Well, this must be simple but I seem to be dense at the moment. I checked heuristically for small $k$ $(2k+1)^{4k+1} = (2k+1) \pmod {4k+2} \tag1 $ but couldn't prove that this is so in general.
For the other quarter of numbers, $(2k)^{4k-1} = 0 \pmod {4k} \tag2 $ I found at least the (even more simple) solution.
So: is (1) true? and if, what is the proof?