(*This is a follow-up to a previous post.*) A *rational Diophantine $m$-tuple* is a set of rationals {$a_1,a_2,\dots a_m$} such that (with $i\neq j$), all $a_i a_j+1$ is a square. **Problem: Find a class of triples that can be extended to sextuples.**

I. Elliptic curve

One solution is described in the paper by Dujella et al. Find rational $u,v$ such that,

$$(-27 - 9 u^2 - 27 u v + u^3 v - 54 v^2) (1 + v^2) (-1 + u v + 2 v^2)=y^2\tag1$$

This is a quartic in $u$ to be made a square. Since it has a rational point, it is birationally equivalent to an *elliptic curve*. Define,

$$\alpha_2 = \frac{-3-6uv-12v^2+u^2v^2}{4+4v^2}\tag2$$

Let $a,b,c$ be the roots of,

$$z^3-uz^2+\alpha_2 z-v =0\tag3$$

Then $a,b,c$ are special rational triples such that $ab+1,\;ac+1,\;bc+1$, and,

$$\begin{aligned} &a^2b^2c^2+1 = p^2\\ &2(a^2+b^2+c^2)-(a+b+c)^2-3 = q^2\\ &2(a^2+b^2+c^2+d^2)-(a+b+c+d)^2-3-6abcbd+(abcd)^2=(pd+q)^2 \end{aligned}\tag4$$

are ** all** rational squares (for any $d$ for the last), implying they obey,

$$\big(2(a^2+b^2+c^2)-(a+b+c)^2-3\big)(1+a^2b^2c^2)=(a + b + c + 3 a b c)^2$$

A ** second** elliptic curve can then yield infinitely many three rational $x_i$ such that $a,b,c,x_1,x_2,x_3$ is a sextuple, but we need not go to that step.

II. Solutions$a,b,c$

Using $v = \frac{t^2-1}{2t}$ and $u = \frac{1 + 130 t^2 - 390 t^4 + 130 t^6 + t^8}{-3 t + 105 t^3 - 105 t^5 + 3 t^7}$, Dujella et al found,

$$\begin{aligned} a &= \frac{18t\,(t^2-1)}{(t^2-6t+1)(t^2+6t+1)}\\ b &= \frac{(t - 1)(t^2 + 6t + 1)^2}{6t\,(t + 1)(t^2 - 6t + 1)}\\ c &= \frac{(t + 1)(t^2 - 6t + 1)^2}{6t\,(t - 1)(t^2 + 6t + 1)} \end{aligned}\tag5$$

Using $v = \frac{t^2-1}{2t}$ and $u = \frac{1 + 1996 t^2 + 102 t^4 + 1996 t^6 + t^8}{-8 t + 2584 t^3 - 2584 t^5 + 8 t^7}$, I found,

$$\begin{aligned} a &= \frac{128t\,(t^2+1)^2}{(t^2-1)(t^2-18t+1)(t^2+18t+1)}\\ b &= \frac{(t^2 - 1)(t^2 + 18 t + 1)^2}{16t\,(t^2 + 1)(t^2 - 18t + 1)}\\ c &= \frac{(t^2 - 1)(t^2 - 18 t + 1)^2}{16t\,(t^2 + 1)(t^2 + 18t + 1)} \end{aligned}\tag6$$

Notice that the triples $(5)$ and $(6)$ have an aesthetically pleasant similarity. And since $(1)$ is an elliptic curve, there should be infinitely many families.

Q:Using $(1)$, can you find another triple $a,b,c$ that has polynomial numerator/denominator of small degree?