Necessary and sufficient conditions for a polynomial in $\mathbb{Z}[t]$ to have an $n$th root in $\mathbb{Z}[[t]]$ 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.
 A: Here is a silly way of coming up with necessary conditions. Let $$p(t)=1+a_1t+a_2t^2+\cdots +a_nt^n \in \mathbb{Z}[t]$$ be a polynomial. Then, using the Taylor expansion for $f(x)=\sqrt{1+x}$, 
$$\sqrt{1+x} = 1 + 1/2x - 1/8x^2 + 1/16x^3 - 5/128x^4 + 7/256x^5 - 21/1024x^6 +\cdots $$
we obtain an expansion of $\sqrt{p(t)}$ in $\mathbb{Q}[[t]]$ as follows:
$$\sqrt{1+(p(t)-1)} = 1 + \frac{(p(t)-1)}{2} - \frac{(p(t)-1)^2}{8} + \frac{(p(t)-1)^3}{16} - \frac{5(p(t)-1)^4}{128} + \cdots$$
$$=1 + \frac{a_1}{2}t + \frac{(-\frac{a_1^2}{4} + a_2)}{2}t^2 + \frac{(\frac{a_1^3}{8} - \frac{a_1a_2}{2} + a_3)}{2}t^3+\cdots$$
Suppose there is a $q(t)=1+\cdots \in \mathbb{Z}[[t]]$ with $q(t)^2=p(t)$. Then $q(t)^2 = \left(\sqrt{p(t)}\right)^2$, and so $q(t) = \pm  \sqrt{p(t)}$ (notice that $\mathbb{Q}[[x]]$ is an integral domain). Since both start with $1$, we must have $q(t) = \sqrt{p(t)}\in \mathbb{Z}[[t]].$
Thus,


*

*$a_1$ must be even,

*$a_2-a_1^2/4$ must be even,

*$\frac{a_1^3}{8} - \frac{a_1a_2}{2} + a_3$ must be even, etc.
