TITLE: $\mathbb Z[\sqrt 3]$ contains infinitely many units
QUESTION [9 upvotes]: I'm asked to show that there are infinitely many units in the ring $\mathbb Z[\sqrt 3]$. 
But I don't really see a good approach to this one, so far. 
Some thoughts: The inverse of $a+\sqrt3 b$ should be given by $\pm(2 - \sqrt 3 b)$, since the norm 
$$N: \mathbb Z[\sqrt{3}] \to \mathbb Z, \qquad N(x+\sqrt{3} y) = x^2 - 3y^2$$
is multiplicative. So if $(a+\sqrt{3}b)^{-1}=x+\sqrt 3 y$, then 
 $$1 = N(1) = N((a+\sqrt{3}b)(x + \sqrt{3}y)) = (a^2 - 3b^2)(x^2 - 3y^2)$$
Hence we must have $\pm 1 = (a^2 - 3b^2) = (a+\sqrt{3}b)(a-\sqrt{3}b)$.
Therefore I need to show that there are infinitely many $a,b$ such that $a^2 - 3b^2 = \pm 1$. 
Here I don't know how to proceed. 
Maybe rewriting as $a = \sqrt{3b^2 \pm 1}$, and now trying to prove that there are infinitely many $b$ for which $3b^2 \pm 1$ is a square?
A hint would be appreciated! =) Thanks.

REPLY [7 votes]: HINT $\ $ If the unit group were finite then all units would be roots of unity ($= {\pm 1}\:$ since it is $\subset \mathbb R$).

REPLY [6 votes]: 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.