * b * int(c): [a, b, int(c)] for (a, b, c) in candidates if c.is_integer() and a * b * c < limit}
pythagorean_triplet = D.__getitem__


def triplet(n):
	for i in range(int(n ** 0.333), 1, -1):
		if n % i:
			continue
		c = n // i
		for j in range(int(c ** 0.5), i, -1):
			if c % j or j == c // j:
				continue
			yield (i, j, c // j)

def pythagorean_triplet(n):
	for (a, b, c) in triplet(n):
		if a ** 2 + b ** 2 == c ** 2:
			return [a, b, c]


def pythagorean_triplet(n):
	for k in range(1, int(n ** (1 / 3))):
		if n % k ** 3:
			continue
		for p in range(1, int((n // k ** 3) ** 0.2) + 1):
			for q in range(1, p):
				(a, b, c) = (p * p - q * q, 2 * p * q, p * p + q * q)
				if k ** 3 * a * b * c == n:
					return sorted([k * a, k * b, k * c])


def pythagorean_triplet(n):
	for z in range(1, int(n ** 0.5) + 1):
		y = n / z
		if y.is_integer():
			x = (z ** 2 + 2 * y) ** 0.5
			if x.is_integer():
				for a in range(1, int(x)):
					b = x - a
					if a * b == y:
						return sorted([a, b, z])


def factors(n, limit):
	return [i for i in range(2, limit + 1) if n % i == 0]

def pythagorean_triplet(n):
	for x in factors(n, int(n ** (1 / 3))):
		nn = n // x
		for y in factors(nn, int(nn ** (1 / 2))):
			z = nn // y
			if x * x + y * y == z * z:
				return [x, y, z]


pythagorean_triplet = lambda n: [[d * round((n / k) ** (1 / 3)) for d in v] for (k, v) in {60: [3, 4, 5], 780: [5, 12, 13], 2040: [8, 15, 17], 4200: [7, 24, 25], 14760: [9, 40, 41], 15540: [12, 35, 37], 40260: [11, 60, 61], 65520: [16, 63, 65], 66780: [28, 45, 53], 92820: [13, 84, 85], 12180: [20, 21, 29], 120120: [33, 56, 65], 192720: [48, 55, 73], 235620: [36, 77, 85], 277680: [39, 80, 89], 328860: [60, 63, 87], 453960: [65, 72, 97]}.items() if round((n // k) ** (1 / 3)) ** 3 == n / k][0]


def pythagorean_triplet(n):
	from math import sqrt

	def divisors(n):
		divs = []
		for i in range(2, int(sqrt(n)) + 1):
			if n % i == 0:
				divs.append(i)
		divs.sort()
		return divs
	divs = divisors(n)
	for a in divs:
		for b in divs:
			c = n // (a * b)
			if a * a + b * b == c * c:
				return [a, b, c]