And here I thought I was done with this problem...

After thinking perhaps I was finally done with the problem, I fell again to trying to construct counterexamples from the n=5 case. At first I just tried random ideas, none of which worked. The quaternions occured to me (over Z₅ or Z₃), and I tried some variations on them, none of which worked.

Finally I gave up on trying random shit and tried to systematically construct a counterexample using what I knew had to be true (squares commuting, for instance). Instead I stumbled on the last step in proving that it's *true* for n=5, upon which I hit myself on the head. Well, I didn't actually. But in my mind I did.

Very well then:
R a ring (not necessarily with identity) where a^5=a ∀a∈R. Then R is commutative.
( Proof )

[0]Well, I did do a computer search for idempotent polynomials of one variable, but only afterwards, and it didn't turn up anything better.
[3]Again, not actually - see above.

...ignore that last entry.

Characteristic p implies (x+y)^p=x^p+y^p *when you have commutativity*. Upon trying an actual calculation with some noncommutative elements, my counterexample does not satisfy a^5=a. It simply does *not* work. Similarly for my generalization and interesting note.

...now I have no idea again.

FWIW, when I was actually trying to prove a^5=a implies commutativity, I was able to get 15xy=15yx and x²y=yx².

Time to go back to trying to prove it, to find a counterexample, or to just stop thinking about it?

In other news, the lounge clock is broken. There are other clocks but I'm just so used to looking at the main one in the lounge that I have to remember it's not actually always 7:45.

-Sniffnoy