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

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](https://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares).