= Solution
First reduce the <polynomial> modulo seven. At residues $0,1,\ldots,6$, its values are respectively $4,2,6,1,0,2,6$. Its only residue root is therefore $4$, and $f'(4)=45\equiv3\pmod7$ is a <unit>. By <Hensel lemma>, this residue root lifts uniquely to the <7-adic integers>; any other root would have to reduce to the same residue root. Thus \b[there is exactly one root in $\mathbb Z_7$, congruent to $4$ modulo seven.]
Modulo five the values at $0,1,\ldots,4$ are $4,2,1,2,1$, so there is no residue root and hence \b[no root in $\mathbb Z_5$.] Modulo three the only possible residue root is $-1$. But writing $x=-1+3t$ gives
$$
f(-1+3t)=6-27t^2+27t^3\equiv6\pmod9.
$$
Thus that residue root cannot lift even modulo nine, and \b[there is no root in $\mathbb Z_3$.] A repeated residue root alone would not establish this obstruction; the second congruence is needed.
Modulo two both residues are roots. The even residue is simple because $f'(0)=-3$ is a <2-adic unit>, so <Hensel lemma> gives exactly one even root. For any odd <2-adic integer>, $x^3\equiv x\pmod4$, whence $f(x)\equiv-2x\equiv2\pmod4$. No odd root can exist. Consequently the full answer is
$$
\boxed{\#\{x\in\mathbb Z_2:f(x)=0\}=1.}
$$
Back to article page
