Does $x^2+y^2=-1$ have solution in $p$ adic number field?

Question

$x^2+y^2=-1$ has solution in $\Bbb{Q}_2(\sqrt{-2})$ since $(3\sqrt{-2})^2+(\sqrt{17})^2=-1$.

But in $\Bbb{Q}_2(\sqrt{-3})$, I cannot find solution $x^2+y^2=-1$ until now and I guess this has no solution. How can I prove $x^2+y^2=-1$ does not have solution in $\Bbb{Q}_2(\sqrt{-3})$?

This can be thought of calculation of Hilbert symbol over quadratic number field. In $\Bbb{Q}_2$, from discussion in Does $17x^4+y^2=-1$ have solution in $\Bbb{Q}_2$?, there is no rational points in $\Bbb{Q}_2$.

If some general theory is know, reference is also appreciated. Thank you for your help.

@Aphelli. Do you mean roots in $\Bbb{Q}_2(\sqrt{-3})$ ? If $z$ is not a unit, taking 2 adic valuation of $z^2=-1-z$, $2v(z)=min{0,v(z)}$.But this proves $z$ is a unit, contradiction. Thus, roots are units. But I cannot proceed from here. — Poitou-Tate, Jul 04 '23 at 14:11
This is a quadratic equation. What is the exact expression for $z$? — Aphelli, Jul 04 '23 at 14:40
$z^2+z+1=(z-\frac{-1+\sqrt{-3}}{2})(z-\frac{-1-\sqrt{-3}}{2} )$ ,thus roots arę in $\Bbb{Q}(\sqrt{-3})$. — Poitou-Tate, Jul 04 '23 at 15:14
You are asking about the Stufe (https://en.m.wikipedia.org/wiki/Stufe_(algebra)) of a $p$-adic field. Highly related: https://math.stackexchange.com/q/3965273/96384. The only case left open there is that of extensions of $\mathbb Q_2$ of degree $2m$ with $m \ge 2$. — Torsten Schoeneberg, Jul 04 '23 at 16:00
@Aphelli. Let $a=\frac{-1+\sqrt{-3}}{2}, -1=a+a^2=a^4+a^2$, thus my question's equation has a solution. — Poitou-Tate, Jul 04 '23 at 16:10

score 1 · Accepted Answer · answered Jul 04 '23 at 23:16

1

See the answer here for more on $x^2 + y^2 = -1$ in extensions of $\mathbf Q_2$. There is a link to that MSE page in one of the comments, but it also provides an answer to your question.

answered Jul 04 '23 at 23:16

KCd

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Does $x^2+y^2=-1$ have solution in $p$ adic number field?

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