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 $2\vert x^2-1$, then $x$ must be odd, so write $x=2y+1, y\in\mathbb{Z}$. 
Then $x^2-1=4y^2+4y=8\frac{y(y+1)}{2}$. 
Note that at least one of $y,y+1$ is even, so the fraction on the right is always an integer. 
Thus we have that $8\vert x^2-1$.
