I am trying to show:

$f(x)=x^2 - 2x$ has precisely two solutions modulo $p^k$ where $p$ is a 0 odd prime, and $k$ is an positive integer.

I'm thinking I need to use Hensel's lemma, and I have shown it is true for $p^2$, but I'm not sure how to continue.

Homework, so hints are what I'm looking for.


I don't think that we need Hensel's lemma.

Note that $f(x)=x(x-2) \equiv 0 \pmod {p^k}$.

However, note that $p$ cannot divide both $x$ and $x-2$ since $2$ is even and $p$ is odd.

  • $\begingroup$ Can you expand a little? $\endgroup$ – Mark Feb 27 '16 at 11:41
  • $\begingroup$ Note that $x(x-2) \equiv 0 \pmod {p^k}$ when $x \equiv 0 \pmod {p^{t}}$ and $x \equiv 2 \pmod {p^{k}}$. However, from my answer, $t$ and $k$ cannot both be positive. One has to be $0$. $\endgroup$ – S.C.B. Feb 27 '16 at 11:42
  • $\begingroup$ I see what you're getting at, but the notation of taking the power of the moduli is different to what i'm used to $\endgroup$ – Mark Feb 27 '16 at 11:45
  • $\begingroup$ @Mark That was just a mistake on my part. $\endgroup$ – S.C.B. Feb 27 '16 at 11:46

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