Units in $\mathbb Z[\sqrt{19}]$ How to find the fundamental unit in $\mathbb Z[\sqrt{19}]$? I have no idea how it may look like.
 A: Observe that $\alpha = 12 + \sqrt{19}$ has norm $5^3$; now look for elements of norm $\pm 5$. You will find $\beta = 9 + 2\sqrt{19}$ is such an element. Thus
$$ \varepsilon = \alpha/\beta^3 = 170 - 39\sqrt{19}$$ 
is a nontrivial unit.
It remains to check that $\varepsilon$ is not a power of another unit. 
It cannot be a square since $39$ is odd. If you set $\varepsilon = \gamma^n$
for $\gamma = c + d \sqrt{19}$, then $d \mid 39$, and you can quickly eliminate all possible cases.
A: To find the fundamental unit of $\mathbf Z\bigl[\sqrt{19}\mkern1mu\bigr]$, you have to find the smallest non-trivial solution in positive integers of the Pell-Fermat equation:
$$x^2-19y^2=1.$$
The tool for this is to find the expansion of $\sqrt{19}$ as a continued fraction. The expansion of a quadratic number is periodic, and if it has period $n$, it is  denoted as 
$$\bigl[a_0;\overline{a_1a_2\dots a_n}\bigr]$$
Lagrange proved the fundamental unit is the convergent of order $n-1$.
Using Pari-GP, we find that 
$$\sqrt{19}=\bigl[4;\overline{213128}\mkern 1mu\bigr],$$
hence the $x$ and $y$ of the fundamental unit are the numerator and denominator of the convergent of order $5$:
$$4+\frac{1}{2+\cfrac{1}{1+\cfrac{1}{3+\cfrac{1}{1+\cfrac{1}{2}}}}}=\frac{170}{39}$$
Thus the fundamental unit (in the first quadrant) is $\;\color{red}{170+39\sqrt{19}}$.
