In trying to solve $a^3 - 2b^3 = 1$ over the integers I came across the need to answer the question: when does $(1+ \sqrt[3]{2} + \sqrt[3]{2}^2)^n$ have no $\sqrt[3]{2}^2$ term in it's expansion (in terms of the basis $\{1,\sqrt[3]{2}, \sqrt[3]{2}^2\}$ of $\mathbb{Z}[\sqrt[3]{2}]$).
Clearly this cannot happen for positive $n$ (create a set of recursions that generate the next coeffs from the previous ones and you can observe all three coeffs are always positive). Of course for $n=0$ it does happen and you get the trivial solution $(a,b)=(1,0)$.
I am struggling to sort out the negative $n$ case. This is the same as studying positive integer powers of $(\sqrt[3]{2} - 1)$.
I guess that for positive $m$ you only ever get a zero coefficient of $\sqrt[3]{2}^2$ in $(\sqrt[3]{2} - 1)^m$ whenever $m=1$ (giving another solution $(a,b)=(-1,-1)$). However I am struggling to prove this using the recursions alone.
Have I missed something easy?
This is a classical problem first solved by Nagell (Solution complète de quelques équations cubiques à deux indéterminées, J. Math. Pures Appl. 4 (1925), 209–270); for an English version see LeVesque's Topics in Number Theory, vol. II, and I have given a brief account in German here (see the appendix).
The question whether elements in a pure cubic field can have squares in which a coefficient with respect to the basis $\{1, \sqrt[3]{m}, \sqrt[3]{m}^2\}$ is $0$ is related to elliptic curves; see this article.