The first paragraph in the following link asserts that the equation $x^3+y^3+z^3=2$ has finite many parametric solutions over $\mathbb{Q}$. In other words, 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 $x^3(t)+y^3(t)+z^3(t)=2$.

Question: What might be the exact evidence for such an assertion? Is it possible that $(1+6t^3,1-6t^3,-6t^2)$ is the unique solution to this equation?

(Edit added after dburde's answer.)

P.S.: I know $$1214928^3+3480205^3-3528875^3=2$$

found by D.R.Heath-Brown and


by D.J.Bernstein. But I think this one


is not reported before.


No, the solution $1214928^3 + 3480205^3 − 3528875^3=2$ is not of the form $(6t^3+1,1-6t^3,-6t^2)$.

  • $\begingroup$ How did you find it? $\endgroup$ – vonbrand Apr 5 '13 at 14:57
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    $\begingroup$ This was found by D. R. Heath-Brown (Oxford) in $1993$, I believe... $\endgroup$ – dburde Apr 5 '13 at 15:05
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    $\begingroup$ @dburde: It is my fault that I did not fully articulate my question. What I want to express is: "Is it possible that $(1+6t^3,1−6t^3,−6t^2)$ is the unique parametric solution to this equation? $\endgroup$ – zy_ Apr 6 '13 at 2:44

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