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This is related to a comment in the post Is the algebraic closure of a $p$-adic field complete.

$\textbf{Q:}$ How to show there are infinite many positive integer $n\geq 2$ s.t. $x^n=2$ does not admit solution over $Q_p$ for any prime $p$? For $p=2$, I could use valuation argument to conclude the assertion easily. For $p\neq 2$, this indicates $x$ is a unit in $Q_p(x)$ extension.

user45765
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    Let $m$ be the order of $2\bmod p$, when does $x^n-2\in \Bbb{F}_p[x]$ have a root ? If it has a root and $p\nmid 2n$ what does it imply about $x^n-2\in \Bbb{Z}_p[x]$ ? – reuns Dec 26 '19 at 19:19
  • @reuns I see. So choose any $n$ s.t. $n$ is not in class of $\frac{p-1}{m}$ in modulo $p-1$. That will get to work by Hensel lifting. Then there can't be a solution over $Q_p$. Thus it has to $[Q_p(x):Q_p]=n$ for such a choice of family of $n$. – user45765 Dec 26 '19 at 19:23
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    For most $n$ it is not true that $[Q_p(x):Q_p]=n$, Hensel lemma (=gradient descent) plus $F_{p^f}^\times$ is cyclic with $p^f-1$ elementsplus $[F_{p^f}:F_p]=f$ shows that (for $p\nmid 2n$) $[Q_p(x):Q_p]=[F_p(x):F_p]=$ the least integer $f$ such that $mn | p^f-1 = $ the order of $p \bmod mn$ – reuns Dec 26 '19 at 19:26
  • @reuns Yes. I think I forgot that I am using simple root Hensel lifting. Thanks for clarification. – user45765 Dec 26 '19 at 19:31

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