let $\displaystyle a$ be a solution of $\displaystyle x^2\equiv 1 \ \mbox{(mod m)}$. Show that $\displaystyle m-a$ is also a solution.

$\displaystyle a^2\equiv 1 \ \mbox{(mod m)}$

$\displaystyle (m-a)^2\equiv 1 \ \mbox{(mod m)}\rightarrow m^2-2am+a^2\equiv 1 \ \mbox{(mod m)}\rightarrow m^2-2am+1\equiv 1 \ \mbox{(mod m)}$

$\displaystyle \rightarrow m-2a\equiv 0 \ \mbox{(mod 1)}$

Not sure if this is going anywhere.