Zero and One are the only Consecutive Perfect Squares

Theorem
Let $$P(n)$$ denote that n is a perfect square.

Then $$P(n) \to \neg P(n+1)$$

Proof
Suppose $$P(n+1)$$. Then $$n+1 = k^2$$ for some $$k$$. Then:

$$ $$

As $$(k+1) \neq (k-1)$$, $$n$$ is not a perfect square, so we have:

$$P(n+1) \to \neg P(n)$$ To which the contrapositive is $$P(n) \to \neg P(n+1)$$