If n is such that every element $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is a root of $x^2-1$. Prove that $n$ divides 24. I have a hard time formulating proofs.  For this problem, I  can see that if $n$ is equal to $8,$ this statement is true.  $(\mathbb{Z}/8\mathbb{Z})^{\times}$ includes elements:  $1,3,5,7$, and all of these are roots of $1-x^2 \pmod 8.$  And obviously $8$ divides $24.$
But how do I prove this without depending on number calculations and only using theorems?  Help Please?  I need a step by step walk through of how to do this proof and what theorems would be appropriate to use.  

Let $x=2m+1$.
$x^2-1=(x+1)(x-1)=(2m+2)(2m)=2^2m(m+1)$.
$m(m+1)$ is divisible by $6$ except when $m \equiv 1 $ or $4$$(mod 6)$, and $x^2-1$ is a multiple of $24$ except for those cases. Let's call the cases in which $m(m+1)$is divisible by 6 as Case I and others as Case II.
$2^2m(m+1)$ must be divisible by $n$ for arbitrary $m$. The periodicity of appearance of factors $2$ and $3$ in $m(m+1)$ needs to be discussed here since they are the factors which frequently appear in $m(m+1)$. If there's no such periodicity, not every odd elements will be divided by $n$. 
Case I:
When $m=0$, $2^2m(m+1)\equiv 0(mod n)$.
When $m=2$, $2^2m(m+1)\equiv 24(mod n)$.
When $m=3$, $2^2m(m+1)\equiv 48(mod n)$.
When $m=5$, $2^2m(m+1)\equiv120(mod n)$.
Case II:
When $m=1$, $2^2m(m+1)\equiv8(mod n)$.
When $m=4$, $2^2m(m+1)\equiv80(mod n)$.
If for every element of $(\mathbb{Z}/n\mathbb{Z})^*$, $x^2-1$ is divisible by $n$, then n must be a divisor of $8$ and divide $24$.
