Find an integer $ m \ge 2 $ so that the equation $ x^2 \equiv 1 $ in $\mathbb{Z}/ m$ has more than two solutions.
In a previous part I proved that there are two solutions $x=1,-1$ when $m$ is prime. I'm not sure if that is of any relevance here?
I'm not sure how to even start this, I was thinking maybe this has something to do with the $\text{gcd}(x, m)$?
Your observation is certainly relevant, since when $m$ is prime there are only 2 solutions - so it means that the example you are looking for cannot have $m$ prime.
So you need to start looking at values of $m$ which are not prime.
A good place to start might be looking at the squares of all elements when $m=8$ ...
Hint: Choose $m$ and $n$ relatively prime such that $x^2\equiv 1\pmod{m}$ and $x^2\equiv 1\pmod{n}$ each have at least $2$ solutions.
