Prove if $2|x^{2} - 1$ then $8|x^{2} - 1$ I have seen this question posted before but my question is in the way I proved it. My books tells us to recall we have proven
if $2|x^{2} - 1$ then $4|x^{2} - 1$ 
Using this and the fact $x^{2} - 1 = (x+1)(x-1)$ and a previous proof in which we have shown if$a|b$ and $c|d$ then $ac|bd$ my proof is as follows. Assume $2|x^{2} - 1$ and $4|x^2 - 1$ then $2|(x-1)(x+1)$ and $4|(x+1)(x-1)$. Assume $2|x-1$ and $4|x+1$ then $(2)(4)|(x+1)(x-1) = 8|x^{2} - 1$. Would this proof be considered valid? I rewrote it and used some other known proofs to help me out but I dont know if by rewriting it, I have proven something else something not originally asked.

If $x$ is even, $x^2-1\not\equiv 0\mod 2$. Hence from the hypothesis you know $x$ is odd. Furthermore, the square of any odd integer is congruent to $1$ modulo $8$.
