1
$\begingroup$

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.

  • 0
    @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

0 Answers 0