# Units of Quadratic Integers over 2

## Theorem

Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:

$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$

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

Let $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$ 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 $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$.

## Proof

For $a + b \sqrt 2$ to be a unit of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$, we require that:

$\exists c, d \in \Z: \paren {a + b \sqrt 2} \paren {c + d \sqrt 2} = 1$

In Quadratic Integers over 2 are Not a Field it is shown that the product inverse of $\paren {a + b \sqrt 2}$ is $\dfrac a {a^2 - 2 b^2} + \dfrac {b \sqrt 2} {a^2 - 2 b^2}$.

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

Hence the result.

$\blacksquare$

## Note

This is not to say that the only units of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$.

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