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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.

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    $\begingroup$ While I believe this assertion is morally true, one has to phrase it up to some sort of equivalence: if $(x(t),y(t),z(t))$ is one such triple, then so is $(x(t+r),y(t+r),z(t+r))$ for any rational number $r$, for example. $\endgroup$
    – Greg Martin
    Commented Apr 1, 2013 at 19:10
  • $\begingroup$ In Manin's book on cubic forms there is, I think, a discussion on the twisted Fermat cube $x^3+y^3+z^3+2w^3=0$, and the parametric solutions (up to changes in the variable t, and permutations). $\endgroup$
    – Dietrich Burde
    Commented Apr 2, 2013 at 20:14
  • $\begingroup$ @D Burde:It is quite easy to find a family of genus 0 curves on the surface $x^3+y^3+z^3=2$. Finding out an intersection of the surface and a tangent plane touching the surface at a rational point could be a easy way to construct a genus 0 curve on the surface. But the curve mentioned in the question is a special genus 0 curve, I think. $\endgroup$
    – Y. Zhao
    Commented Apr 3, 2013 at 11:11
  • $\begingroup$ $x^3+y^3=z^3+w^3$ has indeed infinite many polynomial solutions(as Euler has stated), but we do not know whether $w\vert(x,y,z)$ from such a statement. $\endgroup$
    – Y. Zhao
    Commented Apr 3, 2013 at 13:25
  • $\begingroup$ Crossposting math.stackexchange.com/questions/346332 $\endgroup$
    – Martin Brandenburg
    Commented Apr 5, 2013 at 10:32

2 Answers 2

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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.

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  • $\begingroup$ I have non-constant rational parametrization to $x^3+y^3+z^3=3$. Must x,y,z be positive (not sure my parametrization is)? Is it worth posting the parametrization on MO? $\endgroup$
    – joro
    Commented Aug 7, 2013 at 11:10
  • $\begingroup$ The question asks for integer solutions, and thus necessarily allows for negative variables (else there's nothing to do past $(1,1,1)$). Hence the relevant parametrizations here are by polynomials. Rational (but not polynomial) curves are easy to find: just intersect the surface with the tangent plane at a rational point; e.g. if $(x-3)(x+5)$ is a square then there's a solution $(x,y,3-x-y)$ from the tangent plane at $(1,1,1)$. A complete rational parametrization would be of interest, but might still be off-topic for this question. $\endgroup$
    – Noam D. Elkies
    Commented Aug 7, 2013 at 14:32
  • $\begingroup$ I meant for $d=3$. What do you mean by "there's no nonconstant solution in Q(t) to x^3+y^3+z^3=d unless d is a cube or twice a cube"? Isn't rational parametrization for $d=3$ in Q(t)? $\endgroup$
    – joro
    Commented Aug 7, 2013 at 15:00
  • $\begingroup$ Sorry, that was my typo: I meant ${\bf Q}[t]$, not ${\bf Q}(t)$ ! Let me fix this... $\endgroup$
    – Noam D. Elkies
    Commented Aug 7, 2013 at 15:49
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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.

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