It's pretty easy to generate all the rational points on the given ellipse (a 1-parameter formula), and not too much harder to generate the points on the elliptic curve, using "chord-and-tangent". We might generate these points, and see what fraction have the rational distance property. Of course this only gets a tiny fraction of the full set of solutions. Rich ---------- Quoting James Propp <jamespropp@gmail.com>:
Dan,
You probably already figured this out, but the point (1,1) is treated differently from the other three vertices of the square; this breaks the symmetry (and explains why only bilateral symmetry is observed).
Jim
On Wednesday, May 18, 2016, Dan Asimov <dasimov@earthlink.net> wrote:
This is indeed very interesting!
But
a) I don't understand why, with the problem having the symmetry of the square, the curves seem to have only (x,y) <?> (y,x) symmetry
and
b) that doesn't seem to be the right equation of the cubic curve that I see, both because it lacks (x,y) <?> (y,x) symmetry, and its shape doesn't seem right.
?Dan
On May 18, 2016, at 12:35 PM, Ed Pegg Jr <ed@mathpuzzle.com <javascript:;>> wrote:
http://i.imgur.com/YclyNMC.gif
Take a unit square. Plot a few thousand points (a,b) that are a rational distance from three vertices (all but (1,1)) in black. Plot the inverse point (a,b)/(a^2+b^2) in red.
Can anyone see the ellipse 3 (x^2+ y^2)-4(x+y)+2 x y=0 ?
Or the cubic curve 4 (y^3+ x^3)-3 (y^2+ x^2)+2x y ( 2y+ 2x -1)=0 ?
--Ed Pegg Jr
Points on the ellipse so far:
. . .
. . .
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