 Solution

For $p=2$, the even residue class contains exactly one root, by the simple-root form of Hensel lemma at $a=0$: $f(0)=20$ and $f^{\prime }(0)=-5$ is a unit. For the odd classes use the strong form of Hensel lemma:In both cases $v_2(f(a))>2v_2(f^{\prime }(a))=2$. Each gives exactly one root in $a+4{\mathbb{Z}}_2$. These two balls partition the odd integers, so there are no further roots.

For $p=3$, reduction gives $\overline{f}=X^3+X+2$, whose sole root is $2$. The derivative $3X^2-5$ is always a unit modulo $3$, so that root lifts uniquely. For $p=5$, the only possible residue is zero. But if $x\in 5{\mathbb{Z}}_5$, then $x^3-5x\in 25{\mathbb{Z}}_5$ while $20$ has valuation one, so $f(x)$ cannot vanish. The exact root counts are