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

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.

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@Srivatsan It's not a dup since the OP explicitly asks for a hint, but the prior question does not. – Bill Dubuque Nov 30 '11 at 23:28
@Bill I agree, it's not a duplicate. Alas, it's not possible for me to remove the close vote. Also, this question admits other approaches than the other question: for e.g., your answer. – Srivatsan Nov 30 '11 at 23:33

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 $n$.

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Note: This diophantine equation is an example of Pell's equation. – Srivatsan Nov 30 '11 at 23:08
Firstly: Thank you for taking the time to answer. Secondly: Simply writing HINT in front of a complete solution does not make it a hint! Better would have been something like: "Try to generate new solutions from old ones", or even: "Given a solution $(a,b)$, try to find $m,n,u,v$ such that $(ma + nb, ua + vb)$ is again a solution" or something to that effect. Now I have put in no effort at all to solve this exercise (except of course reducing it to solving Pell's equation for $n=3$), which I don't like very much. Unfortunate. – Sam Nov 30 '11 at 23:16
@Sam Indeed, it was a bit unfortunate. Thanks for the feedback. I will try to do better next time. =) I did consider "Generate new solutions from old ones", but to my ears it sounds too vague to be a productive hint. The second one is a good hint, but I did not think of it. [And, of course, it's not a complete solution: some details are obviously missing.] – Srivatsan Nov 30 '11 at 23:17
Yes, you are right, it is not really a complete solution - it just felt like a bit too much help. :) btw: I sincerely hope the comment did not sound ungrateful - since it was not intended that way. – Sam Nov 30 '11 at 23:27
@Sam, Your criticism is a valid one and I take it constructively. I can assure you I did try, but it's not easy to give the right hint, you know. =) – Srivatsan Nov 30 '11 at 23:33

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.

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Ah, this is a great way of looking at it! The units of a ring form a multiplicative group and therefore having the unit $2 + \sqrt3$, which can indeed be found easily, we immediately obtain infinitely many solutions $(2+\sqrt{3})^n$ for $n\in \mathbb N$ which obviously are all distinct. Very nice! – Sam Nov 30 '11 at 23:31
Only now did I realise, where the $(2a+3b, a+2b)$ from Srivatsan's answer come from: It is because $(2+\sqrt{3})(a+\sqrt{3}b) = (2a+3b) + \sqrt{3}(a + 2b)$. I would upvote again, if only I could. – Sam Nov 30 '11 at 23:37
@Sam Indeed, the point of my answer was to attempt to coax you into viewing this slightly more abstractly using the innate group-theoretic structure (assuming that you know basic group theory - which is a highly recommended prerequisite for studying number theory of quadratic fields). Once you exhibit a unit $\ne \pm 1\:$ then it follows immediately from my hint that there are infinitely many. – Bill Dubuque Nov 30 '11 at 23:41

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

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This answer is great, too. Thank you very much. It seems like your idea highlights yet another reason, why the multiplicative group cannot be finite by relating the ring structure to it being $\subset \mathbb R$. This also generalizes nicely and seems to be worth to keep in mind, cheers. – Sam Nov 30 '11 at 23:44