I know that this question has answers here, but I'm looking for a way to make the following argument work.
Let $(a,b,c)$ be a primitive Pythagorean triple with $a$ even. The goal is to prove that we find coprime integers $x>y>0$ with different parity such that $$a=2xy,\quad b=x^2-y^2,\quad c=x^2+y^2.$$
Those are the equalities that we want to prove, we cannot use them to answer my question!
The idea goes like this: We have $\frac{c+b}{a}=\frac{a}{c-b}$, and we may write this as a reduced fraction $\frac xy$. Surely $x>y>0$ and $x,y$ are coprime by construction. It can now easily be seen that $$\frac{x^2+y^2}{2xy}=\frac ca,\quad \frac{x^2-y^2}{2xy}=\frac ba.$$
If $x$ and $y$ now have different parity, all $4$ fractions above are reduced, and we can conclude.
Surely, $x$ and $y$ are not both even by assumption, but why can they not both be odd?
It is easy to see that one of $a,b$ is odd and the other even. Let us assume that $b$ is odd, so that $c-b$ is even. (if, instead, you chose $b$ to be the even one, it would not be true that one of $x,y$ was even).
Now, say $x,y$ were both odd. It would follow that the numerator $x^2-y^2$ was a multiple of $4$. But of course $2xy$ would not be a multiple of $4$. But this would force $b$ to be even, despite the fact that we assumed it to be odd. Thus it can not be the case that $x,y$ are both odd.