TITLE: If $x ^ 6 = x$, prove that $x ^ 2 = x$, in a ring
QUESTION [7 upvotes]: I found a short and interesting problem:

Given a ring $(R, +, \cdot)$ and knowing that $x ^ 6 = x\ (\forall x\in R)$, prove that $x ^ 2 = x\ (\forall x \in R)$.

While it is short, I cannot figure out how to solve it. If it would be the reverse, then the solution were simple: $(x ^ 2) ^ 3 = x$.
Given this information, can be the problem be solved? If so, which is the simplest way to solve it?

REPLY [2 votes]: Hint $\ j := 2^6\!-2 = 0 = 3^6\!-3=:k \,$ so by Bezout their gcd $\,\color{#c00}2 = (j,k) = nj\!+\!mk = 0$.
Similarly $\ f(x) = x^6\!-x = 0 = f(x\!+\!1)\,$ so over $\,\color{#c00}{\Bbb F_2}\,$ their gcd $= x^2-x = 0.\ \ \ $ QED