Zero and One are the only Consecutive Perfect Squares/Proof 2

Proof
Suppose that $k, l \in \Z$ are such that their squares are consecutive.

That is:


 * $l^2 - k^2 = 1$

Then we can factor the as:


 * $l^2 - k^2 = \paren {l + k} \paren {l - k}$

By Invertible Integers under Multiplication, it follows that:


 * $l + k = \pm 1 = l - k$

Therefore, it must be that:


 * $\paren {l + k} - \paren {l - k} = 0$

That is, $2 k = 0$, from which we conclude $k = 0$.

So if $n$ and $n + 1$ are squares, then necessarily $n = 0$.

The result follows.