Does $17x^4+y^2=-1$ have solution in $\Bbb{Q}_2$?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $-1=...11$.