(*This is a follow-up to a [previous post](https://mathoverflow.net/questions/233367/generalizing-a-pattern-for-the-diophantine-m-tuples-problem).*) A [*rational Diophantine $m$-tuple*](https://web.math.pmf.unizg.hr/~duje/intro.html) 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**
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One solution is described in the [paper](https://web.math.pmf.unizg.hr/~duje/pdf/infsextuples.pdf) 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** $a,b,c$
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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?
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Extending rational Diophantine triples to sextuples
Similar formulas can be obtained by taking multiples $mR$ for the point $R=[0,(t^2+1)^3]$ on the elliptic curve 
$$ y^2=x^3+(3t^4-21t^2+3)x^2+(3t^8+12t^6+18t^4+12t^2+3)x+(t^2+1)^6. $$
The above formulas (5) and (6) correspond to $m=2$ and $m=3$. 
Other possible formulas appear if the rank of this curve is $\geq 2$. E.g. if $t=(T^2-6T+1)/(-1+T^2)$, then we get 
$$ a=\frac{(3T-1)(T-3)(T+1)}{2(T-1)(T^2-6T+1)}, $$ 
$$ b=\frac{8T(T-1)}{(T+1)(T^2-6T+1)}, $$ 
$$ c= -\frac{T^2-6T+1}{2(T-1)(T+1)}, $$
(note that in this case the point $P=[0,1]$ on $y^2=(ax+1)(bx+1)(cx+1)$ has finite order, so this triple does not extends to a sextuple by the method from [mentioned paper](https://web.math.pmf.unizg.hr/~duje/pdf/infsextuples.pdf)), and 
$$ a=\frac{2T(T^2-5T+2)^2(T+1)^3}{(2T^2-T+1)(T^2-6T+1)(T^2-8T-1)(T-1)^2}, $$
$$ b=\frac{2(T-3)(T^2-6T+1)(2T^2-T+1)^2}{(T^2-5T+2)(T^2-8T-1)(T-1)^2(T+1)^2}, $$ 
$$ c=-\frac{(3T-1)(T^2-8T-1)^2(T-1)^3}{2(2T^2-T+1)(T^2-5T+2)(T^2-6T+1)(T+1)^2}. $$


