# Zero and One are the only Consecutive Perfect Squares/Proof 1

## Theorem

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

## Proof

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 $\paren {0 + 1}^2 = 1^2 = 1$

$\blacksquare$