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Let p be an odd prime and k a positive integer. Show that the congruence $x^{2}$ $\equiv 1 \ mod p^{k}$ has exactly two incongruence solutions, namely, $x \equiv \pm 1\mod p^{k}$.

I'm not sure what to do after this: $x^{2}$ $\equiv 1 \ mod p^{k}$ => $p^{k}$ | $x^{2}-1$=(x-1)(x+2)

DJ_
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1 Answers1

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As you have already observed, $p^k\mid (x-1)(x+1)$. In particular $p\mid (x-1)(x+1)$, so either $p\mid x-1$ or $p\mid x+1$. In any case, we cannot have both since $p\not\mid 2$. This implies that $p^k\mid x-1$ or $p^k \mid x+1$. Why?

This works in general: $x^2=n\mod p^k$ has at most two incongruent solutions.

Pedro
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  • How does the p∤2 part fit it? Since we're looking at p^k, can we still use p∣x−1 or p∣x+1 in our reasoning? – DJ_ Nov 29 '13 at 01:46
  • @DJ_ Right, we know that $p\mid x-1$ or $p\mid x+1$. But we cannot have both, since this would imply $p\mid x+1-(x-1)=2$, which is impossible. This means that we can assume WLOG that $p\not\mid x-1$, and? – Pedro Nov 29 '13 at 01:49