Finding the Units in the Ring \mathbb{Z}[t][\sqrt{t^{2}-1}]

Step 1
Given: ${\displaystyle \mathbb{Z}\left[t\right]\left[\sqrt{t^2-1}\right]}$
Write:
${\displaystyle x=t+\sqrt{t^2-1}}$
and
${\displaystyle y=t-\sqrt{t^2-1}}$
Then ${\displaystyle xy=1,\text{i.e.}y=x^{-1},\text{and}t=\frac{(x+y)}{2}}$
${\displaystyle \sqrt{t^2-1}=\frac{(x-y)}{2}}$
Thus we see that
${\displaystyle \mathbb{Z}\left[t\right]\left[\sqrt{t^2-1}\right]\subset \mathbb{Z}\left[\frac{1}{2}\right][x,x^{-1}]}$
Now it is easy to check that if A is an integral domain, then the only units in
${\displaystyle A[x,x^{-1}]}$
are of the form ${\displaystyle a{x}^n}$, where
${\displaystyle a\in A}$
is a unit and n is an integer. Since the units in ${\displaystyle \mathbb{Z}\left[\frac{1}{2}\right]}$ are precisely the elements ${\displaystyle \pm 2^m}$ (for some m) we see that the units in
${\displaystyle \mathbb{Z}\left[\frac{1}{2}\right][x,x^{-1}]}$
are of the form
${\displaystyle \pm 2^m{x}^n=\pm 2^m{(t+\sqrt{t^2-1})}^n}$
It is easy to check that if such an element and its inverse both actually lie in
${\displaystyle \mathbb{Z}\left[t\right][\sqrt{t^2}-1\}]}$,
then necessarily ${\displaystyle m=0}$ (e.g. for norm reasons, or just looking explicitly at their denominators), and so we get the answer that the units are precisely the elements of the form
${\displaystyle \pm {(t+\sqrt{t^2-1})}^n}$
(Here n runs over ${\displaystyle \mathbb{Z}}$; this is the same set as ${\displaystyle \pm {(t\pm \sqrt{t^2-1})}^n}$, where n is now non-negative.)

Step 1
For any non-negative integer n the element
${\displaystyle \pm {(t\pm \sqrt{t^2-1})}^n}$
has norm 1, and there are no elements
${\displaystyle a\left(t\right)+b\left(t\right)\sqrt{t^2-1}}$
of norm -1 because
${\displaystyle a{\left(1\right)}^2-b{\left(1\right)}^2\sqrt{1^2-1}\ge 0}$
Step 2
Suppose
${\displaystyle a\left(t\right)+b\left(t\right)\sqrt{t^2-1}}$
is a unit which is not one of the above units such that ${\displaystyle degb}$ is minimal. Then it has norm 1, so
${\displaystyle a{\left(t\right)}^2-b{\left(t\right)}^2(t^2-1)=1}$
It is not hard to see that the units ${\displaystyle \pm 1}$
are the only units for which ${\displaystyle b=0}$, so WLOG b is nonzero. This implies that
${\displaystyle dega=d+1}$
${\displaystyle degb=d}$
for some non-negative integer d, and moreover the leading terms of a and b must agree up to sign. If the leading terms agree, then
${\displaystyle (a\left(t\right)+b\left(t\right)\sqrt{t^2-1})(t-\sqrt{t^2-1})}$
${\displaystyle =(ta\left(t\right)-b\left(t\right)(t^2-1))+(tb\left(t\right)-a\left(t\right))\sqrt{t^2-1}}$
is a unit with the property that the coefficient of ${\displaystyle \sqrt{t^2-1}}$ has degree strictly less than that of b which is not on the above list, which contradicts the assumption of minimality. Similarly, if the leading terms of a and b are opposite in sign, then
${\displaystyle (a\left(t\right)+b\left(t\right)\sqrt{t^2-1})(t+\sqrt{t^2-1})}$
${\displaystyle (ta\left(t\right)+b\left(t\right)(t^2-1))+(tb\left(t\right)+a\left(t\right))\sqrt{t^2-1}}$
is a unit with the property that the coefficient of ${\displaystyle \sqrt{t^2-1}}$ has degree strictly less than that of b which is not on the above list, again contradicting the assumption of minimality. So no such units exist.

Just a geometric translation of Matts