Pythagoras triples

Question

Regarding the parametrization of the pythagora's triples:

$x=p^2-q^2$

$y=2pq$

$z=p^2+q^2$

When $x=0, p^2=q^2$. Given that $\gcd(p, q)=1$, is there a contradiction? Why(not)?

Answer 1

That is an incomplete parameterization.

The correct form is

$x = r(p^2-q^2), y = 2rpq, z = r(p^2+q^2) $.

Also, if $x=0$ then $y=z$. (and $2pq=p^2+q^2$)

No contradiction.

Answer 2

When we parametrize primitive triples in the usual way, it is the $2pq$ term which is even. In the degenerate case where one of the sides is $0$, that side must be "$y$." We get $p=1$, $q=0$, no problem.