prove or disprove (discrete math) This the question:

Q: Prove or disprove the following statement.
     The difference of the square of any two consecutive integers is odd

This is working step: 
let $m,m+1$ be 2 consective integers:
 $$(m+1)^2-m^2$$
$$m^2+1+2m-m^2$$
$$1+2m$$
If $m$ is odd then $2m=\text{even}$, 
if $m$ is even then $2m=\text{even}$, 
then adding $1$ will make it odd.
Can you please advise me if my working is the right step and could I answer like this? I am new to discrete and not sure if I can ans like this.  Feedback is welcome to improve the presentation of the steps :) 
 A: The fact that ...$2m$ is always even , then $2m+1$ is odd regardless of ...$m$.
A: Your proof looks correct.  You might want to make it more clear that you are saying
$$(m+1)^2-m^2 = (m^2 + 2m + 1) - m^2 = 2m+1$$
when you do your arithmetic.  You also don't need to consider the cases where $m$ is even and odd separately: since $2m$ is a multiple of $2$, it must be even, and so you can conclude that $2m+1$ is not evenly divisible by $2$, so it is odd.
A: This will be very similar to what Strants had to say, but I think it may clarify a thing or two. 
What does it mean for an integer $\ell$ to be odd exactly? It means that we may express $\ell$ in the form $\ell = 2n+1$, where $n\in\mathbb{Z}$ (note that $n$ can be any integer). 
Now consider your question: 

The difference of the square of any two consecutive integers is odd.

Thus, assume (as you did in your first attempt) that $m$ is an arbitrary integer in $\mathbb{Z}$; then, we are dealing with the difference
$$
(m+1)^2-m^2.\tag{1}
$$
Now expand $(1)$ as Strants did in his answer (except with a slight modification):
$$
(m+1)^2-m^2 = (m^2 + 2m + 1) - m^2 = 2m+1 = \ell,
$$
where $\ell\in\mathbb{Z}$. We know $\ell\in\mathbb{Z}$ because adding two integers, namely $2m$ and $1$, yields an integer, namely $\ell$. Here's the important part: notice what form $\ell$ takes. We have that
$$
\ell = 2m+1,
$$
where $m$ is an arbitrary integer in $\mathbb{Z}$. Thus, by definition, we can see that $\ell$ is an odd integer.
Maybe this will clear anything up you did not quite get before.
