I am trying to prove: if $m = p_1p_2\cdots p_r$ with $2 < p_1 < \cdots < p_r$ prime, then
$x^2 \equiv 1\mod m$
has $2^r$ solutions modulo $m$.
I know Euler's Criterion: $p$ is an odd prime s.t $p$ does not divide $b$. Then
$b^{\frac{p-1}{2}} \equiv 1 \mod p \iff b \equiv t^2 \mod p\text{ for some }t \in \mathbb{Z}$
and I know Wilson's Theorem: $n\mid (n-1)! +1 \iff$ $n$ is prime
- How do I put these together?