The following appeared in the problems section of the

March 2015 issue of theAmerican 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?

**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

Solving (1) as a quadratic in b, its discriminant D must be made a square,

D:=a4−12a3+36a2−24a=z2

Using ** any** non-zero solution a0, do the transformation,

a=x+a0

For this curve, let a0=2, and we get,

x4−4x3−12x2+8x+16

Assume it to be a square,

x4−4x3−12x2+8x+16=(px2+qx+r)2

Expand, then collect powers of x to get the form,

p4x4+p3x3+p2x2+p1x+p0=0

where the pi are polynomials in p,q,r. Then solve the system of **three** equations p2=p1=p0=0 using the **three** unknowns p,q,r. One ends up with,

105/64x4+3/8x3=0

Thus, x=−16/35 or,

a=x+a0=−16/35+2=54/35

and you have a new rational point,

a1=54/35=6×32/35

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.

**Attribution***Source : Link , Question Author : Community , Answer Author : Tito Piezas III*