 Solution

A simple-root version of Hensel lemma is: if $R$ is a complete discrete valuation ring, $f\in R[T]$, $f(a_1)\equiv 0(\mathrm{mod}\pi )$, and $f^{\prime }(a_1)$ is a unit, then there is a unique root $a\in R$ congruent to $a_1$ modulo $\pi$.

Suppose $f(a_n)\in {\pi }^{n}R$. The derivative remains a unit, so choose a residue $c_n$ satisfyingand put $a_{n+1}=a_n+c_n{\pi }^n$. Taylor expansion gives $f(a_{n+1})\equiv f(a_n)+c_n{\pi }^n{f}^{\prime }(a_n)(\mathrm{mod}{\pi }^{n+1})$, because $2n\ge n+1$. Thus $f(a_{n+1})\in {\pi }^{n+1}R$. The sequence is a Cauchy sequence and converges by completeness; continuity of polynomial evaluation makes its limit a root.

If roots $a,b$ have the prescribed residue, factor $f(a)-f(b)=(a-b)Q(a,b)$. Modulo $\pi$, the second factor is $f^{\prime }(a_1)$, so it is a unit. Therefore $a=b$. This is the simple-root Hensel lifting by successive residues.