What is the smallest prime which satisfies $p \in {\Bbb{Q}_2(\sqrt{2} )^{\times}}^4$and $p≡3,5$mod8?
The candidate for $p$ is $3,5,11,13,19,29,37,43,53,59,・・・$. To consider which is $4$-th power in $\Bbb{Q}_2(\sqrt{2})$, let $f(x)=x^4-p$ and use some kind of hensel lemma, but I'm not familiar with applying hensel lemma not over $\Bbb{Q}_p$.
Computer calculation is also appreciated. Thank you for your help.
Find prime number which satisfies $p \in {\Bbb{{Q}_2}^{\times}}^2-{{\Bbb{Q}_2}^{\times}}^4$ is related one, but the difference is that we think over quadratic extension of $\Bbb{Q}_2$.
The set whose minimal element you are looking for is empty. In fact, I claim that all odd primes $p \in \mathbb Z$ that are squares in $\mathbb Q_2(\sqrt 2)$ are already squares in $\mathbb Q_2$, i.e. are necessarily $\equiv 1$ modulo $8$; a fortiori this holds for fourth powers.
Namely, each element of $\mathbb Q_2(\sqrt 2)$ is of the form $a+b\sqrt2$ for some (uniquely determined) $a,b \in \mathbb Q_2$. So for its square
$$a^2 +2ab\sqrt2 +2b^2$$
to be some prime $p \in \mathbb Z$ (or just for it to be $\in \mathbb Q_2$), at least one of $a,b$ must be $=0$. Hence, this must be either a square, or $2$ times a square, of an element of $\mathbb Q_2$. Since for odd $p$, $\frac{p}{2}$ is not a square in $\mathbb Q_2$, $p$ itself must be a square in $\mathbb Q_2$.