Units of Quadratic Integers over 2

Theorem
Let $\Z \left[{\sqrt 2}\right]$ denote the set:
 * $\Z \left[{\sqrt 2}\right] := \left\{{a + b \sqrt 2: a, b \in \Z}\right\}$

... that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.

Let $\left({\Z \left[{\sqrt 2}\right], +, \times}\right)$ be the integral domain where $+$ and $\times$ are conventional addition and multiplication on real numbers.

Then numbers of the form $a + b \sqrt 2$ such that $a^2 - 2 b^2 = \pm 1$ are all units of $\left({\Z \left[{\sqrt 2}\right], +, \times}\right)$.

Proof
For $a + b \sqrt 2$ to be a unit of $\left({\Z \left[{\sqrt 2}\right], +, \times}\right)$, we require that:
 * $\exists c, d \in \Z: \left({a + b \sqrt 2}\right) \left({c + d \sqrt 2}\right) = 1$

In Numbers of Type Integer a plus b root 2 are Not a Field it is shown that the product inverse of $\left({a + b \sqrt 2}\right)$ is $\dfrac a {a^2 - 2b^2} + \dfrac {b \sqrt 2} {a^2 - 2b^2}$.

So if $a^2 - 2 b^2 = \pm 1$ it follows that $c$ and $d$ are integers.

Hence the result.

Note
This is not to say that the only units of $\left({\Z \left[{\sqrt 2}\right], +, \times}\right)$.

This simple analysis does not rule out the possibility of there being other $a, b \in \Z$ such that $\dfrac a {a^2 - 2b^2} + \dfrac {b \sqrt 2} {a^2 - 2b^2} \in \Z \left[{\sqrt 2}\right]$.