# Show that the congruence $x^4 - 17y^4 \equiv 2z^2 \pmod p$ has non-trivial solutions for all primes $p$

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.

1. For $p=2$, $(x,y,z)=(1,1,1)$ is a solution.
2. If $(-2/p)=1$ then $(2a,a,a)$ is a solution when $a^2\equiv-2\pmod{p}$.
3. If $(-2/p)=-1$ and $(17/p)=-1$, then $(-34/p)=1$ and $(0,2,2a)$ is a solution when $a^2\equiv-34\pmod{p}$.
4. If $(5/p)=1$, then $(a,1,2)$ is a solution when $a^2\equiv5\pmod{p}$.
5. If $(7/p)=1$, then $(a,1,4)$ is a solution when $a^2\equiv7\pmod{p}$.
Since $p=17$ and $p=5,7$ are covered by case 2 and case 3 respectively, it remains to consider the case that $(17/p)=1$ and $(5/p)=(7/p)=-1$, which implies $(35/p)=1$. So we may take integers $a,b$ such that $a^2\equiv17\pmod{p}$ and $b^2\equiv35\pmod{p}$. Then we complete the proof by the following observations:
• If $(a/p)=1$, then $(c,1,0)$ is a solution when $c^2\equiv a\pmod{p}$.
• If $(b/p)=1$, then $(c,1,3)$ is a solution when $c^2\equiv b\pmod{p}$.
• If $(a/p)=(b/p)=-1$, then $(ab/p)=1$ and $(c,1,17)$ is a solution when $c^2\equiv ab\pmod{p}$.
• Hi, I was searching about this equations and I found this answer. I have to prove that there are solutions mod p for every p, but the equation is a bit different: $x^4-17=2y^2$. How can I generalize your proof? I've done the first two cases but I have difficulties with the third. Thanks! – robbis 14 hours ago