coloring a tetrahedron (abstract algebra) Consider a regular tetrahedron. Each of its faces can either be painted blue or red. Up to rotation how many ways can the tetrahedron be painted?
I am thinking that applying Burnside's counting formula will work, but it has been so long since I have used Burnside's counting formula that I do not quite remember how to use it.

For the general case, we can give an elementary combinatorial solution.
General Case: We have $n$ different colors. The vertices of a regular tetrahedron will be painted using these colors. The colorings obtained from one another as a result of rotating the tetrahedron are considered identical. The number of different colorings is $$ \dfrac{n^4 + 11n^2}{12} .$$
Proof: If we use exactly $4$ colors, there are $2$ different colorings. Because, we apply one of the colors (only $1$ way) to the base. With the remaining $3$ colors, we can paint the faces with $(3-1)!=2$ ways that using circular permutation. The selection number of $4$ out of $n$ colors is $\dbinom{n}{4}$. With the multiplication principle, we get $2\cdot \dbinom{n}{4}$ colorings. Similarly,  if we use exactly $3$ colors, there are $\dbinom{3}{1}\cdot \dbinom{n}{3}$ colorings. If we use exactly $2$ colors, there are $3\cdot \dbinom{n}{2}$ colorings. If we use $1$ color, there are $n$ colorings. In total, we find the number of different colorings is
$$ 2\cdot \dbinom{n}{4} + 3\cdot \dbinom{n}{3} + 3\cdot \dbinom{n}{2} + n = \color{red}{\dfrac{n^4 + 11n^2}{12}} .$$
Especially for $n=2$, we get $\dfrac{2^4 + 11\cdot 2^2}{12} = \dfrac{60}{12} = 5$.
Note: Moreover we can see that, a part of our proof is that it solves the problem. If we use exactly $2$ colors, there are $3\cdot \dbinom{2}{2}$ colorings. If we use $1$ color, there are $2$ colorings. In total, we get $3 + 2 = 5$ different colorings.
