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N.B. All answers just proves it has no solution in $\Bbb{Z}_2$, not in $\Bbb{Q}_2$. But I want to prove it has no solution in $\Bbb{Q}_2$ as title asks.

Does $17x^4+y^2=-1$ have solution in $\Bbb{Q}_2$ ? If we assume $x$ is not a unit, looking $2$ cdic valuation on both side yields $2v(y)=min\{0,4v(x)\}$.If $v(x)$ is positive, $y$ is unit, If $v(x)$ is negaitive, $v(y)=2v(x)$. Such kind of argument does not prove there is no solution in $\Bbb{Q}_2$.

If this has solution, computational approach is also appreciated. Thank you for your help.

Pont
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Note that that $17\equiv1\bmod4$ and all squares in are $\in\{0,1\}\bmod4$. So

$17x^4+y^2\in\{0,1,2\}\bmod 4$.

Thus in $2$-adic integers $17x^4+y^2$ must terminate in $...00,...01$ or $...10$ and can't match $-1=...11$.

Oscar Lanzi
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We can show the more general $17x^2+y^2+z^2=0$ has no solutions, which implies the special case of your equation when $x$ is a perfect square, has no solutions.

Because it's homogeneous, we can assume all of the terms $x,y,z$ are $2$-adic integers with at least one of them having $2$-adic valuation $0$. Reducing modulo $2$ implies that exactly two terms are $1$ modulo 2.

Reducing modulo 4 implies that we have $1+1+0=0 \mod 4$, a contradiction. So there are no solutions.

Merosity
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  • Your answer seems to be correct. Thank you very much.But I'm a bit counfused. In $\Bbb{Q}(\sqrt{-2})$, $(x,y,z)=(1,3\sqrt{-2},-1)$ is obvious solution, but your argument stil works in this $\Bbb{Q}(\sqrt{-2})$ case? – Pont Jul 04 '23 at 11:48
  • Oh, key point is that $x^2+y^2+z^2=0$ does not have solution except for $(0,0,0),(1,1,0),(1,0,1),(0,1,1)$ in $\Bbb{F}_2$. It may have solution in $\Bbb{F}_4$. – Pont Jul 04 '23 at 11:51
  • Yeah, we could have originally scaled it since $17$ is a square in $\mathbb{Q}_2$. Alternatively we could have gotten there by Hilbert symbols $(-17,-1)=(-1,-1)(17,-1)$ with $(17,-1)=1$ because $17*1^2-1^2=4^2$. I don't quite follow your first comment though, I suppose by comparison $x^2+2=0\mod 4$ has a solution now that you've introduced $\sqrt{-2}$, so I don't think that argument works, since before we were stuck with squares only being $0,1 \mod 4$. Now that I'm looking closer I'm not sure if you're conflating $\mathbb{F}_4$ with $\mathbb{Z}/4\mathbb{Z}$ or not. – Merosity Jul 04 '23 at 11:56
  • I mean, in $\Bbb{Q}(\sqrt{-2})$, $17×1^２+(3\sqrt{-2})^2+(-1)^2=0$. Yes!, Hilbert symbol $(-1,-1)=-1$ was a point ! Thank you very much. – Pont Jul 04 '23 at 12:08
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    My $\Bbb{F}_4$ is $\Bbb{F}_2[x]/(x^2+x+1)$. – Pont Jul 04 '23 at 12:13
  • Ok that's good. I still don't know if we're on the same page or not, at least I don't see how this argument would still work in $\mathbb{Q}(\sqrt{-2})$, at least I tried to explain how it loosely fails at the $\mod 4$ step. Additionally, the Hilbert symbol trick isn't necessary since we could substitute $x=\frac{x'}{\sqrt{17}}$ in $\mathbb{Q}_2$ to remove it directly which feels more direct and less high tech to me. – Merosity Jul 04 '23 at 12:25
  • Because this is not on the same page of original question, so I opened new question regarding this issue.https://math.stackexchange.com/questions/4730292/does-x2y2-1-have-solution-in-p-adic-number-field. I would be appreciated if you could comment or answer the question. – Pont Jul 04 '23 at 12:38
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Modulo $16$, $17x^4\equiv x^4\equiv0,1$, and $y^2\equiv0,1,4,9$, so $17x^4+y^2\equiv-1$ has no solution modulo $16$, hence, no solution in the $2$-adics.

Gerry Myerson
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    In the case $v(x)$ is negative, $v(y)$ is also negative. So, $x,y$ are not integral. In this case, why can you take $mod16$ ? – Pont Jul 04 '23 at 08:39