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?
Best Answer
An easier way to approach this would be to observe: If x is odd, either x-1 or x+1 is divisible by 4. Then their product is divisible by 8.