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I'm trying to solve the next problem:

Trying to prove that for every $k$ there is an integer $n=n(k)$ so that for any coloring of the set $\mathbb Z_3^n$ of all $n$-dimensional vectors with coordinates in $\mathbb Z_3$ by $k$ colors, there are three distinct vectors $X$, $Y$, $Z$ having the same color so that $X_i+Y_i+Z_i\equiv 0 \pmod 3$ for all $1 \le i \le n$.

I guess I need to use SCHUR proof in a different way but I don’t know exactly how to determine the coloring function. Any help will be appreciated. Thank you very much!

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    Groovy guy: I've tried to edit your post (by adding LaTeX) for better readability. Please, check, whether I did not unintentionally change meaning of your question.2012-01-07
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    is this a case of hales-jewett? http://en.wikipedia.org/wiki/Hales–Jewett_theorem2012-01-07
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    @yoyo: I think so -- I was wondering why you removed your earlier comment about that. The components add to $0$ iff they're either all different or all the same, so the condition is equivalent to the three vectors being on a line. ([Here's a working link for convenience](http://en.wikipedia.org/wiki/Hales-Jewett_theorem).)2012-01-07
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    More precisely, not all $\{X,Y,Z\}$ as described in the question are "combinatorial lines" as used in the Wikipedia article (a counterexample is $\{12,21,00\}$) but every "combinatorial line" is a valid $\{X,Y,Z\)$, and that is the direction that matters.2012-01-07
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    The Hales-Jewett theorem seems a bit of an overkill for a graph theory (so it seems) question. [Schur's theorem proof](http://www.proofwiki.org/wiki/Schur's_Theorem_(Ramsey_Theory)), looks, indeed, like an easier way, as long as the coloring condition stated right.2012-01-08
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    Thank's for the editing + comments + references I’ll read it all, and try again.2012-01-08
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    @Henning: Thanks for that -- I was suffering from a misconception of what a combinatorial line is.2012-01-10

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