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Let $\mathbb{Z}[\sqrt{3}]$ be the ring $\{a + b\sqrt{3} | a, b \in \mathbb{Z}\}$ with a natural structure. We consider it's norm:

$$a + b\sqrt{3} \mapsto a^2 - 3b^2$$

Then, finding the solution to the Pell's diophantine equation $x^2 - 3y^2 = 1$ amounts to find the multiplicative kernel of the norm map $\mathbb{Z}[\sqrt{3}] \to \mathbb{Z}$. This kernel is easily found as it is contained in $\mathbb{Z}[3]^\times$, the group of units of this ring. I'm stuck at trying to prove the following result, which amounts to the existence of a fundamental solution to the equation:

There is an isomorphism $\mathbb{Z}[\sqrt{3}]^\times \simeq \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}$

Where the torsion part corresponds to the roots of unity in $\mathbb{Z}[\sqrt{3}]^\times$, which are only $\{ \pm 1\}$, and the torsion-free part is given by the powers of the fundamental solution.


Here is what I've done so far: a friend of mine gave me a partial proof of the Dirichlet Unit Theorem (from which the proposition above is a special case). I tried to adapt it, and make it work for this ring using simpler arguments. Let us consider the map: $$f\colon \mathbb{Z}[\sqrt{3}]^\times \to \mathbb{R}$$ $$z \mapsto \log|z|$$ where $| \cdot |$ here represents the absolute value function on the real numbers. This application is a group homomorphism, from the multiplicative group of the ring to the additive group of the reals. It's kernel is precisely the torsion part of the group of units, so I just need to show that it's image is cyclic. Now, let $z_1$ and $z_2$ be two units in $\mathbb{Z}[\sqrt{3}]^\times$. Then, it is enough to show that there exists a rational number $\frac{p}{q}$ such as that $z_1 = \pm(z_2)^{\frac{p}{q}}$. Now, because we have $\dim_\mathbb{R}\mathbb{R} = 1$ as a real vector space, we know that there exists real numbers $k_1$ and $k_2$ that satisfy: $$k_1\log |z_1| + k_2\log |z_2| = 0$$ This implies that $z_1 = \pm (z_2)^{\frac{k_1}{k_2}}$ inside $\mathbb{R}$. Now here comes the tricky part: though $z_1$ and $z_2$ are (possibly) irrational numbers, they are very controlled irrational numbers. Seems intuitive that for a real power of $z_2$ to still lie in $\mathbb{Z}[\sqrt{3}]$, this power must actually be a rational power (and a very special one at that, given that roots are scarce in this ring). This however, seems very non-trivial and outside the realm of the number theory. Is this true? Is there some "elementary" proof of this fact? Otherwise, how can we characterize the units of $\mathbb{Z}[\sqrt{3}]$ without resorting to the full power of the Dirichlet Unit Theorem?

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I think it is worth pointing out how close you were to finishing off your own argument!

The key insight is that is any non-trivial subgroup of $\mathbb R$ is either dense or isomorphic to $\mathbb Z$. (See here and here.) So we must prove that the image of your homomorphism $f : \mathbb Z[\sqrt 3]^\times \to \mathbb R$ is not dense in $\mathbb R$.

To prove this, pick any $\delta > 0$. It is enough to prove that the set $f^{-1}(-\delta, \delta)$ is a finite set.

If $z = a + b\sqrt {3}$ is a unit such that $\log |z| \in (-\delta, \delta)$, then $$ | a + b \sqrt{3}|<e^\delta, \ \ \ \ \ \ \ \ | a - b \sqrt 3 | < e^\delta.$$ [Note that $z$ is a unit, so $|a^2 - 3b^2| = |N(z)| = 1$, hence $-\log |z| = \log|z^{-1}| = \log|a - b \sqrt{3}|$.]

Therefore, $ | a | < e^\delta$ and $|b| < \frac{e^\delta}{\sqrt 3},$ by the triangle inequality. Since $a, b \in \mathbb Z$, there are only finitely many possible $a$ and $b$ satistfying these inequalities, and we're done.

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