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Can you describe, in parametric form or in any other explicit way, all rational solutions to any of the following equations: $$ y^2 + z^2 = x^3+1, $$ $$ y^2 + z^2 = x^3-1, $$ $$ y^2+x^2y+z^2+1=0. $$ Please consider each equation separately, this is not a system.

For the first two equations, I earlier asked the same question for integer solutions, see How to describe all integer solutions to $x^2+y^2=z^3+1$? , and no way more explicit than try all $x$ and solve the resulting equation in $(y,z)$ has been suggested. However, sometimes rational solutions are easier to parametrize than integer ones.

The last equation has no integer solutions, but has rational ones.

The motivation for this question is that there three equations are the smallest ones (in a sense defined here: What is the smallest unsolved Diophantine equation?) for which the problem of describing all rational solutions is non-trivial.

Of course, the problem is equivalent to describing all coprime integers (or all integers, or all rationals) solutions to the homogeneous equations $$ y^2t + z^2t = x^3+t^3, $$ $$ y^2t + z^2t = x^3-t^3, $$ $$ y^2t+x^2y+z^2t+t^3=0, $$ respectively.

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    $\begingroup$ For what it's worth, Magma says that each of these three surfaces in $\mathbb{P}^{3}$ is birational to $\mathbb{P}^{2}$ over $\overline{\mathbb{Q}}$, but not over $\mathbb{Q}$. $\endgroup$
    – Jeremy Rouse
    Aug 10 at 12:31
  • $\begingroup$ Thank you. Can you please (a) clarify whether this means that it is not possible to find a parametrization of all solutions in the form $x=R_1(u)$, $y=R_2(u)$, $z=R_3(u)$ where $u$ is a vector of rational parameters and $R_i$ are ratios of polynomials in $u$ with rational coefficients. And (b) present your Magma code. In addition to the yes/no answer, can the code actually find the corresponding map for surfaces that are birational to ${\mathbb P}^2$ over ${\mathbb Q}$ ? $\endgroup$
    – Bogdan Grechuk
    Aug 10 at 12:52
  • $\begingroup$ I think your question is a little bit vague. Would you be satisfied if the points given by $(R_{1},R_{2},R_{3})$ didn't hit every rational point, but missed finitely many (or missed points on some union of curves)? Do you want the function $(x,y,z) \mapsto (R_{1},R_{2},R_{3})$ to be one-to-one? If the answer to both of those questions is no, then it might be enough to know whether your cubic surfaces are unirational. Segre proved in 1943 that smooth cubic surfaces are unirational if they have a rational point, but yours are singular. $\endgroup$
    – Jeremy Rouse
    Aug 10 at 16:45
  • $\begingroup$ Because the problem is to explicitly describe all rational solutions, we do not need the function to be one-to-one: some solutions may be covered many times by parametrization. Missing finitely many points is not an issue if we can explicitly list the exceptions. Missing finitely many curves is acceptable if we can list these curves and find rational points on each of them. Because finding rational points on curves is often easier, this would reduce the problem to a finite number of potentially easier problems. $\endgroup$
    – Bogdan Grechuk
    Aug 10 at 17:02

1 Answer 1

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Equivalent eqn given by OP is shown below:

$x^3+t^3=y^2t+z^2t=t(y^2+z^2)$ --(1)

Eqn. (1) has parametric solution given below:

$x=(260k^2-72k+8)$

$t=(65k^2-18k+2)$

$y=(520k^2-130k+2)$

$z=(65k^2-130k+16)$

for, $k=2$ we have:

$(452^3+113^3)=113(911^2+8^2)$

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    $\begingroup$ While this provides an infinite family of rational points, it does not give a parametrization of all points. For example, the point $(x,y,z,t) = (3,-1,5,1)$ is missed by this family. $\endgroup$
    – Jeremy Rouse
    Aug 15 at 15:33

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