All rational solutions to $px^4-4y^4=z^2$ with $p\equiv 3\pmod8$ satisfy $xyz=0$

Question

Let $p$ be a prime with $p\equiv 3\pmod 8$ , then all rational solutions to $px^4-4y^4=z^2$ satisfy $xyz=0$.

Apparently, the argument to prove this is that $-1$ is not a quadratic residue of $p$.

I know that $-1$ is not a quadratic residue of $p$ because of the law of quadratic reciprocity and $p\equiv 3mod4$. However, I fail to see how this proves that all solutions satisfy xyz=0. If I take the equation modulo $p$, then I'm stuck with $-4y^4\equiv z^2 mod p$. What am I missing?

Answer 1

Assuming that $x,y,z$ are non-zero, it's clearly equivalent to the following : $px^4=4y^2+z^2=(2y^2)^2+z^2$. Hence $px^4$ can be written as sum of two perfect squares, which is a contradiction noting that $p\overset{4}{\equiv}3$ and $\nu_p(px^4)\overset{2}{\equiv}1$. For more information on the theorem check the following link:
https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem