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

Theorem
If $n$ is a perfect square other than $0$, then $n+1$ is not a perfect square.

 Suppose that $k, l \in \Z$ are such that their squares are consecutive, i.e.:


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

Then we can factor the left-hand side as:


 * $l^2 - k^2 = \left({l + k}\right) \left({l - k}\right)$

By Invertible Integers under Multiplication, it follows that:


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

Therefore, it must be that:


 * $\left({l + k}\right) - \left({l - k}\right) = 0$

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.