solving $x^3-2y^3=1$ using cubic number field

Solution 1:

First, let me point out that $E:x^3-2y^3=1$, together with the point $[1,0,1]$, is an elliptic curve. Its Mordell-Weil group can be calculated by finding a Weierstrass model, which can be chosen to be $y^2=x^3-27$, and then verify that $E(\mathbb{Q})\cong \mathbb{Z}/2\mathbb{Z}$. Thus, there are only two rational points on $E$, namely $[1,0,1]$ and $[-1,-1,1]$ in projective coordinates, which correspond to $(1,0)$ and $(-1,-1)$ in affine coordinates.

To calculate the integral points using the method you mention, you first need to notice that if $K=\mathbb{Q}(\sqrt[3]{2})$, then the norm from $K$ down to $\mathbb{Q}$ of $a+b\sqrt[3]{2}$ is $a^3+2b^3$, for any integers $a,b\in\mathbb{Z}$. Thus, if $x,y$ are integers such that $$x^3-2y^3=1,$$ then $x-y\sqrt[3]{2}$ has norm $1$. Since $K/\mathbb{Q}$ has one real embedding, and two complex embeddings, Dirichlet's unit theorem says that the unit group has rank $1$. Moreover, the only roots of unity are $\pm 1$, and a generator of the free part of the unit group is $-1+\sqrt[3]{2}$. Hence, $x-y\sqrt[3]{2}$ is of the form $$\pm (-1+\sqrt[3]{2})^n$$ for some $n\in\mathbb{Z}$. When $n=0$, we notice that $-1$ has norm $-1$, so we only get a possible solution, namely $x=1$, $y=0$. When $n=1$, we only get one solution, namely $x=-1$, $y=-(1)=-1$. In order to finish the proof, you need to show that for any other $n$, the power $(-1+\sqrt[3]{2})^n$ is not of the form $a+b\sqrt[3]{2}$, i.e., there is a non-zero term in $\sqrt[3]{4}$. For example, $$(-1+\sqrt[3]{2})^{-1} = 1+\sqrt[3]{2}+\sqrt[3]{4},$$ so this does not yield a point in $x^3-2y^3=1$.