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$\begingroup$ Boris, thanks for your comments! I like your proof of the k=6 case. It seems, however, that the k=6 argument can't (at least naively) be extended to the case of k=8. To see this, find the 4-AP with a maximal largest element and (of these) a minimal smallest element. Now pass to the sub-progression containing this 4-AP (and relabel the progression as {0,1,2,3}). The intersection of our set with this progression could look like {-3,0,1,2,3,7}, which doesn't contain a 8-half-AP. Establishing the k=8 case would be very interesting. $\endgroup$

Mark Lewko
– Mark Lewko

2009-11-06 22:52:39 +00:00
Commented Nov 6, 2009 at 22:52

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