A $p$-adic unit is an $n^{th}$ power if it is congruent to a $n^{th}$ power mod $m$ (Eisenbud exercise)

Question

Exercise 7.27 in Eisenbud's Commutative Algebra poses the following problem: Give a criterion for a $p$-adic unit $u$ to be a $n^{th}$ power for any $n$. The hint states: "The form of the criterion will be as follows: $u$ is congruent to an $n^{th}$ power mod $m$, for an integer $m$ that you should determine." This is in the chapter on Hensel's lemma, so I've tried using that, but I haven't found a general answer.

Note: We assume "$p$-adic unit" means a unit of the ring $\mathbb{Z}_p$. See the comments below the first answer about what "for any $n$" means. This is why the first answer does not answer the question I intended to ask.

Answer 1

The question as quoted makes no sense to me.
Here’s the story, at least if by “$p$-adic unit” you mean a unit of the ring $\Bbb Z_p$, or indeed if you mean the units of the ring of integers of a finite extension of $\Bbb Q_p$.

The group of principal units, namely those congruent to $1$ modulo the maximal ideal, is uniquely divisible by all $n$ prime to $p$, that is (in $\Bbb Z_p$), if $u\in1+p\Bbb Z_p$, then $u$ has a unique $n$-th root in $1+p\Bbb Z_p$. But the only elements of the units of $\Bbb Z_p$ that are divisible by all powers of $p$ are the roots of unity (except for when $p=2$). Indeed, the group of roots of unity of $\Bbb Z_p$ for $p>2$ is cyclic of order $p-1$, so that for such a root $\zeta$, you get $\zeta^{p^n}=\zeta$

Answer 2

A form of Hensel's Lemma states that if $f(x)\in\mathbb{Z}_p[x]$ and $a\in\mathbb{Z}_p$ satisfy $v_p(f(a)) > 2v_p(f'(a))$, then $a$ can be lifted to a root of $f$. Here we are interested in roots of $f(x)=x^n-u$, where $n$ is a positive integer and $u\in\mathbb{Z}_p^\times$. An $n$-th root of $u$ modulo $p$ has valuation $0$, so to apply Hensel's Lemma, we need to find $a\in\mathbb{Z}_p$ such that $v_p(a^n-u) > 2v_p(n)$, or in other words, $u$ needs to have an $n$-th root modulo $p^{2v_p(n) +1}$.