Zero and One are the only Consecutive Perfect Squares
Theorem
If $n$ is a perfect square other than $0$, then $n+1$ is not a perfect square.
Proof 1
Let $x$ and $h$ be integers such that:
- $x^2 + 1 = \paren {x - h}^2$
Then:
| \(\ds x^2 + 1\) | \(=\) | \(\ds \paren {x - h}^2\) | ||||||||||||
| \(\ds 1\) | \(=\) | \(\ds -2 x h + h^2\) | ||||||||||||
| \(\ds 2 x h\) | \(=\) | \(\ds h^2 - 1\) | ||||||||||||
| \(\ds 2 x h\) | \(=\) | \(\ds \paren {h - 1} \paren {h + 1}\) |
We have that Consecutive Integers are Coprime.
However by the Fundamental Theorem of Arithmetic, both sides must have the same unique prime decomposition.
Therefore $h$ cannot have any prime factors since they cannot be shared by $\paren {h - 1} \paren {h + 1}$.
This leaves $h = -1$, $h = 0$, or $h = 1$ as the only possibilities since they are the only integers with no prime factors.
If $h = -1$ then $h + 1 = 0$, so $2 x h = 0$.
It follows that $x = 0$.
If $h = 1$ then $h - 1 = 0$, so $2 x h = 0$.
It follows that $x = 0$.
If $h = 0$, then $2 x \cdot 0 = \paren {-1} \paren 1$, which is a contradiction.
Therefore the only pairs of consecutive squares are:
- $0^2 = 0$ and $\paren {0 + \paren {-1} }^2 = \paren {-1}^2 = 1$
and:
- $0^2 = 0$ and $\paren {0 + 1}^2 = 1^2 = 1$
$\blacksquare$
Proof 2
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.
$\blacksquare$