Proof that the square of an odd number is of the form $8m+1$ for some integer $m$

$$n^2 = (2k+1)^2 = 4k^2+4k+1 = 4k(k+1)+1$$

At least one of $k$ and $k+1$ is divisible by $2$, so let $k(k+1)=2m$. Then, $n^2 = 4(2m)+1 = 8m+1$.

Okay, we want $4k^2 + 4k + 1 = 8m + 1$ so we want $m = \frac {4k^2 + 4k}8=\frac {k^2+k}2$

So w have to prove that $k^2 + k$ is even. Can you do that.