If x is odd then x^2 - 1 is divisible by 8

I know you have to replace x with (2n + 1) to make it odd, but I can't seem to factor (2n + 1)^2 - 1 into being divisible by 8. Can anyone help me?

I haven't done proofs in two months, so this may a be a bit rusty. However, I hope this will meet your needs.

Let x be an odd integer. That is, let x = 2n + 1 where neZ.

(2n+1)^2 - 1 ≡ 0 (mod 8)
4n^2 + 4n + 1 - 1 ≡ 0 (mod 8)
4n^2 + 4n ≡ 0 (mod 8)

So:

8|(4n^2 + 4n)

Hence,

4n^2 + 4n = 8k for some keZ
4n^2 + 4n - 8k = 0

Multiplying both sides by 2, we have:

8n^2 + 8n - 16k = 0

8(n^2 + k - 2k) = 0.

Since n^2 + k - 2keZ, the result follows.

(2n + 1)^2 - 1 = 4n^2 - 4n = 4*(n(n-1))
So if x is odd, then x^2 - 1 is divisible by 4. We don't know about 8.
But if you look at (n(n-1)), we know that its divisible by 2.
Thus n(n-1) is of the form 2u, where u is an integer.
So (2n + 1)^2 - 1 = 4n^2 - 4n = 4*(n(n-1)) = 4*2u = 8*u.

Note:
odd*odd = odd, (I)
even*even = even, (II)
even*odd = odd*even = even. (III)
n(n-1) takes the form of (III), which is why its even.