Ring such that $x^4=x$ for all $x$ is commutative Let $R$ be a ring such that $x^4=x$ for every $x\in R$. Is this ring commutative?
 A: Here is an interesting observation:
$2x=(x+x)=(x+x)^4=16x^4=16x$. Thus $14x=0$.
Also $3x=(3x)^4=81x^4=81x$. Thus $78x=0$. Since gcd $(78,14) =2$ we get $2x=0$ for all $x \in R$.
Now one probably needs to look at $x+y=(x+y)^4$ and probably $(x+y+z)=(x+y+z)^4$ or $(x+y+1)=(x+y+1)^4$. 
P.S. Actually characteristic 2 is also obtained the following way: $x=x^4=(-x)^4=-x$. 
A: Here's an old post of mine from Yahoo! Answers:
First, note $-x = (-x)^4 = x^4 = x$, so $x+x = 0$ for any $x$ in $R$. Then
$(x^2+x)^2 = x^2 + x + x^3 + x^3 = x^2+x$. Thus $x^2+x$ is idempotent, and it is easy to see idempotent elements are central in this ring. [I give a proof of this at the end.]
Now let $x=a+b$, where $a$ and $b$ are arbitrary. From above, for any $c$ in $R$,
$c(x^2 + x) = (x^2 + x)c$, and expanding this out and cancelling terms we get
$c(ab + ba) = (ab + ba)c$. Setting $c=a$, we get, after cancelling again,
$a^2b = ba^2$. Thus, for any $x$ in $R$, $x^2$ is central. Then of course
$x = (x^2+x)-x^2$ is central.
To prove that idempotents are central, first note that if $xy=0$, then
$yx = (yx)^4 = y (xy)(xy)(xy)x = 0$. So now if $z^2 = z$, then
$z(y - zy) = 0$, so $(y-zy)z = 0$, or $yz = zyz$. Similarly,
$(yz - y)z = 0$, so $z(yz-y) = 0$, or $zy = zyz$. Thus $yz = zy$.
