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?

Alternatively, you could prove this by induction on $x$.
Base case: Consider the case where $x = 1$. Clearly, $(1)^2 - 1 = 0$ is divisible by 8. Hence, the base case is true
Inductive Hypothesis: Assume that the statement is true for some $x = k$ where $k$ is an odd integer greater than 1.
Inductive Step: Since $k$ is odd, we must show that the statement holds for the next odd integer, $k + 2$. Plugging $x = k + 2$ into $x^2 - 1$,
$$
(k + 2)^2 - 1 = k^2 + 4k + 3 = (k^2 - 1) + 4(k + 1)
$$
$k^2 - 1$ was divisible by 8 by the inductive hypothesis. Thus, the statement holds for $k + 2$ if and only if $4(k + 1)$ is divisible by 8. Observe that since $k$ is odd, is can be written as $k = 2q + 1$ for some integer $q$. This gives us
$$
4(k + 1) = 4((2q + 1) + 1) = 4(2q + 2) = 8q + 8 = 8(q + 1)
$$
Thus, we have shown that 8 divides $4(k + 1)$ which means that it also divides $(k + 2)^2 - 1$, completing the proof.
