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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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    @Bill I agree, it's not a duplicate. Alas, it's not possible for me to remove t$h$e close vote. Also, t$h$is question admits ot$h$er approaches than the other question: for e.g., your answer.2011-11-30

3 Answers 3

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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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    For your second approach, the main point is to find u>1. Otherwise, you could have a finite group. Now, it is easy to find u>1 by inspection.2011-11-30
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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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    @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.2011-11-30
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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.2011-11-30