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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?

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  • $\begingroup$ Has anyone managed to exploit the recurrences yet? $\endgroup$ – fretty Dec 30 '13 at 13:26
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    $\begingroup$ $a^3-2b^3=1$ is an example of what's called a Thue equation. I'm not sure it can be solved along the elementary lines you are pursuing --- it may need something a bit more advanced. Anyway, I have given you a search term. $\endgroup$ – Gerry Myerson Dec 31 '13 at 17:42
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    $\begingroup$ By the way there are no other integer solutions: $a^3-2b^3=1$ is equivalent to an elliptic curve $E$. $E$ has rank $0$ and the torsion is $\cong \Bbb Z_2$, so you get only 1 point out of it. The torsion point corresponds to $(-1,-1)$. The infinity point $\mathcal O$ corresponds to the case when $a=1$, but in that case $b=0$, so you get $(1,0)$ and there are no other points. So if a solution corresponds to no $\sqrt 8$ in expansion then it must be true that $\sqrt 8$ is never $0$. $\endgroup$ – Yong Hao Ng Dec 31 '13 at 18:31
  • $\begingroup$ Thanks. I know how to solve the Diophantine using these methods but was really searching for a non-trivial use of Dirichlet's unit theorem (for a possible talk). Now I must confess this strategy was mentioned in an answer to a question on this site somewhere (and so I assumed it was doable this way) but cannot for the life of me find the question. $\endgroup$ – fretty Jan 1 '14 at 8:55
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    $\begingroup$ Maybe math.stackexchange.com/questions/61014/solve-x3-1-2y3 $\endgroup$ – Gerry Myerson Jan 2 '14 at 3:48
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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.

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