= Solution
A simple-root version of <Hensel lemma> is: if $R$ is a complete <discrete valuation ring>, $f\in R[T]$, $f(a_1)\equiv0\pmod\pi$, and $f'(a_1)$ is a unit, then there is a unique root $a\in R$ congruent to $a_1$ modulo $\pi$.
Suppose $f(a_n)\in\pi^nR$. The derivative remains a unit, so choose a residue $c_n$ satisfying
$$
\frac{f(a_n)}{\pi^n}+c_nf'(a_n)\equiv0\pmod\pi,
$$
and put $a_{n+1}=a_n+c_n\pi^n$. Taylor expansion gives $f(a_{n+1})\equiv f(a_n)+c_n\pi^nf'(a_n)\pmod{\pi^{n+1}}$, 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 $f'(a_1)$, so it is a unit. Therefore $a=b$. This is the <simple-root Hensel lifting by successive residues>.
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