What is $17$ adic value of $3+2\sqrt{-2}\in \Bbb{Q}_{17}$ ? I expanded $\sqrt{-2}=7+a_117+・・・$, so $3+2\sqrt{-2}=17+{a_1}17+・・・=17(1+a_1+a_{2}17＋・・・)$, thus $ord_{17}(3+2\sqrt{-2})\ge 1$.
If $ord_{17}(3+2\sqrt{-2})\ge 2$, $a_1$ should be $16$, but if so, $(7+16・17＋・・・)^2≡-2mod289$, then $7×17×2×16≡-51mod289$. This is contradiction.
Thus, I conclude $ord_{17}(3+2\sqrt{-2})= 1$.
Am I thinking too hard? I would appreciate it if you could tell me if there is a simpler way to calculate it.
After knowing $\sqrt{-2}=7+a_1p+\cdots\ \ (p=17)$, we see $$\nu_{17}(3+2\sqrt{-2})=\nu_{17}(17+2a_1p+\cdots)\geqq 1$$ and $$\nu_{17}(3-2\sqrt{-2})=\nu_{17}(-11-2a_1p-\cdots)=0,$$ but these two numbers multiply to $17=p$, so $$\nu_{17}(3+2\sqrt{-2})+\nu_{17}(3-2\sqrt{-2})=\nu_{17}((3+2\sqrt{-2})(3-2\sqrt{-2}))=\nu_{17}(17)=1,$$ hence $\nu_{17}(3+2\sqrt{-2})=1$ follows.
EDIT: In fact, the computation of $\nu_{17}(3-2\sqrt{-2})$ was unnecessary, just knowing it is $\geq 0$ (which follows from $\sqrt{-2}\in \mathbb{Z}_{17}$, at once) was enough.