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I am interested in finding every rational solution of $x (y - z) y (z - x) z (x - y) = t^2$ (expressed in homogenous form, to show its symmetry).

Among other approaches I am pursuing, it is clear that letting $t = x y z u$ (or a similar transform that flips all of x, y, z to the other side, each to the power 1, such as $t = x (y - z) u$) results in an equation which is quadratic in each of, x, y, z. So, holding two of these fixed, say x, y, and with u also fixed, one can in general leapfrog from one solution (x, y, z) to another (x, y, z') where z, z' are the two roots of the quadratic in z.

But naturally the next question is how many disjoint "3D lattices" of solutions formed in this way are needed to include every rational solution. Since u is fixed for every solution in any given lattice, I would imagine there must be an infinite number of these lattices, but that one can hop between them via the transformations that flip x, y, z to the right hand side and back again, if that makes sense.

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    $\begingroup$ The leapfrogging technique you describe is known as "Vieta jumping" (en.wikipedia.org/wiki/Vieta_jumping). $\endgroup$ Jul 28, 2012 at 23:03
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    $\begingroup$ This is a K3 surface, so it cannot be rationally parametrized; and the technique you describe (together with an initial solution such as $(x,y,z,t)=(2,-3,5,60)$) must yield a set of points that's Zariski-dense, which on a surface means it's not contained in any union of curves. So, no finite description exists along the lines you seem to want. $\endgroup$ Jul 29, 2012 at 0:44
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    $\begingroup$ ...and I see it's a rather special K3, the "singular K3 surface" [Picard number $20$, which is maximal in characteristic zero] and discriminant $-4$, obtained as the twist of the Legendre curve $Y^2 = X(X-1)(X-\lambda)$ by $\sqrt{\lambda^2-\lambda}$ (take $(X,\lambda) = (y,x)$). $\endgroup$ Jul 29, 2012 at 2:59
  • $\begingroup$ Many thanks for the interesting and informative comments $\endgroup$ Jul 29, 2012 at 9:02

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