Some already asked on this site whether the natural numbers can be partitioned to finite number of arithmetic progresions with distinct differences, with the condition that the intercection between them is the empty set. My question is, removing that assumption - if you let the intercection to not be empty, is it still false?
2026-04-18 01:32:19.1776475939
Partitioning the natural numbers to finite number of arithmetic progressions
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After taking $a_n=2n+1$, $b_n=4n+2$ we are left only with multiples of $4$. Each of the progressions $c_n=3n$, $d_n=6n+4$, $e_n=12n+8$ covers one third of these.