How to check if a solution to the cubic Pell's equation is a fundamental unit? For the quadratic case, the fundamental solution of the Pell equation $x^2-dy^2 = 1$ is the smallest solution in the ordinary sense, which means that the values of both $x$ and $y$ in the fundamental solution are smaller than in any other solution to this Pell equation. This means that once we have found a solution $(x_0,y_0)$ to Pell equation, which is indeed a unit of the quadratic field $\mathbb{Z}[\sqrt{d}]$, one can check if it is a fundamental unit by a finite number of steps: this algorithm is very computationally inefficient, but one can simply check all pairs $(x,y)$ such that $|x|<x_0, |y|<y_0$, and see if $(x_0,y_0)$ arises as a power of one of the smaller pairs. If it is not, than $(x_0,y_0)$ is a fundamental unit.
Of course, what I wrote above is perhaps the slowest possible algorithm to check if a unit is a fundamental unit, but it explains that, in principle, there isn't an "essential" problem in the quadratic case.
The cubic Pell equation, which arises as the norm form of the pure cubic field $\mathbb{Q}[d^{1/3}]$, is:
$$x^3+dy^3+d^2z^3-3dxyz=1$$
Since a fundamental solution now involves a triple of numbers $(x_0,y_0,z_0)$, the fundamental solution need not be such that every other solution has $|x|>|x_0|,|y|>|y_0|,|z|>|z_0|$, so it is not nessecerily a "smallest solution". Therefore, there must be some other criterion for a solution to be a smallest unit in a suitable sense.
My questions are therefore:

*

*What is a suitable criterion for a unit to be a fundamental unit of a pure cubic field?

*How can one check if a given unit satisfies this criterion in a finite number of steps?

 A: You refer to  "a unit of the quadratic field $\mathbf Z[\sqrt{d}]$", but that is not a field. Do you mean a unit in that quadratic ring? It need not be the full ring of integers of the quadratic field $\mathbf Q(\sqrt{d})$.
For cubefree integers $d$, the ring of integers of $\mathbf Q(\sqrt[3]{d})$ need not be $\mathbf Z[\sqrt[3]{d}]$: it isn't when $d \equiv \pm 1 \bmod 9$. Since you are looking at integral solutions of the norm equation
$$
x^3 + dy^3 + d^2z^3 - 3dxyz = 1
$$
are you only interested in units in $\mathbf Z[\sqrt[3]{d}]$ even if it is not the full ring of integers of $\mathbf Q(\sqrt[3]{d})$, or will you allow rational solutions with denominator $3$ when $d \equiv \pm 1 \bmod 9$ in order to include units in the full ring of integers all the time?
(Edit: When I said the denominator is at worst 3, I was thinking of the basis $\{1,\sqrt[3]{ab^2}, \sqrt[3]{a^2b}\}$ instead of $\{1,\sqrt[3]{d},\sqrt[3]{d^2}\}$, where $d = ab^2$ with $(a,b) = 1$.)
Based on numerical data, it is reasonable to ask if the fundamental unit $x_0 + y_0\sqrt[3]{d} + z_0\sqrt[3]{d^2}$ in $\mathbf Z[\sqrt[3]{d}]$ that is greater than $1$ has $\boxed{x_0 > y_0 > z_0 > 0}$ when $d \geq 3$. (For $d = 2$, the fundamental unit greater than $1$ has $x_0 = y_0 = z_0 = 1$.) The formula for coefficients in Corollary 2.1 of L.C. Zhang's paper On the units of cubic and bicubic fields Acta Math. Sinica 1 (1985), 22-34 might allow you to prove that boxed inequality, which then could give you a tedious but deterministic way of finding the fundamental unit: let $x_0$ runs through the positive integers, and for each value sample over all $y_0$ and $z_0$ fitting the boxed inequality. The first time you find a solution would then give you the least solution greater than $1$.
