Ring with four solutions to $x^2-1=0$ I am looking for a ring $R$ in which $2$ is invertible and there are four solutions to $x^2-1=0$.
$R=\Bbb Z/8\Bbb Z$ has the four solutions $1,3,5,7$ to $x^2-1=0$, but $2$ is not invertible.

The ring $R=\Bbb{Z}/15\Bbb{Z}$ seems to fit the bill. The solutions are (the cosets of) $\pm1$ and $\pm4$. A process finding the solutions amounts to finding the solutions modulo $3$ and $5$ (two for each), and then combining them using the Chinese Remainder Theorem. There are many examples of how to do that on our site.
