Exercise 2.4.9 (If $x^2\equiv 1 \pmod m$, but $x \not \equiv \pm 1 \pmod m$, then $1<(x-1,m) < m$)

Proof. Suppose that $x^2\equiv 1(\mathit{mod}m)$, but $x\not\equiv \pm 1(\mathit{mod}m)$.

If $m\wedge (x-1)=m$, then $m\mid x-1$, this is contrary to the hypothesis. Moreover, $m\wedge (x-1)\le m$, so

and similarly $m\wedge (x+1)<m$.

Since $x^2\equiv 1(\mathit{mod}m)$, $m\mid (x-1)(x+1)$. If $m\wedge (x-1)=1$, then $m\mid x+1$, and this is contrary to the hypothesis. Therefore

and similarly $1<m\wedge (x+1)$.

So, if $x^2\equiv 1(\mathit{mod}m)$, but $x\not\equiv \pm 1(\mathit{mod}m)$, then $1<(x-1,m)<m$, and that $1<(x+1,m)<m$. □