Prove that if $x$ is odd then $x^2 -1$ is divisible by $8$. 
If $x$ is odd then prove that $x^2-1$ is divisible by $8$.

I start by writing: $x = 2k+1 $  where $k\in\mathbb{N}$.
Then it follows that:
$(2k+1)^2 -1 = 4k^2 +4k + 1 -1 $  
Therefore:
$$\frac{4k^2 +4k}{8} = \frac{k(k+1)}{2}$$ 
At the end part I can see that for what $k$ is, the number on top is divisible by 2. I was expecting the end result to be a number, not a fraction. Or is it "divisible" from the definition $\text{even} = 2k$ that completes the proof?

Are you familiar with the mod operator?
We want to show
$x^2 - 1 \equiv 0\mod8$
$x^2 \equiv 1\mod8$ for $x = 2k+1$ for some $k$
$(2k+1)^2 \equiv 1 \mod 8$
$4k^2 + 4k + 1 \equiv 1 \mod 8$
$4k^2 + 4k \equiv 0 \mod 8$
$4 (k^2 + 4k) \equiv 0 \mod 8$
This is what we want to show.
If $k$ is odd then $k^2$ is odd. And $k^2 + k$ is even.
If $k$ is even then $k^2 + k$ is even
Hence $k^2 + k$ is always of the form $2a$, And hence
$4 (k^2 + 4k) \equiv 4(2a) \equiv 8a \equiv 0 \mod 8$
