In a cyclic group, say $(\mathbb{Z}/p\mathbb{Z})^*$ where $p$ is a prime, why does the equation $x^2=1 \mod p$ only have two solutions?


  • $\begingroup$ Do you mean when r=2 there is only two solutions? If so, it is because $\mathbb{Z}/(p\mathbb{Z})$ is an integral domain. $\endgroup$ – NickC Oct 20 '15 at 20:57
  • $\begingroup$ Yes I mean for x^2=1 $\endgroup$ – jmsac Oct 20 '15 at 21:06
  • $\begingroup$ Have you learned that a cyclic group of order $n$ has exactly one subgroup of order $d$ for each $d$ dividing $n$? $\endgroup$ – pjs36 Oct 20 '15 at 21:10
  • $\begingroup$ @Rramiro de la Vega: : How that, the set of units is not a group? $\endgroup$ – Bernard Oct 20 '15 at 21:18
  • $\begingroup$ @Bernard: Right, I misinterpreted the notation. $\endgroup$ – Ramiro de la Vega Oct 20 '15 at 21:22

Because a quadratic equation in a field (more generally in an integral domain) has at most two roots. This is because $\alpha$ is a root of $p(x) \iff p(x) $ is divisble by $\;x-\alpha\;$ and $\;\deg p(x)q(x)=\deg p(x)+\deg q(x)$.

  • $\begingroup$ To add to this, observe that $x=\pm 1$ are solutions and since there can be at most two, these are the only two. $\endgroup$ – Oiler Oct 20 '15 at 21:25
  • $\begingroup$ Yes. I think it was implicit in the O.P.'s question. Do you think I should add this fact? $\endgroup$ – Bernard Oct 20 '15 at 21:27
  • $\begingroup$ It can't hurt! Maybe O.P. Is a diligent comment reader and it won't be necessary. $\endgroup$ – Oiler Oct 21 '15 at 1:56

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