Prove the difference between two consecutive square numbers is always odd

Take an integer n. The next consecutive integer is n + 1

Subtract the difference of the squares:
(n + 1)^2 - n^2
n^2 + 2n + 1 - n^2

n^2 terms cancel, we get:
2n + 1

2 is even. For n, if we use an even:
we have even * even = Even
Add 1 we have Odd

2 is even. For n, if we use an odd:
we have even * odd = Even
Add 1 we have Odd

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