Let $A$ be a ring such that for any $x \in A, x^2=0$ implies $x=0$. Show that if $a \in A$ satisfies $a^3=a$ then $2a$ is in the center of $A$.
I have no idea how to use the $x^2=0$ implies $x=0$ thing. My progress is as follows: note that if $a^3=a$ then $6a = 0$.
Denote by $B$ the set of $a$'s in $A$ for which $a^3=a$. I suspect that $B$ actually induces a subring. First of all, $B$ is closed under taking inverses: if $a^3=a$ then $(-a)^3 = - (a^3) = -a$ so $-a \in A$. Also $B$ is closed under addition: if $a \in B$ then $(2a)^3 = 8a^3 = 8a = 2a; (3a)^3 = 27a^3 = 3a$ and $4a = - 2a, 5a = -a$.
Let's say this actually induces a subring. Then the subring is of characteristic at most $6$, so it is of characteristic $1,2,3$ or $6$. If it is of characteristic $1$ or $2$ it is trivial that $2B$ is in the center of $A$. Otherwise I dont know how to proceed.
Edit: indeed I don't think this forms a subring. I proved it: here is a quick outline: Since the ring is reduced, any idempotent element is central. If $a=a^3$ then $a^2=a^4$ so $a^2$ is idempotent and thus central. We can show that if $a^3=a$ then $(a+1)^2$ is idempotent and thus central. Then since the center of the ring is a subring, we have that $(a+1)^2-a^2-1$ is in the center so $2a$ is in the center.
The problem now asks to show that any two elements of $B$ commute; That is, if $a^3=a$ and $b^3=b$ then $ab=ba$. Haven't cracked this one yet so ideas are welcome.