From time to time I ask about units in Cubic fields. I [noticed](http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/tracenorm.pdf) for $\mathbb{Z}[\sqrt[3]{2}]$ I get an analogue of the Pell equation: $$ \det \left[ \begin{array}{ccc} a & 2c & 2b \\ b & a & 2c \\ c & b & a \end{array} \right] = a^3 + 2b^3 + 4c^3 - 6abc = 1 $$ without citing the Dirichlet unit theorem. --- Clearly the answers are $a + b\sqrt[3]{2} + c\sqrt[3]{4} = (1 + \sqrt[3]{2} + \sqrt[3]{4})^n$ with $n \in \mathbb{Z}$ (since we this number is unit we can have negative exponent. Except, I have no way of ruling out other families of solutions. And is there any constructive way of solving this, i.e. without taking a wild guess? In the $\text{deg}=2$ case (such as $x^2 - 2y^2 = 1$) there is an answer using the pell eq and the continued fraction $\sqrt{2} = [1;\overline{2}]$. Can we do something analogous here? --- There are continued fractions you can do on triples of numbers. I think the first step here is: $$ (1, \sqrt[3]{2}, \sqrt[3]{4}) \to (1, \sqrt[3]{2}-1, \sqrt[3]{4}-1) \to \dots $$ not sure what the smallest number is here. I wonder if the familiar story from Pell works here? Does this euclidean algorithm repeat? Maybe this does not lead to finding units. --- Since I am basically asking for a <del>walk-through</del> explication of the Dirichlet unit theorem, there are numerous resources that discuss this in generality. http://mathoverflow.net/questions/19021/avoiding-minkowskis-theorem-in-algebraic-number-theory [What is your favorite use of the pigeonhole principle?](https://math.stackexchange.com/questions/62565/what-is-your-favorite-application-of-the-pigeonhole-principle) Another one that comes to mind is Hasse's **Lectures on Number Theory** if you have an English copy of the book. --- The equation $z^3 - 2 = (z - \sqrt[3]{2})(z - \omega\sqrt[3]{2})(z - \omega\sqrt[3]{2}) = 0$ has a real root and a complex conjugate pair of roots. So the field $\mathbb{Q}(\sqrt[3]{2})$ has a real embedding and a complex embedding. Then define an embedding $K=\mathbb{Q}(\sqrt[3]{2}) \to \mathbb{R}^3$ $$ (a+b\sqrt[3]{2}+c\sqrt[3]{4})(a+ b\omega\sqrt[3]{2}+c\omega^2\sqrt[3]{4}) (a+b\omega^2\sqrt[3]{2}+c\omega\sqrt[3]{4}) = a^3 + 2b^3 + 4c^3 - 6abc$$ This is not a familiar norm since it's cubic. And then define a region: \begin{eqnarray} |a+b\sqrt[3]{2}+c\sqrt[3]{4}| &=& c_1 \\ (a+ b\omega\sqrt[3]{2}+c\omega^2\sqrt[3]{4}) (a+b\omega^2\sqrt[3]{2}+c\omega\sqrt[3]{4}) &\leq & c_2^2 \end{eqnarray} The second equation defines the interior an conic section. $$ a^2 + \sqrt[3]{2}b^2 + 2\sqrt[3]{4}c^2 + ab + ac + 4bc \leq c_2^2 $$ I am following arguments [here](http://wstein.org/129/lectures/day10/day10.pdf) and [here](http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/unittheorem.pdf) trying to get the case $K= \mathbb{Q}(\sqrt[3]{2})$. If $c_1 c_2^2 < A$ can we show there is a lattice point in $\mathcal{O}_K$ inside this 3D region? This should work in a similar way for $\mathbb{Q}(\sqrt[3]{m})$ maybe the constant $A$ changes.