TITLE: Proof that the square of an odd number is of the form $8m+1$ for some integer $m$
QUESTION [9 upvotes]: 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 $4(2m)+1$?

REPLY [5 votes]: 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.