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MJD
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Coloring of $K_17$$K_{17}$

For any 3-coloring of $K_17$$K_{17}$ I have to show there exists either a red, blue or green triangle. To start, can I use proof by contradiction with color red, blue, green? So $(0,0,136)$ means all 136 edges are green. Clearly this has green triangle. If we color the "outside" of the graph all one color, say red, then we have 17 edges that are red and no edge of the interior can be colored red in order to avoid red triangles. Then I try to obtain a contradiction?

Coloring of $K_17$

For any 3-coloring of $K_17$ I have to show there exists either a red, blue or green triangle. To start, can I use proof by contradiction with color red, blue, green? So $(0,0,136)$ means all 136 edges are green. Clearly this has green triangle. If we color the "outside" of the graph all one color, say red, then we have 17 edges that are red and no edge of the interior can be colored red in order to avoid red triangles. Then I try to obtain a contradiction?

Coloring of $K_{17}$

For any 3-coloring of $K_{17}$ I have to show there exists either a red, blue or green triangle. To start, can I use proof by contradiction with color red, blue, green? So $(0,0,136)$ means all 136 edges are green. Clearly this has green triangle. If we color the "outside" of the graph all one color, say red, then we have 17 edges that are red and no edge of the interior can be colored red in order to avoid red triangles. Then I try to obtain a contradiction?

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Coloring of $K_17$

For any 3-coloring of $K_17$ I have to show there exists either a red, blue or green triangle. To start, can I use proof by contradiction with color red, blue, green? So $(0,0,136)$ means all 136 edges are green. Clearly this has green triangle. If we color the "outside" of the graph all one color, say red, then we have 17 edges that are red and no edge of the interior can be colored red in order to avoid red triangles. Then I try to obtain a contradiction?