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Oct 24, 2016 at 7:18 vote accept CommunityBot
Oct 24, 2016 at 6:01 comment added Robert Z @user.3710634 Yes, excluding $(0,0,\sqrt{3})$ and adding points $(0,0,\frac{2m}{m^2-1})$ is fine.
Oct 23, 2016 at 21:08 comment added user370634 Fine, but let to generalize this solution somehow. By adding the points $(0,0,\frac {2m}{m^2-1})$, $m≠±1$ where $m$ is an integer, to our set and excluding $(0,0,\sqrt{3})$, does we get a more general solution? If yes, then Is this the most general form of the solution? I ask this question based on the comments of 'zyx' on this page: math.stackexchange.com/questions/1978452/…
Oct 22, 2016 at 10:18 history edited Robert Z CC BY-SA 3.0
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Oct 22, 2016 at 9:36 history edited Robert Z CC BY-SA 3.0
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Oct 22, 2016 at 9:25 history edited Robert Z CC BY-SA 3.0
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Oct 22, 2016 at 9:09 history edited Robert Z CC BY-SA 3.0
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Oct 22, 2016 at 8:45 history answered Robert Z CC BY-SA 3.0