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.

To answer your question without resorting to any explanation beyond what you have provided:
No. This part is seriously off "Assume 2|x−1 and 4|x+1" without any explanation. When you say "assume this", "assume that", you have to cover all options.
2 is prime so it must divide either (x+1) or (x-1), that essential part is missing.
Now, if (x+1) divides 2, then either (x+1) divides 4, or (x-1) divides 4. If (x-1) divides 2, then either (x-1) divides 4, or (x+1) divides 4.
Now make all combinations


*

*(x+1) divides 4 (here you need to add that (x-1) divides 2)

*(x+1) divides 2 and (x-1) divides 4

*(x-1) divides 4 (here you need to add that (x+1) divides 2)

*(x-1) divides 2 and (x+1) divides 4


Now you can conclude.
