Rational solutions to a+b+c=abc=6a+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?

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(6ab)=6

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

D:=a412a3+36a224a=z2

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

a=x+a0

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

x44x312x2+8x+16

Assume it to be a square,

x44x312x2+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

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