$x^2+y^2=-1$ 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.
Best Answer
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.