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

Question

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.

Answer 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.

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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 $n$.

Answer 2

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.

Answer 3

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