How many ways are there to color the five vertices of a regular 17-gon either red or blue, such that no two adjacent vertices of the polygon have the same color?
The answer is zero! Call the polygon $A_{1} A_{2} \ldots A_{17}$. Suppose for contradiction such a coloring did exist.

If we color $A_{1}$ red, then $A_{2}$ must be blue. From here we find $A_{3}$ must be red, then $A_{4}$ must be blue; thus $A_{5}$ must be red, $A_{6}$ must be blue, Proceeding in this manner, we eventually find that $A_{15}$ is red, $A_{16}$ is blue, and then $A_{17}$ is red. But $A_{1}$ and $A_{17}$ are adjacent and both red, impossible.

The exact same argument holds if we started by coloring $A_{1}$ blue. Therefore, there are no colorings at all with the desired property.