TITLE: When is a $p$-adic unit an $m$-th power
QUESTION [5 upvotes]: Given $x\in\mathbb{Q}_p^*$, we can write $x=p^nu$ where $n\in\mathbb{Z}$ and $u\in\mathbb{Z}_p^*$. Then we can decide whether or not $x$ is a square by looking at $n$ and $u$. If $p\neq 2$ then $x$ is a sqaure if and only if the "first component" of $x$ is a square as a member of $\mathbb{F}_p^*$ and $n$ is even. If $p=2$ then $x$ is a square if and only if $n$ is even and $u\in1+8\mathbb{Z}_p$.

Is there a similar characterization of when $x\in\mathbb{Q}_p^*$ is an $m$-th power?

REPLY [4 votes]: Let me start from @Cantlog’s response, which points out that the difficulty is in knowing when $1+pa$ is an $m$-th power. The answer depends strongly on the divisibility of $m$ by $p$. In particular, if $m$ is indivisible by $p$, then $1+pa$ is always an $m$-th power. Indeed, as @Jyrki points out, $(1+pa)^z$ has a well-defined meaning for any $p$-adic integer $z$. For, if $\{n_i\}$ is a sequence of positive integers with $p$-adic limit equal to $z$, one shows (it’s a nice exercise) that $\{(1+pa)^{n_i}\}$ is Cauchy, and you define $(1+pa)^z$ to be the limit.
When $m$ is indivisible by $p$, $1/m$ is a $p$-adic integer, and by the remarks above, you’re good to go. So the question reduces to what the $pa$’s are such that $1+pa$ is a $p$-th power. When $p>2$, you see that $(1+pb)^p=1+p^2bu$, where $u$ is a $p$-adic unit. This also is an easy consequence of the Binomial Theorem. So, as long as $a$ itself is divisible by $p$, $1+pa$ is a $p$-th power. The story is just a little more complicated for $p=2$, because the expansion of $(1+z)^2$ involves so few terms. But if $a$ is a $p$-adic unit, then $1+pa$ is definitely not a $p$-th power.