I want to show that $$x^4 - 17y^4 \equiv 2z^2 \pmod p$$ has non-trivial solutions for all primes $p$.
I haven't been able to progress much, but one attempt was to consider the case that $p$ is such that $2$ is a square (a quadratic residue) mod $p$, and thus $2z^2$ is a square, therefore there is a solution if and only if $x^4 - 17y^4$ is congruent to a square for some $x$ and $y$.
Then we can consider when $17$ is a square, then show there exist $x$ and $y$ such that $x^4 - 2z^2$ is congruent to a square. Then I suppose we will have to consider the case that neither $17$ nor $2$ is a square mod $p$, so then we have to show $2z^2 + 17y^4$ is congruent to a square.
So far I have not been able to get anywhere with this approach. Any help is appreciated.