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?
Best Answer
(Edit at the bottom.) Here is an elementary way (known to Fermat) to find an infinite number of rational points. From $a+b+c = abc = 6$, we need to solve the equation,
$$ab(6-a-b) = 6\tag1$$
Solving $(1)$ as a quadratic in $b$, its discriminant $D$ must be made a square,
$$D := a^4-12a^3+36a^2-24a = z^2$$
Using any non-zero solution $a_0$, do the transformation,
$$a=x+a_0\tag2$$
For this curve, let $a_0=2$, and we get,
$$x^4-4x^3-12x^2+8x+16$$
Assume it to be a square,
$$x^4-4x^3-12x^2+8x+16 = (px^2+qx+r)^2$$
Expand, then collect powers of $x$ to get the form,
$$p_4x^4+p_3x^3+p_2x^2+p_1x+p_0 = 0$$
where the $p_i$ are polynomials in $p,q,r$. Then solve the system of three equations $p_2 = p_1 = p_0 = 0$ using the three unknowns $p,q,r$. One ends up with,
$$105/64x^4+3/8x^3=0$$
Thus, $x =-16/35$ or,
$$a = x+a_0 = -16/35+2 = 54/35$$
and you have a new rational point,
$$a_1 = 54/35 = 6\times 3^{\color{red}2}/35$$
Use this on $(2)$ as $x = y+54/35$ and repeat the procedure. One gets,
$$a_2 = 6\times 4286835^{\color{red}2}/37065988023371$$
Again using this on $(2)$, we eventually have,
$$\small {a_3 = 6\times 11838631447160215184123872719289314446636565357654770746958595}^{\color{red}2} /d\quad$$
where the denominator $d$ is a large integer too tedious to write.
Conclusion: Starting with a "seed" solution, just a few iterations of this procedure has yielded $a_i$ with a similar form $6n^{\color{red}2}/d$ that grow rapidly in "height". Heuristically, it then suggests an infinite sequence of distinct rational $a_i$ that grow in height with each iteration.
$\color{blue}{Edit}$: Courtesy of Aretino's remark below, then another piece of the puzzle was found. We can translate his recursion into an identity. If,
$$a^4-12a^3+36a^2-24a = z^2$$
then subsequent ones are,
$$v^4-12v^3+36v^2-24v = \left(\frac{12\,e\,g\,(e^2+3f^2)}{(e^2-f^2)^2}\right)^2$$
where,
$$\begin{aligned} v &=\frac{-6g^2}{e^2-f^2}\\ \text{and,}\\ e &=\frac{a^3-3a^2+3}{3a}\\ f &=\frac{a^3-6a^2+9a-6}{z}\\ g &=\frac{a^3-6a^2+12a-6}{z} \end{aligned}$$
Starting with $a_0=2$, this leads to $v_1 = 6\times 3^2/35$, then $v_2 = 6\times 4286835^2/37065988023371$, ad infinitum. Thus, this is an elementary demonstration that there an infinite sequence of rational $a_i = v_i$ without appealing to elliptic curves.