Let $(A, \cdot)$ be an abelian semigroup with $|A| \geq 2$ such that there exists $a, b \in A$ for which $a^2x^3b^2 = x^2, \forall x \in A$. Prove that $x^3 = x^2, \forall x \in A$.
I first tried playing along with the given equality by substituting $x \rightarrow a$ and $x \rightarrow b$, with no real results. However, I found it particularly useful to consider $x \rightarrow ab$, and then I obtained $a^2(ab)^3b^2 = (ab)^2$, but since $A$ is abelian, we can write that $(ab)^5 = (ab)^2$, for the fixed $a, b \in A$. Then I set $x \rightarrow abx$, and I obtained that $a^2(abx)^3b^2 = (abx)^2$, which after rearranging with respect to the commutativity of the semigroup, I obtained $(ab)^5x^3 = (ab)^2x^2$. However, we know from the previous result that $(ab)^5 = (ab)^2$, then the new equality becomes $(ab)^2x^3 = (ab)^2x^2, \forall x \in A$.
Edit. Setting $x \rightarrow a, x \rightarrow b$ and multiplying the two resulting equalities, we can also obtain $(ab)^7 = (ab)^2$. Do not know if that can be utilized in any useful way.
This was as far as I could get, and I am completely lost because I cannot find any meaningful way of reducing the $(ab)^2$ terms from the left. Any hints or answers would really help me out a lot figuring this out. Thanks a lot!
From your equality $(ab)^2x^3=(ab)^2x^2$, we can do this:
$$x^2=(ab)^2x^3=(ab)^2x^2\ \ \ \text{(From hypothesis)}$$ Multiplying $x$ to the first and third term, $$x^3=(ab)^2x^3.$$
Is this alright?