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Say that a rational number $a$ is good iff there is a rational number $b$ with $ab(6-a-b)=6$, or equivalently iff $a^4 - 12a^3 + 36a^2 - 24a$ is the square of a rational number. Denote by $G$ the set of all good numbers.

It is known that $G$ is infinite ; moreover, elaborate algebraic manipulation shows that $G$ is invariant by the transformation

$$f(a)=\frac{-54a(-6+12a-6a^2+a^3)^2} {−216+1296a^2−2160a^3+1296a^4−108a^5−234a^6+108a^7−18a^8+a^9}$$

(see Aretino's comment on Tito Piezas' answer here)

Also, $G$ clearly contains $1,2,3,8$.If we denote the roots of $a^3 - 12a^2 + 36a - 24$ by $t_1<t_2<t_3$, then clearly $G\subseteq (-\infty,0] \cup [t_1,t_2] \cup [t_3,\infty)$.

Can more be said about the structure of $G$ ? For example, is $G$ bounded ? Does $G$ have an accumulation point ? What can be said about the closure of $G$ ?

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Ewan Delanoy
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    well if i followed everything correctly, the curve has positive rank and so its rational points are either dense in both real components, or dense in the component of the identity, so after quotienting / looking at the $x$ component, your set will be either dense in $[t_3,0]$ or dense in $[t_3,0] \cup [t_1,t_2]$ – mercio Aug 09 '15 at 17:59
  • @mercio I'm not very familiar with elliptic curves and I must be missing something, but how does a positive rank garantee density ? If I understood correctly, here we have an elliptic curve of rank 1, so the group has a subgroup isomorphic to $\mathbb Z$. But ${\mathbb Z}\times \lbrace 0 \rbrace$ is isomorphic to $\mathbb Z$ and yet is not dense anywhere – Ewan Delanoy Aug 10 '15 at 07:53
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    @EwanDelanoy: You may be interested to know that the irreducible nonic denominator has a solvable Galois group. I have no idea why. Pls see this post. – Tito Piezas III Aug 10 '15 at 09:43
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    "Isomorphic to Z" as an abstract group, but you're putting it in a compact topological group that's the union of two circles. Within each circle it's like the multiples of $\sqrt 2 \mod 1$, which is dense (the closure is a closed subgroup, and any proper closed subgroups of the circle is finite). – Noam D. Elkies Aug 13 '15 at 16:52
  • @NoamD.Elkies thanks for your clear explanation. – Ewan Delanoy Aug 14 '15 at 07:42

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