Zero and One are the only Consecutive Perfect Squares/Proof 2
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Theorem
If $n$ is a perfect square other than $0$, then $n+1$ is not a perfect square.
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 left hand side 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.
$\blacksquare$