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Solving $a+b+c = abc = 6$ in the rationals entails solving,

$$-24a+36a^2-12a^3+a^4=z^2\tag1$$

which is birationally equivalent to an elliptic curve. It can be shown that if $a$ is a solution, then so is the transformation,

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

The irreducible nonic has a solvable Galois group with order $54$. (A related discussion is in this MSE post.) More generally, if $a+b+c = abc = n$ so,

$$-4na+n^2a^2-2na^3+a^4=z^2\tag2$$

Define $u=\large{\frac{-a^3+na^2+n}{a}}$. If $a$ is a solution to $(2)$, then so is,

$$f(a) = \frac{n(n^2-3u)^2}{u^3-9n^2u+n^2(n^2+27)}$$

Expanding the denominator, one gets a nonic in $a$ with a solvable Galois group. (The particular example was just the case $n=6$.) Even more generally, given any elliptic curve,

$$F(x) = y^2\tag3$$

and the transformation,

$$f(x) = \frac{g(x)}{h(x)}$$

Question 1: If both $x$ and $f(x)$ are rational solutions to $(3)$, does that imply $g(x)$ and $h(x)$ have solvable Galois groups?

In response to a comment:

Question 2: Define $(3)$ as an elliptic curve with an infinite number of distinct rational points. Assume the set $S$ of transformations $f_i(x) = \frac{g_i(x)}{h_i(x)}$ such that both $x$ and $f_i(x)$ are rational solutions to $(3)$. In this set $S$, is there always at least one non-trivial transformation such that its $g_i(x)$ and $h_i(x)$ have solvable Galois groups?

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  • $\begingroup$ Not in general. Your $f(x)$ is the $x$-coordinate of an endomorphism of the elliptic curve and $h$, for instance, has roots giving the $x$-coordinates of the kernel of the endomorphism and the Galois group is a subgroup of the automorphism group of said kernel. Sometimes this is solvable but if the endomorphism is multiplication by $m$ for some large $m$, it won't be. $\endgroup$
    – Felipe Voloch
    Aug 12, 2015 at 12:14
  • $\begingroup$ @FelipeVoloch: Thanks. Would you have at hand an example of an explicit elliptic curve and a transformation with $h(x)$ having a non-solvable group? (P.S. I've also added a question 2. Maybe the answer to the second is a "yes".) $\endgroup$
    – Tito Piezas III
    Aug 12, 2015 at 12:58
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    $\begingroup$ Take a random elliptic curve and use multiplication by 5. $\endgroup$
    – Noam D. Elkies
    Aug 12, 2015 at 13:20
  • $\begingroup$ Sorry, I meant multiplication by 5 in the group law, as I see that Felipe Voloch already suggested. (And you must have meant g and h to have a non-solvable group.) $\endgroup$
    – Noam D. Elkies
    Aug 12, 2015 at 14:37
  • $\begingroup$ @NoamD.Elkies: Oh, I think I understand. I believe I just used $m=3$ in the group law in my post, but using $m=5$ will be a different matter. So I guess the answer to my second question is indeed "yes", as long as I stick to $m\leq4$. $\endgroup$
    – Tito Piezas III
    Aug 12, 2015 at 14:49

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