Does $(3+2\sqrt{-2})y^2=1+12\sqrt{-2}-4・17x^4$ have solution in $\Bbb{Q}_{17}$ ? I feel this question is very difficult.
Here, $\sqrt{-2}$ denotes one of root of $x^2+2=0$ in $\overline{\Bbb{Q}_{17}}$ whose $17$adic valuation is $ord_{17}(\sqrt{-2})=1$(Its $17$ adic expansion stars with $7$).
Looking at $17$ adic valuation of both side gives $1+2v(y)=min \{2, 1+4v(x)\}$(because $2$ and $1+4v(x)$ cannot coincide). Thus, $v(y)=2v(x)$ and $v(x)\leq 0$. So, both $x$ and $y$ are not integral elements. Substitute $x=1/X$, $y=1/Y$. Then $(3+2\sqrt{-2})X^4=X^4Y^2(1+12\sqrt{-2})-4・17Y^2$. Divide both side by $17$ gives $uX^4=X^4Y^2v-4Y^2$($u,v$ are units). But I cannot proceed from here.
Recently I'm calculating this kind of $p-$adic equation. If possible, I want to calculate these by my hand, but some computational approach is also appreciated. Thank you for your help.
Related: Does $2y^2=4+17x^4$ have solutions in $\Bbb{Q}_2(\sqrt{-3})$? (This is much more easier one than this, I believe)
It's known that $\mathbb{Q}_{17}^{\times}=\mathbb{Z}\oplus\mathbb{F}_{17}^{\times}\oplus\mathbb{Z}_{17}$ . which indicates $t=x^4,\, x\in\mathbb{Q}_{17}^{\times}$ iff $val(t)\equiv 0\bmod 4$ and $t/17^{val(t)}\equiv \bar x^4\bmod 17$ for some $\bar x\in \mathbb{F}_{17}$ (you can also use Hensel lemma).
Now we reduce to $\mathbb{F}_{17}$, take any $y\equiv 5\bmod 17$, you'll get $x^4$ be a number that $\equiv 1\bmod 17$, which is always solvable.