Pythagorean triples with the same c value $a^2 + b^2 = c^2$ 
There are, Primitive Pythagorean Triples, that share the same c value. For example, 
$63^2 + 16^2 = 65^2$ and $33 ^2 + 56^2 = 65^2$.
I have been trying to figure out why the following theorem for finding such triples works.
Take any set of primes. Ex: $5,13,17$. 
Now take their product, $1052$, this is the new $c^2$ value. You can express that $c^2$ value as a triple by factoring the product of primes as Gaussian Integers:
$(2-i)(2+i)(3+2i)(3-2i)(4+i)(4-i)$
Now if you take three of those Gaussian Integers and solve for their product:
ex, $(2+i)(3+2i)(4+i) = 9+32i$ and you have found an $a$ value $9$ and a $b$ value $32$ that works in the theorem. $9^2 + 32^2 = 1105$
You can continue with this:
$(2-i)(3+2i)(4+i) = a + bi$
$(2+i)(3-2i)(4+i) = a + bi$
$(2+i)(3+2i)(4-i) = a + bi$
In fact, the number of triples with this method is $2^{||p|| - 1}$ where $||p||$ is the number of primes used.
Can someone please explain why this method of finding these "c stuck triples" works the way it does?
EDIT: It appears that you cannot take any prime, $p$ but rather must use primes congruent to $1mod4$ according to Fermat's theorem on sums.

1) $(a + bi)(a - bi) = a^2 + b^2$ always.
2) $(a + bi)(c + di)(e + fi) = g + hi \ne (a + bi)(c + di)(e - fi) = j + ki$
yet $(a - bi)(c - di)(e - fi) = g - hi \ne (a - bi)(c - di)(e + fi) = j - ki$
so 
3) $(a + bi)(c + di)(e + fi)(a - bi)(c - di)(e - fi) = (g+hi)(g - hi) = g^2 + h^2 =K$
and
4) $(a + bi)(c + di)(e - fi)(a - bi)(c - di)(e + fi) = (j+ki)(j - ki) = j^2 + k^2 = pqr$
So $pqr = g^2 + h^2 = j^2 + k^2$.
You do have to choose primes that a gaussian factorable which seems to me like begging the question.
