The other day, Fred wrote:
No proper integer case has emerged with any edge of length unity; just one with edge 2 has been found in 2 dimensions, and only 12 out of 63 in 3 dimensions.
Suppose a proper set of points P has all pairwise distances integers, and P contains A,B with d(A,B)=1. Then for all C, d(C,A)=d(C,B), since by the proper triangle inequality d(C,A) - d(C,B) < 1. So the n-2 points in P other than {A,B} all live in some n-3-dimensional subspace, and the unit-length line segment AB has that plane as its perpendicular bisector. That makes it easy to see that there will be no 4-point solution. Let A=(0,1/2), B=(0,-1/2), then we would need two more points C, D on the x-axis, say at respective distances c and d from A. The x-coord of C is then +-sqrt( c^2 - 1/4 ) = +-sqrt(4c^2-1)/2, and likewise D is at +-sqrt(4d^2-1)/2. That makes it clear that the x-coords of C and D are both "irrational but sqrt(rational)", and the sum or difference of two such numbers is always irrational (just look at (sqrt(n)+m)^2 to see why). I have a niggling feeling that a decent number theorist might immediately see how to generalize that argument to higher dimensions, but perhaps I'm being (sorry) irrational. Even if that doesn't lead to a proof of impossibility, it certainly means it's not surprising that a search for general integer-separation point sets would have a hard time finding ones with a unit distance: almost none of the sets you consider will meet the distance pairing constraint. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.