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What is the least $n$ such that the set of all integers can be partitioned into $n$ disjoint subsets, none of which contain any infinite arithmetic progressions (arbitrarily long but finite arithmetic progressions are allowed)? In particular, can it be done for $n=2$? I tried e.g. splitting the integers into perfect powers and non-powers but then $4n+2$ is never a perfect power.

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    How about creating 2 subsets A and B as follows: put 1 into A, then 2,3 into B, then 4,5,6,7 into A etc. Where the number going into each set doubles at each step. Could A or B contain an infinite arithmetic progression?2012-10-11
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    Are you disallowing uni-directional arithmetic series, or just bi-directional arithmetic series?2012-10-11
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    Disallowing uni-directional arithmetic series. Why did I miss something as simple as $...-5,-4,1,4,5,6,7...$?2012-10-11
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    If you are only forbidding bi-directional arithmetic series, then you can just choose $A_1=\{0,2,4,...,-1,-3,-5,...\}$ and $A_2=\{-2,-4,-6,...,1,3,5,...\}$2012-10-11
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    @Thomas: It's useful to use a semicolon (;) or a pipe (|) as a separator, to indicate that "$\ldots$" isn't supposed to "connect" to the numbers on one side. e.g. {0,2,4,...|-1,-3,-5,...}2012-10-11

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