Here is the problem:

For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $\dfrac{\binom{n}{2}}{3}$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle?

Trivially $n\equiv 0 \text{ or } 1 \pmod{3}$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks

  • $\begingroup$ Even assuming "polyhpn" in the title is a typo for polygon, it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. $\endgroup$ – hardmath Aug 6 '18 at 21:20
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    $\begingroup$ $n$ needs to be odd; focusing on a vertex, $v$, the $n-1$ edges out of $v$ are divided into same-colored pairs. $\endgroup$ – Mike Earnest Aug 6 '18 at 21:58
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    $\begingroup$ See site.uottawa.ca/~lucia/courses/7160-17/slides/… for a construction $\endgroup$ – Mike Earnest Aug 6 '18 at 22:16

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