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.

The system of equations:   
$$\left\{\begin{aligned}&x^2+y^2=z^2\\&q^2+t^2=z^2\end{aligned}\right.$$
Formulas you can write a lot, but will be limited to this. Will make a replacement.  
$$a=p^2+s^2-k^2$$
$$b=p^2+s^2+k^2-2pk-2ks$$
$$c=p^2+k^2-s^2+2ps-2kp$$
$$r=s^2+k^2-p^2+2ps-2ks$$ 
The solution then is.  
$$x=2ab$$  
$$y=a^2-b^2$$  
$$q=2cr$$  
$$t=c^2-r^2$$  
$$z=a^2+b^2$$
$p,s,k$ - integers.
