There are 120 differences , so some of them must coincide. But a pair that coincide need not use four distinct integers. It could happen that , with . But suppose that an integer is the middle integer of more than one such triple, say and . Then the four distinct numbers satisfy |a-a'| = |b-b'|.

So we may assume that there are at most 14 such triples, one for each of the 16 available numbers except for the smallest and the largest (which cannot be the middle element of a triple). That still leaves more than 100 differences . So there must be at least one pair of equal differences using four distinct integers.