<s>[INST] Number of solutions to $x^3 - c^2x^2 + p^2 \equiv 0 \mod p^3$ I would like to find the number of solutions ($\mod p^3$) to $x^3 - c^2x^2 + p^2 \equiv 0 \mod p^3$, where $p$ is prime and $c$ is an integer not divisible by $p$.  I think Hensel's Lemma will be useful, but I'm not entirely sure how to apply it.
Help would be greatly appreciated.
 [/INST] If $x$ satisfies $x^3 - c^2x^2 + p^2 \equiv 0 \pmod {p^3}$, $x$ also satisfies
$$x^3 - c^2x^2 + p^2 \equiv 0 \pmod {p^2}$$
So $x^2(x-c^2)\equiv 0 \pmod {p^2}$. So $x$ satisfies  $p\mid x$ or $p^2\mid(x-c^2)$ and $x$ satisfies only one formula. (Because $c$ is not disible by $p$.)
If $p\mid x$, then there exists $y\in\mathbb{Z}$ s.t. $x=py$. Substitute $x$ with $py$ then we get
$$(1-c^2y^2)p^2\equiv 0\pmod{p^3}$$
This formula is equivalent to
$$1-c^2y^2\equiv 0\pmod{p}$$
So the number of $y$ is 2, namely $y\equiv c^{-1} \pmod{p}$ and $y\equiv-c^{-1} \pmod{p}$. So $x\equiv \pm c^{-1} p + kp^2 \pmod{p^3}$ ($k$ is arbitary integer) is solution of this equation, therefore the number of this equation is $2p$. 
If $p^2\mid(x-c^2)$, then there exists $z$ s.t. $x=c^2+zp^2$. Substitute $x$ with $c^2+zp^2$ and rearrange it then we get
$$c^4 zp^2 +p^2\equiv 0 \pmod{p^3}$$
This equation is equivalent to
$$c^4 z + 1\equiv 0 \pmod{p}$$
and it has only one soultion $z\equiv 1 \pmod p$, so the number of solutions of this equation is $2p+1$.
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