$\mathbb Z[\sqrt 3]$ contains infinitely many units

Solution 1:

A sketch. Show that if $(a,b)$ is a solution to the equation $x^2 - 3y^2 = 1$, then $(2a+3b, a+2b)$ is a larger solution. Iterating this infinitely many times gives us infinitely many solutions.

Another approach is the following. I presume it is a variation of Bill's hint.

Hint: If $u$ is a unit, then $u^n$ is a unit for all integers $n$.

Solution 2:

HINT $\ $ If the unit group were finite then all units would be roots of unity ($= {\pm 1}\:$ since it is $\subset \mathbb R$).

Solution 3:

Another (equivalent) Hint: Look at $(2\pm \sqrt{3})^n =a_n \pm \sqrt{3}\,b_n$. Show that $a_n \pm \sqrt{3}\,b_n$ is a unit.