A set of elements in a reduced unity ring

Question

Let $(A,+,\cdot)$ be a unity ring with the property that if $x \in A$ and $x^2=0$ then $x=0$. Consider the set $M=\{a\in A | a^3=a\}$. Prove that:

a) $2a\in Z(A)$, $\forall a\in M$, where $Z(A)$ denotes the centre of the ring $A$;
b) $ab=ba$, $\forall a,b\in M$.

My attempts revolved around the fact that an idempotent element in a reduced ring is central.
So, since for $a\in M$ we have that $(a^2)^2=a^2$, it follows that $a^2\in Z(A)$, $\forall a\in M$.
The next thing I wanted to use in order to solve a) was that $Z(A)$ is a subring of $A$, so if I had proved that $(a+1)^2 \in Z(A)$, $\forall a\in M$, then we would have reached the desired conclusion. However, I couldn't prove this and I honestly doubt that it is true.
Another idea that I had was to prove that $M$ is a subring of $A$. Of course, this didn't work out because I cannot even prove that $M$ is closed under addition. Again, I don't know if this is true and it most likely isn't.
As for b), I think that a) should be of use, but I don't know how. It is a well-known problem that a ring with $x^3=x$ for any $x$ in that ring is commutative, but since $(M,+,\cdot)$ is almost definitely not a ring, this doesn't help.
EDIT: Is there any chance that this question is simply wrong? I tended to believe this before asking it here too, but since nobody has made any progress on it until now I am even more inclined to think so.

Answer 1

We show $M \subseteq Z(A)$. This immediately gives (a) and (b).

Lemma 1: $yx = 0 \implies xzy = 0$ for any $z$.

Proof: $(xzy)(xzy) = xz(yx)zy = 0$.

Lemma 2: $x^2 = x$ implies $x \in Z(A)$.

Proof: For any $y \in A$, a short computation shows $(xy-xyx)(xy-xyx) = 0 = (yx-xyx)(yx-xyx)$.

Lemma 3: $a \in M \implies a^2 \in Z(A)$.

Proof: $(a^2)^2 = a^4 = a^2$, so use Lemma 2.

Claim: For any $a \in M$, $a \in Z(A)$.

Proof: Since $(a-1)[a(a+1)]=0$, Lemma 1 implies that for any $b \in A$, $$0 = a(a+1)b(a-1) = (a^2b+ab)(a-1).$$ Also, $a(a+1)(a-1)=0$ implies $$0 = ba(a+1)(a-1) = (ba^2+ba)(a-1) = (a^2b+ba)(a-1),$$ where the last equality used Lemma 3. Subtracting gives: (1) $0 = (ba-ab)(a-1)$. The exact same argument shows: (2) $0 = (a-1)(ba-ab)$. (1) immediately implies $0 = (ba-ab)(a-1)b = (ba-ab)(ab-b)$, and (2) with Lemma 1 implies $0 = (ba-ab)b(a-1) = (ba-ab)(ba-b)$. Subtracting the two results gives $(ba-ab)^2 = 0$.