Prove that (n+1)^2 - (n-1)^2 is always odd for all positive integer values of n.

Let's try it this way.
Do you know how to expand a binomial squared? If you do here's one way to do it.
(n +1)^2 = n^2 + 2n + 1
(n - 1)^2 = n^2 - 2n + 1

Now put them together
(n^2 + 2n +1) - (n^2 - 2n + 1) Remove the brackets.
(n^2 + 2n + 1 - n^2 + 2n - 1
2n + 2n = 4n is what is left. What you have given us is always even.

To show (but not to prove) that the result is even, let n = 10
(10 +1)^2 - (10 -1 )^2
(11)^2 - (9^2)
121 - 81 = 40

Maybe if you start out with (say) n = 7 (an odd number) you might get an odd number.
(7 + 1)^2 - (7 - 1)^2
8^2 - 6^2 (doesn't matter what the answer is. It will be even.
64 - 36 = 28 and that is also even.

Nice question. Maybe because you are in middle school mathematics, this last proof is good enough. In any case, you will never get an odd result. Isn't that something? Thanks for posting.