SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Consider the cases: 1. Use 1 color: $\left(\genfrac{}{}{0ex}{}{n}{1}\right)=n$,

2. Use 2 colors: $4\times \left(\genfrac{}{}{0ex}{}{n}{2}\right)=2n(n-1)$,

3. Use 3 colors: $9\times \left(\genfrac{}{}{0ex}{}{n}{3}\right)=3\frac{n(n-1)(n-2)}{2}$,

4. Use 4 colors: $6\times \left(\genfrac{}{}{0ex}{}{n}{4}\right)=\frac{n(n-1)(n-2)(n-3)}{4}$

The sum up will give: $$\frac{n^4+n^2+2n}{4}$$

Remark. We can rotate the coloring with respect to the angles $0,\frac{2\pi }{4},2\frac{2\pi }{4},3\frac{2\pi }{4}$. We get $4\times$ (the number $S$ of the rotationally distinct colorings). The arbitrary rotation of the colorings with only 1 color gives the same coloring, thus we have to add $3\times \left(\genfrac{}{}{0ex}{}{n}{1}\right)$ and similarly $2\times \left(\genfrac{}{}{0ex}{}{n}{2}\right)$. Thus, we get $$4\times S=n^4+3\times \left(\genfrac{}{}{0ex}{}{n}{1}\right)+2\times \left(\genfrac{}{}{0ex}{}{n}{2}\right)=n^4+n^2+2n\Rightarrow S=\frac{n^4+n^2+2n}{4}.$$