The roots of $X^2-1$ in the ring $(\mathbb{Z}/n\mathbb{Z})[X]$

The roots of $X^2-1$ in the ring $(\mathbb{Z}/n\mathbb{Z})[X]$

I don't know how to prove the following.
Let $n$ be an odd natural number with $t$ different prime divisors. Then
the polynomial $X^2-1$ has exactly $2^t$ roots.
Own tries
First of all: $X^2-1=(X-1)(X+1)$. We say that $a \in \{1, 2, \cdots, n \}$
is a root if and only if $(a-1)(a+1)=kn$ for some integer $k$. Now observe
that $p|(a-1) \iff p\nmid(a+1)$, except when $p=2$. If $a$ is odd, $2$
divides both $a+1$ and $a-1$. If $a$ is even, $2$ divides neither. Luckely
$n$ is odd. This means that a prime divisor $p$ from $n$ only can appear
in maximal one of the numbers $a+1, a-1$.
It looks like we only need to prove that there exists $k \in \mathbb{Z}$
so that $p$ appears in at least one of these numbers: $kn+a-1, kn+a+1$,
because, be previous observations, there are exactly $2^t$ was to
"arrange" prime divisors in the factorisation of $\bar{a}-\bar{1}$ and
$\bar{a}+\bar{1}$. Is this correct?