roots of a polynomial mod $p^n$ Let $f = X^2 + uX + v$ be an irreducible polynomial in $\mathbb{Z}[X]$ and let $\alpha$ and $-(u+\alpha)$ be it's two distinct roots. 
If we consider $f$ in $(\mathbb{Z}/p\mathbb{Z})[\alpha]$ it has again only two roots (i.e. $\alpha$ and $-(\overline{u} + \alpha)$ where $\overline{u}$ is reduction of $u$ modulo $p$)
What can we say about the roots modulo $p^n$ ?
 A: Consider $x^2 - p$ with $p$ prime, which is irreducible because it is Eisenstein at $p$. If $x^2 - p = 0 \pmod{p^2}$ for some integer $x$, then $p^2 | x^2-p$, which means $ap^2 = x^2 - p$ and $x^2 - p - ap^2 = 0$. Since $x^2-p - ap^2$ is again Eisenstein at $p$, it is irreducible, hence has no root, which contradicts the fact that $x$ solved the equation in the first place. 
So I guess what I'm saying is that in general you cannot assume you will have roots mod $p^2$. 
The theorem that works though in this case is called Hensel's lemma ; it allows you to lift roots of a polynomial mod $p$ to roots mod $p^n$ for any integer $n$ in a unique way, assuming some non-degeneracy (i.e. that if $x$ is a root of $f(X)$ mod $p$, then $f'(x) \neq 0 \pmod p$. Note that this is the case for the roots of $x^2 - p$ (which is a double root at $0$ modulo $p$), because if $f(X) = X^2 - p$ then $f'(X) = 2X$ which is zero when evaluated at $0 \pmod p$. 
If you want to understand better the situation you should read up on Hensel's lemma. Start by the wiki page and if you need more, elementary number theory books that cover the standard topics will explain how it works in detail.
Hope that helps,
