Prove that if $n^{2} - \left(n - 2\right)^{2}$ is not divisible by $8$ then $n$ is even

Question

Let $n$ be an integer. Prove that if $n^{2} -\left(n - 2\right)^{2}$ is not divisible by $8$ then $n$ is even.

Can anyone help me step by step to understand this?

Answer 1

We have $n^2-(n-2)^2 = n^2-(n^2-4n+4) = 4n-4$. $$\text{$8$ does not divide $4(n-1)$ $\iff$ $2$ does not divide $(n-1)$ (Why?)}$$ Now conclude what you want.

Answer 2

Hint: expand $(n-2)^2$. After you subtract, factor the result.

Answer 3

It may be easier to prove the contrapositive: if $n$ is odd, then $n^2 - (n-2)^2$ is divisible by $8$.

You can do this using casework on the remainder when $n$ is divided by $8$. There are four possibilities.