1. You can show using trig identities that the set S of points on the unit circle {x^2+y^2=1} at angle t with tan(t/4)=rational, is dense and has the property that all point coordinates are rational and that all interpoint distances are rational. 2. Now in 4-dimensional space (x,y,u,w) consider the radius=3 circle in the xy plane and the radius=4 circle in the uw plane. [This is based on the pythagorean triple 3,4,5, but you can use any other.] We thus get a dense set of points on this circle-pair with all interpoint distances rational and all coordinates rational. 3. Similarly in 6-dimensional space (x,y,u,w,p,q) we can use three circles in the xy, uw, and pq planes with radii 240, 117, 44 [Any integer 3-tuple forming an "Euler brick" can be used, http://en.wikipedia.org/wiki/Euler_brick ]. 4. Can this be extended to 8 dimensions? For that we'd need a "4-dimensional Euler brick": a 4-tuple of integers a,b,c,d with all 6 sums of two-squared, equal to squares, i.e. a^2+b^2=square, a^2+d^2=square, ..., c^2+d^2=square. I do not know whether such a 4brick exists, but none exist with max(a,b,c,d)<=99999.