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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.

Poitou-Tate
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    By a fondamental property we have $$|3+2\sqrt{-2}|{17}\le \max{|3|{17},|2|{17}\cdot|\sqrt{-2}|{17}}=\max{1,|\sqrt{-2}|{17}}$$ so your true problem is to calculate $|\sqrt{-2}|{17}$ and you are done. – Ataulfo Jun 04 '23 at 17:36
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    Note that the answer depends on making the choice that the symbol $\sqrt{-2}$ in $\mathbb Z_{17}$ should denote specifically the square root that is $\equiv 7$ mod $17$. A priori there is no reason why we should not call $\sqrt{-2}$ the one which is $\equiv 10$ mod $17$, and for that we get a different answer. Compare no. 3 in https://math.stackexchange.com/a/4007515/96384. – Torsten Schoeneberg Jun 04 '23 at 22:56

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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.

Ayaka
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