The numbers $\{1,2,...,2005\}$ are divided into $6$ disjoint subsets. Prove that for one of them we can find $x,y,z$ elements in it, not necessarily distinct such that $x + y = z$.
I have no idea how to solve this or where to start. Also I can't find this online anywhere. Does someone know how to solve this?
What you want to prove is that the Schur number $S(6)$ is less than $2005$.
It is a known fact that $S(n)\leq R(n,n)-2$ where $R$ denotes the ramsey number.
Using the inequality $R(n,m)\leq \binom{n+m-2}{n-1}$
We get:
$S(6)\leq R(6,6)-2\leq \binom{10}{5}-2=250$.
So we did a little bit better than aked :)
If you want to find a step by step solution without so much theory, this is problem $857$ of putnam and beyond.