How many variations of new primitive Pythagorean triples are there when the hypotenuse is multiplied by a prime? When I was looking at all the Pythagorean triples from 1 to 1000, I noticed a certain trend which I cannot seem to explain. Given $a, b, c$ are integers where $c$ is the hypotenuse, and $a < b$, the trends were


*

*For all $(a, b, c)$s where $c$ is a prime, there is only one pair possible. Here, not all primes have this property. There are primes where there are no triplets corresponding to it yet I could not find a consistent property between these primes.

*When c is multiplied by another prime that has a property of having a triplet, if that prime is a different prime than the first prime, there will be two new primitive Pythagorean pairs formed. For example, for (3, 4, 5) and (5, 12, 13), when we look at c = 65 which is 5 times 13, there will be 4 pairs which are (52, 39), (56, 33), (60, 25), (63, 16) where (65, 56, 33) and (65, 63, 16) are new primitive Pythagorean triples.

*If c is multiplied by the same prime, c, there will only be 1 new primitive Pythagorean triple produced.


Can anyone explain to me why these trends are here? Thanks in advance! Below I will document my failed approach just in case it will help clarify my question.
My(unsuccessful) approach
For 1, I tried to use a proof by contradiction where I tried to prove that
$$c^2 = a_1^2 + b_1^2 = a_2^2 + b_2^2$$
was impossible. Yet sadly I could not find any contradiction.(I assume there is one though)
For 2, where $c_0, c_1$ are primes, when looking at $c_0c_1$,
$$c_0^2c_1^2 = (a_0^2 + b_0^2)(a_1^2 + b_1^2) = a_0^2c_1^2 + b_0^2c_1^2
= a_1^2c_0^2 + b_1^2c_0^2$$
where the other two triples which will be primitive will come from the expansion of it.
For 3, the basic approach was the same as 2 except for the fact that $c_0 = c_1$ I failed to find any primitive triplets.
 A: For the first statement. 
If $c=4n+1$, where $n$ is natural and $c$ is a prime number then it's true.
Also, we have $$c_0^2c_1^2=(a_0^2+b_0^2)(a_1^2+b_1^2)=a_0^2a_1^2+a_0^2b_1^2+b_0^2a_1^2+b_0^2b_1^2=$$
$$=(a_0a_1+b_0b_1)^2+(a_0b_1-a_1b_0)^2=(a_0a_1-b_0b_1)^2+(a_0b_1+a_1b_0)^2.$$
A: 
Can anyone explain to me why these trends are here?

Let $n$ be a Natural number that factors as $\def\mod{\operatorname{mod}}$
$$
n=\!\!\prod_{p\text{ prime}} p^{m_p}
$$


*

*If $m_p>0$ for some prime with $p\neq1\mod 4$, then there is no primitive Pythagorean triple $(a,b,n)$.

*Let $n$ be composed only of primes $p=1\mod 4$, and $m$ be the number of such primes (without multiplicity). Then there are exactly $2^{m-1}$ primitive Pythagorean triples of the form
$$(a,b,n)\quad\text{ where }\quad 0<a<b $$
For example, $65=5·13$ gives rise to $2^{2-1}=2$ such triples, same for $85=5·17$. Thus, $c=5525=5^2·13·17$ gives rise to 4 primitive triples, namely:


*

*$(1036, 5427, c)$

*$(2044, 5133, c)$ 

*$(2163, 5084, c)$ 

*$(3124, 4557, c)$
To see / compute this, factor $c$ over $\Bbb Z[i]$, the Gaussian integers. $c$ will decompose into $2m$ prime factors (not counting multiplicity) of $m$ conjugate pairs. To get all triples, take a prime of the first$^1$ pair to it's appropriate power $m_p$, and multiply it with whatever conjugate of the remaining pairs to their's powers $m_p$.$\def\cc{\mathrm{c.c}}$
For example, up to units we have the factorizations  $17=(4+i)_\cc$, $5=(2+i)_\cc$ and $13=(2+3i)_\cc$ where "c.c." means to multiply with the according complex conjugate. This gives – up to order, units and complex conjugation – the 4 distinct products of norm 5525
$$ (2+i)^2·(2\pm3i)·(4\pm i)$$ 

$^1$The order does not matter, just fix one order at your preference.
