Using Schur Theorem I have to prove that for any $n\geq 1$ there is a number $f(n)$ such that if
$A=${$2,3,4,...f(n)$} is partitioned into n sets $A_1$, $A_2$...$A_n$ there is a $j\in${1,2,...,n} and $x,y,z\in A_j$ such that $z=xy.$
I have tried to find a certain coloring that would give this result and take f(n)-1=R(3,3,..3) (with n times 3) but I do not manage to use all the colors (I am not sure I need to use all the colors). Any idea would be helpful. Thank you
Schur's theorem tells us that there is an $s(n)$ such that if $\{1,2,\dots,s(n)\}$ is partitioned into $n$ pieces, in one of the pieces we find $a,b,c$ with $a+b=c$. Take $f(n)=2^{s(n)}$, and note that if $A=\{2,3,\dots,f(n)\}$ is partitioned into $n$ pieces, then in particular $B=\{2^1,2^2,\dots,2^{s(n)}\}$ is partitioned into $n$ pieces. This induces a partition of $C=\{1,2,\dots,s(n)\}$ into $n$ pieces: Put $k$ in the $j$th piece of $C$ iff $2^k$ is in the $j$th piece of $B$. By definition of $s(n)$, there are $a,b,c$ in the same piece of $C$ with $a+b=c$. But then $2^a,2^b,2^c$ are in the same piece of $B$, and therefore in the same piece of $A$, and also $2^a2^b=2^{a+b}=2^c$.
(Of course, $2^{s(n)}$ is only an upper bound, and a more careful analysis would give better estimates for the smallest possible value of $f(n)$.)