## Special Pythagorean triplet

Let’s consider the general case $$a + b + c = s$$, where $$s$$ can only be even. All the pythagorean triplets can be generated with the Euclid’s formula $$a = k*(m^2 - n^2)$$, $$b = k*(2*m*n)$$, $$c = k*(m^2 + n^2)$$. When substituting the above in $$a + b + c = s$$ we get $$2*k*m*(n + m) = s$$ and $$k = \frac{s'}{m*(n+m)}$$ with $$s' = s/2$$. The conditions for the Euclid formula are: $$m > n$$, $$m - n$$ odd and $$m$$ and $$n$$ coprime. So we just need to generate all the pairs $$(m,n)$$ which satisfy these conditions until we find one that evenly divides $$s'$$ at which point we calculate $$k$$. This can be done, starting with $$m = 2$$ and $$n = 1$$, by applying the following formulas at each iteration: $$(2m - n, m)$$, $$(2m + n, m)$$ and $$(m + 2n, n)$$ (see Generating all coprime pairs).

def euclid_formula(k, m, n):
return k*(m**2 - n**2), k*(2*m*n), k*(m**2+n**2)
def coprimes():
from collections import deque
pairs = deque([(2, 1)])
while True:
(m, n) = pairs.popleft()
yield m, n
pairs.append((2*m - n, m))
pairs.append((2*m + n, m))
pairs.append((m + 2*n, n))

def k(s):
s1 = s/2
for (m, n) in coprimes():
k, rem = divmod(s1, m*(n+m))
if rem == 0:
return euclid_formula(k, m, n)

a, b, c = k(1_000)
int(a*b*c)
31875000

Source code of the solution(s):