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:


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?
  • 1
    $\begingroup$ Look it up in the proof of Dirichlet's unit theorem; you will need to understand the archimedean valuations of a number field and the regulator. $\endgroup$ Jan 9 at 13:34
  • $\begingroup$ @franzlemmermeyer - thanks! But I'm not a proffesional mathematician so I cannot make a deep study of famous theorems and their proofs (like Dirichlet unit theorem) by myself. It will be more useful to me if you will cite some known results... maybe there is a result that gives an upper bound on the analytic norm of the fundamental unit of $\mathbb{Q}[n^{1/3}]$ in terms of $n$. $\endgroup$
    – user2554
    Jan 9 at 14:57
  • $\begingroup$ I mean - is there a result that limits our search of fundamental unit of $\mathbb{Q}[n^{1/3}]$ to a certain parallelopipe in the $xyz$ space? $\endgroup$
    – user2554
    Jan 9 at 15:02
  • $\begingroup$ If you're interested in the more elementary aspects of Pell equations I suggest getting hold of Barbeau's book "Pell's equation". $\endgroup$ Jan 9 at 19:11
  • 1
    $\begingroup$ See: Jeans, N.S. and Hendy, M.D., 1978. Determining the fundamental unit of a pure cubic field given any unit. Mathematics of Computation, 32(143), pp.925-935. $\endgroup$
    – O. S. Dawg
    Jan 14 at 17:41

2 Answers 2


We might consider a slightly more efficient method to identify a fundamental unit for the quadratic case.

Let us pick the integers in $\mathbb Q[\sqrt{7}]$. We know that any unit will solve one of the equations


and since $-1$ is nonquadratc $\bmod 7$ we can safely limit our trials to the $+$ sign in the constant term. By brute force or using continued fractions as a guide, we find $a=\pm16$ is the absolutely smallest candidate for which the roots are in the required field. Thereby we identify any $\pm8\pm3\sqrt7$ (independent choices of the signs) as a unit in this integer domain. That this is the fundamental unit in connected, in the quadratic case, with the coefficient $a$ being absolutely smallest among all nonzero possibilities giving roots in the required domain. Powers of this fundamental unit and their negatives correspond to absolutely larger $a=\pm254,\pm2048,...$.

The proposal is that a unit in the cubic domain will satisfy an equation of the form


where the odd degree allows us to specify the $+$ sign in the constant term; roots with the $-$ sign there are just negatives of those using the $+$ sign. We expect that raising any unit to a power will tend to increase the magnitudes of $a$ and $b$. On this argument a unit that minimizes $|a|+|b|$ is expected to be fundamental.

We try this with the integers of $\mathbb Q[\sqrt[3]{7}]$. One such unit is $-2+\sqrt[3]{7}$ (there is no $\sqrt[3]{49}$ term in this case) giving the polynomial equation

$x^3+6x^2+12x+1=0, |a|+|b|=18.$

Squaring this gives roots of

$x^3-12x^2+132x-1=0, |a|+|b|=144>18,$

and higher powers blow up the "norm" even further as with the quadratic case.

What if we try out roots of $x^3+ax^2+bx+1$ where $|a|+|b|\le18$? We have $685$ candidates:

$x^3-18x^2+0x+1=0\to\text{ casus irred.}$

$x^3-17x^2-1x+1=0\to\text{ casus irred.}$

$x^3-17x^2+0x+1=0\to\text{ casus irred.}$

(Casus irreducibilis equations will not have roots in a real integer field and are not further analyzed.)


$x^3-6x^2+9x+1=0\to x=2-\phi^{2/3}-\phi^{-2/3} (\phi=(1+\sqrt5)/2)$


$x^3-6x^2+12x+1=0\to x=2-\sqrt[3]{9}$


and it turns out that the only cases having irrational roots in the required domain are $x^3+6x^2+12x+1=0$ $x^3+12x^2+6x+1=0$, the latter derived from the former through multiplicative inversion.

If this reasoning is correct, then the fundamental unit in the integers of $\mathbb Q[\sqrt[3]{7}]$ is $\sqrt[3]7-2$ or any number derived from it by some combination of additive and multiplicative inversions.

The polynomial equation approach can be used systematically to search for units and identify units that would he fundamental accordingvyo the $|a|+|b|$ criterion above.

Start with the equation


where $a$and $b$ are integers. Without loss of generality we may take $a$ positive, $b<a$, and for roots in a real cube-root field the discriminant criterion for only one real root holds:


Thus for any choice of $a$ only a finite range of $b$ values are allowed, and we try out each of these to find cases where $(-D)/3$ is a perfect square. In these cases the root of the cubic will be a unit less than $-1$ in $\mathbb Q[\sqrt[3]{m}]$. The unit with minimal $a$ will then be fundamental, and its negative should correspond to the entry in the Project Euclid table given here.

For example, we try $a=12$. Then the discriminant criterion becomes


and the only values of $b$ less than $12$ meeting this condition are $-7,-6,-5,...+6$.

We render $-D$ for eachbof these trials and find two where the result is three times a perfect square:







Solving the equation via Cardano's method for $b=+6$ gives the root


where $56=2^3×7$ and $49=7^2$ imply that the root is in $\mathbb Q[\sqrt[3]{7}]$. The above root us the reciprocal os $\sqrt[3]7){7}-2$ treated above. A similar treatment for the $b=-6$ case gives a negative unit in $\mathbb Q[\sqrt[3]{3}]$:


Neither field admits a solution for positive integral $a<12$, so both units given above are fundamental. While this example seems straightforward, most values of $m$ require much larger polynomial coefficients $a$ and $|b|$ to identify solutions in this way.

  • $\begingroup$ thanks! [+1], your answer is helpful, but your coclusion that $7^{1/3}-2$ is a fundamental unit is not in accord with the table in [this article] (projecteuclid.org/journals/proceedings-of-the-japan-academy/…). But maybe I am misunderstanding this table/your answer. $\endgroup$
    – user2554
    Jan 9 at 20:49
  • $\begingroup$ In fact Project Euclid and I agree. If $x$ is my choice, which will be negative since I am setting the constant in my cubic to $+1$, then the entry in the table is $-1/x$. Multiply out and see. Since $-1/x$ has the same algebraic relation to other units as $x$, either one may be picked with equal validity. (And so can their additive inverses.) $\endgroup$ Jan 9 at 20:58
  • $\begingroup$ Thanks (I multiplied the saw you are right)! if the criterion for a unit to be a fundamental unit is that simple (the sum of absolute values of two coefficients in its minimal polynomial should be minimized), why did I read that this is relatively an advanced and deep topic? $\endgroup$
    – user2554
    Jan 9 at 21:09
  • $\begingroup$ What's advanced is finding the unit to begin with. That part is easy for $\mathbb Q[\sqrt[3]{7}]$, not so much (if I believe the table, which I do) for $\mathbb Q[\sqrt[3]{15}]$. One little lesson I've learned through my calculations is $\sqrt[3]{28}-3$ is not fundamental in its domain. Incredibly, you need its square root! $\endgroup$ Jan 9 at 21:26
  • $\begingroup$ do you have some insights regarding my mathoverflow question? - mathoverflow.net/questions/322954/…. Notice that the case $\mathbb{Q}[15^{1/3}]$, which you cited as a particular example of a difficult case (computationally) is included in Gauss's table, and I verified its correctness by multiplying it in the value of the table from project euclid article. $\endgroup$
    – user2554
    Jan 9 at 21:57

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 $1$.

  • $\begingroup$ A denominator of $3$ is not enough for rational solutions to always capture the fundamental units. $\mathbb Q[\sqrt[3]{12}]$ gives a denominator of $2$; $\mathbb Q[\sqrt[3]{28}]$ gives a denominator of $6$. $\endgroup$ Jan 10 at 1:01
  • $\begingroup$ @OscarLanzi I have added an edit to the paragraph where I wrote about the possible denominators. $\endgroup$
    – KCd
    Jan 10 at 2:19
  • $\begingroup$ You may want to look at the latest edit to my answer, which combines our two concepts to identify the units. $\endgroup$ Jan 10 at 15:09

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