<s>[INST] how to parameterize diophant triples on the elliptic curve? How to parameterize Diophantine triples on the elliptic curve?
Example: In the article of the Andrej Dujella , 
The parameterized state of the diophantine triples $(a, b, c)$ on the 
$$E : y^2 = x^3 + (3t^4 − 21t^2 + 3)x^2+(3t^8 +12t^6 +18t^4 +12t^2 +3)x+(t^2+1)^6$$
is
$$a = \dfrac{18t(t−1)(t+1)}{(t^2 − 6t + 1)(t^2 + 6t + 1)}\\ 
b=\dfrac{(t−1)(t^2 +6t+1)^2}{6t(t+1)(t^2 −6t+1)}\\ c=\cdots$$
How are these parameters?
 [/INST] The idea in the paper 
A. Dujella, M. Kazalicki, M. Mikic and M. Szikszai, There are infinitely many rational Diophantine sextuples, Int. Math. Res. Not. IMRN 2017 (2) (2017), 490-508.
is to construct rational Diophantine triples $a,b,c$ (i.e. $ab+1$, $ac+1$, $bc+1$) such that 
on the induced elliptic curve $y^2=(x+ab)(x+ac)(x+bc)$ the point with $x$-coordinate equal to $1$ has order $3$. 
The condition can be written in the form 
$$ \sigma_2=(\sigma_1^2 \sigma_3^2-12\sigma_3^2-6\sigma_1 \sigma_3-3)/(4+4\sigma_3^2), $$
where $\sigma_1=a+b+c$, $\sigma_2=ab+ac+bc$, $\sigma_3=abc$. 
Inserting this in the condition that $(ab+1)(ac+1)(bc+1)$ is a perfect square, 
we get $1+\sigma_3^2$ is a square, i.e. $\sigma_3=\frac{t^2-1}{2t}$. 
The polynomial $X^3-\sigma_1X^2+\sigma_2X-\sigma_3$ should have rational roots, 
so its discriminant has to be a perfect square.
From this condition we got the quartic in $\sigma_1$
$$ (t^6-2t^4+t^2)\sigma_1^4+(23t^3-23t^5+t^7-t)\sigma_1^3+(-45t^2-45t^6+126t^4)\sigma_1^2
+(162t^5-162t^3-54t^7+54t)\sigma_1-27-54t^4+108t^2+108t^6-27t^8=w^2, $$
which is equivalent to the elliptic curve written in the question. 
The curve in question has an obvious point $R$ with $x$-coordinate equal to $0$. 
We compute the point $2R$, which has $x$-coordinate $-3/4(t^2-6t+1)(t^2+6t+1)$. 
Transfering it back to the quartic, we get 
$$ \sigma_1 = \frac{t^8+130t^6-390t^4+130t^2+1}{3(t-1)(t+1)(t^2-6t+1)(t^2+6t+1)t}. $$
If we insert obtained values of $\sigma_1,\sigma_2,\sigma_3$ in $X^3-\sigma_1X^2+\sigma_2X-\sigma_3=0$, 
we get three rational solutions for $X$, which are exactly $a,b,c$ written in the question. 
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