If a set N₉ = {1, 2, 3, 4, 5, 6, 7, 8, 9} of 9 numbers is split into two subsets, then at least one of them contains three terms in arithmetic progression.
The statement is not true for a set N₈ of 8 integers.

Seems obvious?

Prove it.

The N₉ partition must have a subset with at least 5 numbers.  If I didn't make any mistakes, there are four possibilities where this subset has no APs.  In each case the other subset has an AP in its 4 numbers instead.

Edited on March 25, 2017, 9:30 am

Posted by Jer on 2017-03-24 12:39:01