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.

  • $\begingroup$ To apply Hensel's lemma, suppose $f(x) = x^3 - c^2x^2 + p^2 \equiv 0\ (p)$. It follows that $x^2(x - c^2) \equiv 0\ (p)$, so either $p \mid x$ or $p \mid x - c^2$. Assume $p \nmid x$, then $3x - 2c^2 \equiv 3x - 2x \equiv x \not\equiv 0\ (p)$. This will correspond to a solution modulo $p^3$ according to the lemma. $\endgroup$ – J.H. Mar 18 '13 at 1:48
  • $\begingroup$ @J.H. So in the case that $p \nmid x$, is the number of solutions $\mod p^3$ equal to the number of solutions of $f(x) \equiv 0 \mod p$? $\endgroup$ – Jack Mar 18 '13 at 2:21

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 $2p+1$.

  • $\begingroup$ Can you explain what you mean by $\bigg(\frac{c}{p}\bigg)$? You say the number of $x$ is equal to $1 + \bigg(\frac{c}{p}\bigg)$ but this is not an integer. $\endgroup$ – Jack Mar 18 '13 at 1:04
  • $\begingroup$ @Jack See here $\endgroup$ – Hanul Jeon Mar 18 '13 at 1:27
  • $\begingroup$ The second line should read $x^2(x - c^2) \equiv 0\ (p^2)$? $\endgroup$ – J.H. Mar 18 '13 at 1:30
  • $\begingroup$ @J.H. What does it mean ? $\endgroup$ – Hanul Jeon Mar 18 '13 at 1:34
  • $\begingroup$ I think it should be $c^2$ instead of $c$. $\endgroup$ – J.H. Mar 18 '13 at 1:43

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