Checkboard coloring problem If there are $q$ colors available and $n$ is odd, prove that there are $\frac14(q^{n^2}+2q^{(n^2+3)/4}+q^{(n^2+1)/2})$ distinct colored $n\times n$-chessboards. (Adjacent fields need not be differently colored.) I know that we can use Burnside's lemma, but I want to know if there is other way to prove it. Thank you!

There are $q^{(n-1)^2/4 + (n+1)/2}$ different chessboards that have fourfold rotational symmetry. There are $q^{(n+1)/2 + n(n-1)/2}$ with twofold rotational symmetry, but each that is not fourfold symmetric is counted twice here, so that must be corrected.

There are $q^{n^2}$ colorations of the chess board, but the non-rotationally symmetric ones are counted four times each, and the strictly twofold ones are counted twice.

With exclusion-exclusion, you should get to your answer.