# Solutions to the equation $x^2=1$ in a cyclic group

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?

Thanks!

• 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. – NickC Oct 20 '15 at 20:57
• Yes I mean for x^2=1 – jmsac Oct 20 '15 at 21:06
• Have you learned that a cyclic group of order $n$ has exactly one subgroup of order $d$ for each $d$ dividing $n$? – pjs36 Oct 20 '15 at 21:10
• @Rramiro de la Vega: : How that, the set of units is not a group? – Bernard Oct 20 '15 at 21:18
• @Bernard: Right, I misinterpreted the notation. – 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 $\;\deg p(x)q(x)=\deg p(x)+\deg q(x)$.
• To add to this, observe that $x=\pm 1$ are solutions and since there can be at most two, these are the only two. – Oiler Oct 20 '15 at 21:25