Rational solutions to $a+b+c=abc=6$ The following appeared in the problems section of the 
March 2015 issue of the American Mathematical Monthly.

Show that there are infinitely many rational triples 
  $(a, b, c)$ such that $a + b + c = abc = 6$.

For example, here are two solutions $(1,2,3)$
and $(25/21,54/35,49/15)$. 
The deadline for submitting solutions was July 31 2015,
so it is now safe to ask: is there a simple solution?
One that doesn't involve elliptic curves, for instance? 
 A: A rational triple $(a,b,c)$ satisfies $a+b+c=abc=6$ if and only if
$$c=6-a-b\qquad\text{ and }\qquad ab(6-a-b)=6,$$
where the latter is equivalent to
$$a\cdot b^2+a(a-6)\cdot b+6=0,$$
which shows that $b=\tfrac{6-a}{2}\pm\tfrac{1}{2a}\sqrt{a^2(a-6)^2-24a}$. Because $a$ and $b$ are rational, the expression
$$a^2(a-6)^2-24a=a^4-12a^3+36a^2-24a,$$
must be a rational square. So it suffices to show that the curve
$$x^4-12x^3+36x^2-24x=y^2,$$
has infinitely many rational points. I don't see an elementary way to do so. It may be worth noting that it can be seen to have the rational points $(0,0)$ and $(2,2)$.
A blunt way to finish is to note that this curve is birational to the elliptic curve
$$y^2=x^3-9x+9,$$
of positive rank, so indeed there are infinitely many rational solutions.
A: We just need to prove that for infinite values of $q\in\mathbb{Q}$ the polynomial
$$ p(x)=x^3-6x^2+qx-6 $$
completely splits over $\mathbb{Q}$. That is the same as requiring that
$$ p(x+2) = x^3+(q-12)x+(2q-22) $$
completely splits over $\mathbb{Q}$. Assuming that $u,w,w$ are the roots of the above polynomial, then $u+v+w=0$, $uv+uw+vw=(q-12)$, $uvw=22-2q$, so we just need to show that there is an aperiodic rational map $\phi:(u,v)\to(\tilde{u},\tilde{v})$ that preserves:
$$-2=uvw+2(uv+uw+vw) = -2u^2-2uv-2v^2-u^2 v- v^2 u$$
but honestly I do not know how to find it without invoking the group structure for an elliptic curve.
