Prove or disprove (discrete math)Prove or disprove the following statement. The difference of the square...

Step 1
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.
Step 2
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.

Step 1
What does it mean for an integer ℓ to be odd exactly? It means that we may express ℓ 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 Z; then, we are dealing with the difference $\begin{array}{}\text{(1)}& (m+1{)}^2-m^2.\end{array}$
Step 2
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 ℓ. Here's the important part: notice what form ℓ takes. We have that $\ell =2m+1,$ where m is an arbitrary integer in Z. Thus, by definition, we can see that ℓ is an odd integer.
Maybe this will clear anything up you did not quite get before.