$17x^4+y^2=-1$ does not have solution in $\Bbb{Q}_2$ (https://math.stackexchange.com/questions/4730095/does-17x4y2-1-have-solution-in-bbbq-2).

But does $17x^4+y^2=-1$ have solution in $\Bbb{Q}_2(\sqrt{-5})$?

Calculating Hilbert symbol $(-17,-1)=(17,-1)(-1,-1)=1 ((-1,-1)=1$ because $-(\sqrt{-5})^2-2^2=1$ ) yields nothing.

Taking $2-$ adic valuation yields no contradiction, this is because $\Bbb{Q}_2(\sqrt{-5})/\Bbb{Q}_2$ is ramified extension.

This question raised when I tried to calculate $2$-Selmer group of elliptic curve $E:y^2=x^3+17x$ over $\Bbb{Q}(\sqrt{-5})$.

$17x^4+y^2=-1$ does not have solution in $\Bbb{Q}_2$ (https://math.stackexchange.com/questions/4730095/does-17x4y2-1-have-solution-in-bbbq-2).

But does $17x^4+y^2=-1$ have solution in $\Bbb{Q}_2(\sqrt{-5})$?

Calculating Hilbert symbol $(-17,-1)=(17,-1)(-1,-1)=1 ((-1,-1)=1$ because $-(\sqrt{-5})^2-2^2=1$ ) yields nothing.

Taking $2-$ adic valuation yields no contradiction, this is because $\Bbb{Q}_2(\sqrt{-5})/\Bbb{Q}_2$ is ramified extension.