If there is an integer such that and then there are 3 idempotents, namely 0, 1 and Otherwise 0 and 1 are the only idempotents.
Is that right?
Okay, this is actually very simple. I can’t believe I’ve taken such a long time to twig it.
It is clear that is an idempotent of if and only if divides Let where are the distinct prime factors of If is an idempotent then each prime must divide or but not both – otherwise it would divide So whichever factor it divides, the whole lot will divide that factor. And since each either divides or (2 choices), it emerges that the number of idempotents is precisely
Hence the number of idempotents of where is where is the number of distinct prime factors of
This is one of the most wonderful problems I’ve ever tried – thank you ever so much, NonCommAlg. It’s a shame I can only thank you once – I would thank you more times if I could.