If $x ^ 6 = x$, prove that $x ^ 2 = x$, in a ring 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?

See Theorem $2.6$ in the lecture notes here. If $x^6=x$ for all $x$, then it is easy to show that $x^3+x^5=0$ for all $x$. Also $-x=(-x)^6=x^6=x$, so that $x=x^6=x\cdot x^5=x\cdot x^3=x^4$. A similar step shows that $x^2=x$ and that $R$ is commutative.
