Let be a ring, which may or may not have We proved in here that if for all then is commutative. A similar approach shows that if for all then is commutative.

**Problem**. Prove that if for all then is commutative.

**Solution**. Clearly is reduced, i.e. has no nonzero nilpotent element. Note that for all because Hence is an idempotent for every because

Thus is central for all by Remark 3 in this post. Therefore is central for all But

and hence is central. Thus which gives

But What about the following questin?

If R is a ring such that a^5=a for every a in R. Is it commutative?

Also What about the following question?

If R is a ring such that a^n=a for every a in R where n is a natural number. Is it commutative?

See this post: https://ysharifi.wordpress.com/2018/03/14/rings-satisfying-xn-x-are-commutative/

The same for rings where x^2=x for every x in the ring.

Yes, but, as I’m sure you know, the case is trivial.