Showing that there are infinitely many integer solutions to $x^2+y^2=z^2$ Let $x,y,z \in \mathbb{N}$ with $x,y$ relatively prime, $x$ even and $x^{2}+y^{2}=z^{2}$. Show that there are infinitely many $(x,y,z)$ triplets which satisfy these conditions.

I we use Euclid's formula
$ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$
we can see that
$$
A^2+B^2=\big((m^2-k^2)+(2mk)^2\\
=(m^4-2m^2k^2+k^4)+4m^2k^2\\
=(m^4+2m^2k^2+k^4)\\
=(m^2+k^2)^2=C^2
$$
One problem with Euclid's formula is that it generates trivial triples
like $f(1,0)=(1,0,1)\quad f(3,0)=(9,0,9)\quad \cdots\qquad$
I developed a formula that generate non-trivial Pythagorean triples for every pair of natural number $(n,k)$ and produces only the subset of triples where $GCD(A,B,C)$ is an odd square -- meaning it includes all primitives where
$GCD(A,B,C)=1^2$.
\begin{equation}
  A=(2n-1)^2+2(2n-1)k\quad B=2(2n-1)k+2k^2\quad C=(2n-1)^2+2(2n-1)k+2k^2
\end{equation}
$$A^2=(2n-1)^4+4(2n-1)^3 k+4(2n-1)^2 k^2$$
$$B^2=4(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
$$C^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$
$$A^2+B^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4=C^2$$
