# Solutions mod p of the equation $x^4-17=2y^2$

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!

• All $p$ such that Kroneckers symbol is $\Biggl(\dfrac{-34}{p}\Biggr)=1$: 5, 7, 19, 23, 29, 31, 37, 43,.... Example: $a^2\equiv-34\pmod{43}\quad\Longrightarrow\quad a\equiv\pm3\pmod{43}$. – Dmitry Ezhov Jun 19 '19 at 22:43

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.