# partitioning the integers without arithmetic progressions

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? – Old John Oct 11 '12 at 17:53
Are you disallowing uni-directional arithmetic series, or just bi-directional arithmetic series? – Thomas Andrews Oct 11 '12 at 17:54
Disallowing uni-directional arithmetic series. Why did I miss something as simple as $...-5,-4,1,4,5,6,7...$? – Angela Richardson Oct 11 '12 at 17:58
If you are only forbidding bi-directional arithmetic series, then you can just choose $A_1=\{0,2,4,...,-1,-3,-5,...\}$ and $A_2=\{-2,-4,-6,...,1,3,5,...\}$ – Thomas Andrews Oct 11 '12 at 17:59
@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,...} – Hurkyl Oct 11 '12 at 19:48