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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?

Gerry Myerson
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fretty
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  • Has anyone managed to exploit the recurrences yet? – fretty Dec 30 '13 at 13:26
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    $a^3-2b^3=1$ 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. – Gerry Myerson Dec 31 '13 at 17:42
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    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 $0$. – Yong Hao Ng Dec 31 '13 at 18:31
  • 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. – fretty Jan 01 '14 at 08:55
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    Maybe http://math.stackexchange.com/questions/61014/solve-x3-1-2y3 – Gerry Myerson Jan 02 '14 at 03:48
  • I think it was elsewhere, I definitely recall an answer mentioning this fundamental unit explicitly and then suggesting this strategy. Anyway, I was doubtful I would get any further in this vein! – fretty Jan 02 '14 at 10:36
  • Update: Not that I need the answer now but I stumbled upon a simple proof of this fact by considering the expression mod $1+\sqrt[3]{2}$ and finding binomial relationships between the coefficients of $\sqrt[3]{2}^2$. – fretty Jul 04 '16 at 07:24

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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.