Odd number proof 
Prove that for every odd number $n$, it holds that $n^2+3$ is not not divisible
  by $8$.

My idea:
Let $n=2k+1$ for $k \in \mathbb{N}$, which implies
$$n^2+3=(2k+1)^2+3=4k^2+4k+1+3=4k^2+4k+4$$
How can I conclude that I cannot divide $4k^2+4k+4$ by $8$?

Just for fun, a slightly different approach:
If $n^2+3$ is divisible by $8$, then so is $(n-4)^2+3=n^2+3-8n+16$, hence, by induction, $8$ divides $n^2+3$ for at least one $n$ between $-2$ and $2$. But $0^2+3=3$, $(\pm1)^2+3=4$, and $(\pm2)^2+3=7$ are not divisible by $8$.
