Necessary and sufficient conditions for a polynomial in $\mathbb{Z}[t]$ to have an $n$th root in $\mathbb{Z}[[t]]$

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.