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From time to time I ask about units in Cubic fields. I noticed 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 walk-through explication of the Dirichlet unit theorem, there are numerous resources that discuss this in generality.

Avoiding Minkowski's theorem in algebraic number theory.

What is your favorite use 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 and here 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.

From time to time I ask about units in Cubic fields. I noticed 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 walk-through explication of the Dirichlet unit theorem, there are numerous resources that discuss this in generality.

Avoiding Minkowski's theorem in algebraic number theory.

What is your favorite use 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 and here 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.

From time to time I ask about units in Cubic fields. I noticed 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 walk-through explication of the Dirichlet unit theorem, there are numerous resources that discuss this in generality.

Avoiding Minkowski's theorem in algebraic number theory.

What is your favorite use 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 and here 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.

From time to time I ask about units in Cubic fields. I noticed 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 walk-through explication of the Dirichlet unit theorem, there are numerous resources that discuss this in generality.

Avoiding Minkowski's theorem in algebraic number theory.

What is your favorite use 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 and here 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.

From time to time I ask about units in Cubic fields. I noticed 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 walk-through of the Dirichlet unit theorem, there are numerous resources that discuss this in generality.

Avoiding Minkowski's theorem in algebraic number theory.

What is your favorite use 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. Otherwise I found Artin-Whaples, and I have to work out the real and complex embedding of $K = \mathbb{Q}(\sqrt[3]{2})$ - and I have trouble deciding if their Axiom 2 and Axiom 2a are related to more modern terminology.

I could work through as much and show where I got stuck?

From time to time I ask about units in Cubic fields. I noticed 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 walk-through explication of the Dirichlet unit theorem, there are numerous resources that discuss this in generality.

Avoiding Minkowski's theorem in algebraic number theory.

What is your favorite use 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 and here 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.