If you turn only two adjacent faces of the Rubik's cube, is it possible to reach a state where only three corner pieces are out of place (and all other pieces are in the original places)?
This was a question I thought of many years ago. At that time I was able to prove it by writing a computer program to count for me. But now I am in fact curious to find out what elegant solutions there are.
Firstly, this webpage gives two different proofs that the answer is "no", and that there are exactly $120$ possible permutations of the corner pieces.
Secondly, I give a proof below that my brother and I found. My question is whether there are other ways to prove this that are fundamentally different, and hopefully more elegant than these proofs. We are also curious if there is any deeper reason our proof works at all, because intuitively it has no reason to work since the final parity argument only excludes some 3-cycles.