Proof that the square of an odd number is of the form $8m+1$ for some integer $m$ 
Show that $n$ is odd $\rightarrow \exists m \in \mathbb Z,n^2  = 8m + 1$

Let $k$ be an integer so that $n = 2k+1$. Then $n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2\left(2k^2+2k\right) + 1$. 
But this isn't equal to $8m+1$, so can I change that into $4(2m)+1$? 

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