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I'm trying to prove that for every $p$ there are solutions $\pmod{p}$ for this equation. I've tried to follow this answer: Show that the congruence $x^4 - 17y^4 \equiv 2z^2 \pmod p$ has non-trivial solutions for all primes $p$.

The equation is a bit different, and I have problem in the third point because I can't find a solution when $-34 \equiv a^2 \pmod{p}$ for some $a$. Can anyone help me? Thanks!

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  • $\begingroup$ All $p$ such that Kroneckers symbol is $\Biggl(\dfrac{-34}{p}\Biggr)=1$: 5, 7, 19, 23, 29, 31, 37, 43,.... Example: $a^2\equiv-34\pmod{43}\quad\Longrightarrow\quad a\equiv\pm3\pmod{43}$. $\endgroup$ – Dmitry Ezhov Jun 19 '19 at 22:43
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If $p$ is an odd prime and $-34\equiv a^2\pmod{p}$ for some $a$ then $$0^4-17\equiv2\left(\tfrac{a}{2}\right)^2\pmod{p}.$$


The linked question gives solutions to the congruence $$a^4-17b^4\equiv2c^2\pmod{p},$$ for every prime number $p$. Moreover, all solutions given there have $b\not\equiv1\pmod{p}$. So for $$x\equiv ab^{-1}\pmod{p} \qquad\text{ and }\qquad y\equiv cb^{-2}\pmod{p},$$ we have the following chain of congruences mod $p$: $$x^4-17\equiv a^4b^{-4}-17\equiv b^{-4}(a^4-17b^4)\equiv b^{-4}(2c^2)\equiv 2(cb^{-2})^2\equiv2y^2\pmod{p}.$$ This shows that every solution $(a,b,c)$ to the linked question with $b\not\equiv0\pmod{p}$ yields a solution $(x,y)=(ab^{-1},cb^{-2})$ to your question.

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  • $\begingroup$ thanks a lot, your answer is very clear! To prove that there are no non trivial rational points, I'm following Cassel's "Lectures on Elliptic Curves", chapter 18, which is the same argument as pdfs.semanticscholar.org/5b9c/…. I'm not sure if the last observation that 5 and 10 are not quaratic residue mod 17 is enough. Can you help me? Thanks! $\endgroup$ – robbis Jun 20 '19 at 14:44
  • $\begingroup$ I'm not sure what you mean exactly; perhaps this is matter for a new question? $\endgroup$ – Servaes Jun 21 '19 at 7:33

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