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

Let $x$ and $h$ be integers such that $x^2 + 1 = (x - h)^2$

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x^2 + 1\)	\(=\)	\(\displaystyle (x - h)^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 1\)	\(=\)	\(\displaystyle -2xh + h^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 2xh\)	\(=\)	\(\displaystyle h^2 - 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 2xh\)	\(=\)	\(\displaystyle (h - 1)(h + 1)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Consecutive Integers are Coprime, but both sides must have the same unique prime factorization by the Fundamental Theorem of Arithmetic, so $h$ cannot have any prime factors since they cannot be shared by $(h - 1)(h + 1)$.

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 $2xh = 0$. It follows that $x = 0$.

If $h = 1$ then $h - 1 = 0$, so $2xh = 0$. It follows that $x = 0$.

If $h = 0$, then $2x\cdot 0 = (-1)(1)$, a contradiction.

Therefore the only pairs of consecutive perfect squares are $0^2 = 0$ and $(0 + (-1))^2 = (-1)^2 = 1$, and $0^2 = 0$ and $(0 + 1)^2 = 1^2 = 1$.

$\blacksquare$