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Timeline for Diophantine question

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Feb 8, 2013 at 2:14 answer added Aaron Meyerowitz timeline score: 3
Feb 7, 2013 at 19:19 answer added Zack Wolske timeline score: 3
Feb 7, 2013 at 16:52 comment added Zack Wolske There are still quite a lot besides those. For example, take $x=2$, and any $z$. Then the equation simplifies to $n_z((y-1)^2 -1)=q^2$, which is Pell's equation if $n_z$ is not a square. Choose $z$ so that $6(z^2-z)$ is not a square (another Pell equation, but this time we want non-solutions) and you'll get infinitely many more solutions. I think the same tricks will work with any $x$ value: complete squares, get a Pell-type equation for $z$ in terms of $x$ and choose a non-solution $n_z$, then write the Pell equation for $y$ in terms of that, and solve.
Feb 7, 2013 at 16:37 comment added Noam D. Elkies Well the $x=1$ family wasn't that obvious... (Though maybe it's obvious in the "standard E8 setup".) The usual heuristics suggest that there should be a sparse but infinite set of "random" solutions (if $\max(x,y,z)\in[H,2H)$ there are about $H^3$ numbers of size $H^6$ so we expect about $H^0$ of them to be squares), and a list of solutions up to say $10^3$ might also turn up new some parametric families.
Feb 7, 2013 at 10:00 comment added Hauke Reddmann @Peter: No, not necessary. @Noam: Obviously (although: zero dimension...not sure if want)...the "besides" is the interesting part. (I hoped for a general applicable method.) Note that negative y or z is fishy - I'm still pondering if negative dimensions on the RHS of a Clebsch-Gordon expansion make any sense at all.
Feb 6, 2013 at 20:56 comment added Noam D. Elkies Numerical search suggests there's lots of solutions: any of the LHS factors can be set to zero, or $x=1$ and $q = \pm 6yz$, and quite a bit besides that.
Feb 6, 2013 at 20:24 history asked Hauke Reddmann CC BY-SA 3.0