Prove that the difference between the squares of two consecutive numbers is always an odd number
For two consecutive numbers $p, q \in R$ where $q=p+1$, it follows that $p^2 - q^2 = (p-q)(p+q)$. Moreover, $$p^2 - q^2 = (p-(p+1))(p+(p+1))$$ which gives $$p^2 - q^2 = -1(2p+1)$$ Hence the number is in fact odd.
My question is whether I should consider the case where $q=p-1$ too and is there a way to generalize it.
First of all, you probably meant $p,q \in \textbf{Z}$. Now, if the smallest of two numbers is $n$, then you are interested in $(n+1)^2-n^2=n^2+2n+1-n^2=2n+1$ - which is odd.
Note: to avoid considering all cases, as you asked you can say (without lost of generality) that $n$ is the smallest among two of them, which will give you the fact that $n+1$ is the seocnd number.