sum of three cubes and parametric solutions The first paragraph in the following link asserts that the equation $x^3+y^3+z^3=2$ has finite many parametric solutions over $\mathbb{Q}$, i.e., there are finite many polynomial triples $(x(t),y(t),z(t))$ with $x(t),y(t),z(t)\in\mathbb{Q}[t]$ satisfying the equation $x^3+y^3+z^3=2$.
Question: What might be the exact evidence for such an assertion?
Edit: Complementary materials on this problem:
Segre, Beniamino. "A note on arithmetical properties of cubic surfaces." Journal of the London Mathematical Society 1.1 (1943): 24-31.
Bremner, Andrew. "On diagonal cubic surfaces." manuscripta mathematica 62.1 (1988): 21-32.
The first paper states that a genus 0 curve on such diagonal cubic surface must be the complete intersection with another surface. The second one states that a genus 0 curve corresponding to a parametric solution should have some unusual properties at infinity. Although they are quite strong confinements, it is still ambiguous that why such a statement(finite many parametric solutions) tends to be reasonable.
 A: Seeing that the link is to a 1996 announcement that I posted to 
sci.math.research, I suppose I should explain.  Yes, the formulation 
in that announcement is not clearly stated $-$ one doesn't polish 
USENET posts like a published paper, and can't even edit after the fact 
(as is possible on mathoverflow) to correct blatant typos like the stray 
"+1" in the formula "(x,y,z)=(1+6t^3+1,1-6t^3,-6t^2)".
As it happens here there is a
 published paper
that appeared only a few years later:

Elkies, Noam D.: Rational points near curves and small nonzero 
  $|x^3-y^2$| via lattice reduction, 
  Lecture Notes in Computer Science 1838 (proceedings of ANTS-4, 2000; 
  W.Bosma, ed.), 33-63 (arXiv:math.NT/0005139).

but the relevant section (3.2) doesn't address parametrizations of
$x^3+y^3+z^3=2$.  The answer to the present question is that
D.Burde is basically right: the correct statement was, and still is,
that all known solutions of $x^3 + y^3 + z^3 = 2$ in ${\bf Q}[t]$
come from the identity
$$
(1+6t^3)^3 + (1-6t^3)^3 + (-6t^2)^3 = 2
$$
by permuting  $x,y,z$ and substituting some polynomial for $t$.
The substitution need not be linear, but nonlinear substitutions like
$(x,y,z) = (1+6t^6, 1-6t^6, -6t^4)$ give no new $(x,y,z)$ solutions either.
I don't think any method is known that would prove that there are
no other nonconstant solutions, or that there's no nonconstant solution in 
${\bf Q}[t]$ to $x^3+y^3+z^3=d$ unless $d$ is a cube or twice a cube.
All that can be said is that if there were such a solution that had
small enough degree and coefficients then it would have turned up in 
searches for integral solutions such as the searches described in that ANTS-4 paper
and also on this page.
A: I found a proof of the following fact in the article of G. Payne and L. Vaserstein,
"Sums of three cubes", contained in the book "The arithmetic of function fields" (1992):
The set of integral solutions to $x^3+y^3+z^3=1$ cannot be covered by a finite set
of polynomial solutions $x(t),y(t),z(t)\in \mathbb{Q}(t)$. The case $x^3+y^3+z^3=2$
there is the open question $4$. As far as I know (but I may be wrong), 
your question here is still an open problem. Elkies only says, that there are only
finitely many polynomial solutions (known) in this case, but it is not clear to me
whether this is proved. In fact, one only knows one parametric family, coming from the more general equation $x^3+y^3+z^3=2w^3$, which
has the quadruple $(6t^3+s^3,s^3-6t^3,-6st^2,s^3)$ as parametric solution. For $s=1$
one obtains the parametric triple for $x^3+y^3+z^3=2$. I know that this does not 
really answer your question, but perhaps the reference to Vaserstein's articles will be helpful, and other people know more.
