Let $p(t) = \sum p_k t^k$ be a polynomial in $\mathbb{Z}[t]$, with $p_0=1$. Is there a necessary and sufficient condition (congruence or other) on the coefficients $p_k$ such that $p(t)$ admits a square root in $\mathbb{Z}[[t]]$?
More generally, is there a necessary and sufficient condition for $p(t)$ to have an $n$-th root in $\mathbb{Z}[[t]]$?
Based on the fact that $(1+n^2z)^{1/n} \in \mathbb{Z}[[z]]$ for any $n \in \mathbb{Z}^+$, we obtain a sufficient condition, namely that $n^2 \mid p_k$ for all $k>0$. This is, however, very weak.
Edit: Clarified notation as per one of the comments. Also, a recent paper `On the Integrality of $n$th Roots of Generating Functions' by Heninger, Rains, and Sloane provides a criterion that may be of some use:
"Let $\mu_n = n \prod_{q \mid n} q$ where $q$ ranges over the primes dividing $n$. Then $f= \sum u_n t^n \in \mathbb{Z}[[t]]$ admits an $n$th root in $\mathbb{Z}[[t]]$ if and only if $f \mod \mu_n$ does as well."
As such, we can hope to get lucky and have some reduction that is obviously an $n$th power (for example, if $f \mod \mu_n$ is the power of some polynomial, instead of just a power series. Note of course that this does not resolve our original question.
Edit II: This question has been untouched for a long while, but perhaps this will help another as a reference. For completeness, I'd like to add that the condition that $p$ reduce modulo $\mu_n$ to the $n$th power of a polynomial in $\mathbb{Z}[t]$ is actually both sufficient and necessary for $p$ to admit an $n$th root in $\mathbb{Z}[[t]]$ (given, of course, that $p(0)=1$). The proof relies on the following combinatorial lemma, that $\mu_n \mid (\mu_n/n)^k \binom{n}{k}$ for $k=1,\ldots, n$. One may then show that our $n$th roots of $p(t)$, say $\sum a_n t^n \in \mathbb{Z}[[t]]$, reduce to polynomials $\mod \mu_n/n$, by showing that $n a_n$ is eventually divisible by $(\mu_n/n)^k \binom{n}{k}$ for some $k$, through a symmetry/combinatorial argument.