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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.

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1 Answer 1

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  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 $c^2\equiv ab\pmod{p}$.
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  • $\begingroup$ 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! $\endgroup$
    – robbis
    Jun 19, 2019 at 21:22
  • $\begingroup$ @robbis $x^4-17y^4=2z^2$ nearly implies $(x/y)^4-17=2(z/y^2)^2$, so you only need to check that $y$ is invertible mod $p$ in each case. $\endgroup$
    – Mercury
    Jun 20, 2019 at 14:46
  • $\begingroup$ Right, just by change of variables! I haven't thought of this smple trick! thanks! $\endgroup$
    – robbis
    Jun 20, 2019 at 14:48
  • $\begingroup$ do you also know about non-solvability over the rationals? $\endgroup$
    – robbis
    Jun 20, 2019 at 14:49

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