Prove that the difference between alternate consecutive squares as always even

Take an integer n. The next alternate consecutive integer is n + 2

Subtract the difference of the squares:
(n + 2)^2 - n^2
n^2 + 4n + 4 - n^2

n^2 terms cancel, we get:
4n + 4

Factor out a 4:
4(n + 1)

If n is odd, n + 1 is even. 4 * even is always even
If n is even, n + 1 is odd. 4 * odd is always odd

Since both cases are even, we've proven our statement.