Suppose $\epsilon = a + b \sqrt d$ a unit in $R=\Bbb Z [\sqrt d] = \{a+b\sqrt d : a,b \in \Bbb Z\}$ . Proof that $\pm a\pm b \sqrt {d}$ are units. I'm trying to solve some problems for my ring theory exam. I have problems solving this one:

Let $d\in \Bbb Z_{>0}$ such that $d$ is not a square. Suppose
   $\epsilon = a + b \sqrt d$ a unit in $$R=\Bbb Z [\sqrt d] = \{a+b\sqrt
 d : a,b \in \Bbb Z\}$$
Proof that $\pm a\pm b \sqrt {d}$ are units. Show that $\Bbb Z[\sqrt
 5]$ contains infinite many units.

Hints or a detailed proof are both really appreciated. 
 A: $a + b\sqrt{d}$ is a unit means we can find $a', b' \in \mathbb{Z}$ such that
$(a + b\sqrt{d})(a' + b'\sqrt{d} ) = 1$. However, this implies
$$\begin{cases}a a' + b b' d & = 1\\ a b'+ b a' & = 0\end{cases}
\quad\implies\quad
(a - b\sqrt{d})(a' - b'\sqrt{d} ) = 1
$$
and hence $a - b \sqrt{d}$ is a unit. One can show other combination $\pm a \pm b \sqrt{d}$ are units using similar arguments.
About the part $\mathbb{Z}[\sqrt{5}]$ contains infinitely many units. You need to use/show two things:


*

*$a + b\sqrt{d}$, $a' + b'\sqrt{d}$ are units $\quad\implies\quad (a + b\sqrt{d})(a' + b'\sqrt{d})$ is a unit.

*find a unit $a + b\sqrt{5}$ of $\mathbb{Z}[\sqrt{5}]$ and verify $(a + b\sqrt{5})^n, n\in \mathbb{Z}$ are all distinct.

A: Here is a possible line of attack: (1) Suppose $a + b\sqrt{d}$ is a unit, i.e., has a multiplicative inverse in $\mathbb{Z}[\sqrt{d}]$. How would you find the inverse $\frac1{a + b\sqrt{d}}$? Remember "rationalizing the denominator". (2) Based on this, what criteria must be satisfied for an inverse to exist? (3) Show these criteria are met for $a - b\sqrt{d}$ if they are met for $a + b\sqrt{d}$. 
A: Another approach for 1st task would be to use the norm on $Z[\sqrt{5}]$ (denote  $d(\epsilon) =d(a+b\sqrt{5})=|a^2 -5b^2|$):
$d(\epsilon)\cdot d(x)=(*)=d(\epsilon\cdot x)=d(1)=1$. 
This implies $d(\epsilon)=1$, thus $|a^2 -5b^2|=1$, thus $a^2 -5b^2=\pm 1$.
a) $a^2 -5b^2=1$ implies $(a+b\sqrt{5})(a-b\sqrt{5})=1$, i.e. $a-b\sqrt{5}$ is unit  with inverse $a+b\sqrt{5}$
b) $a^2 -5b^2=-1$ would mean that the unit is $-(a-b\sqrt{5})=-a+b\sqrt{5}$.
Clearly, the remaining $a$ and $b$ combinations would result in both cases (a) and b)) just from the change of sign.
Step (*) can be easily proven.
