Commutative ring and zerodivisor 
Let $a \ne 0$ belong to a commutative ring. Prove that $a$ is a zero divisor if and only if $a^2b=0$ for some $b \ne 0$.

I know that to be a zero divisor there has to be a non zero element $b$ such that $ab=0$ but I am stuck on how to prove this.
A walk through would be very helpful
 A: Suppose $a \ne 0$ is a zero divisor. Then by hypothesis, there exists some $b \neq 0$ such that $ab=0$. Therefore $a(ab)=a0=0$. On the other hand, $a(ab)=(aa)b=a^2b$. Thus $a^2 b=0$ for some $b \ne 0$.
Conversely, suppose $a \ne 0$, $b \ne 0$, and $a^2 b =0$. Now, either $ab=0$ or $ab \ne 0$. If $ab=0$, then clearly $a$ is a zero divisor. If $ab \ne 0$, then $a (ab) =0$ shows that $a$ is again a zero divisor. Therefore in both cases, $a$  must be zero divisor.
Thus the theorem is proved.
A: If $b$ witnesses the fact that $a$ is a zero divisor, simply compute $a^2 b$.
Otherwise, try regrouping $0 = a^2 b$ as
$$0 = a (ab)$$
Now consider two cases: $ab = 0$ and $ab \ne 0$.
A: It is easier to work with the negations, i.e. we want to show that


*

*$a$ is regular (i.e. $ab=0 \Rightarrow b=0$)

*$a^2$ is regular
are equivalent. For 1. => 2., assume $a^2 b = 0$ and write this as $a(ab)=0$. Now use 1., and after that once again 1., to obtain $b=0$. For 2. => 1., assume $ab=0$, multiply with $a$ and use 2. to conclude $b=0$.
A: $$a\;\;\text{zero divisor}\;\;\implies\;\exists\; b\in R\;\;s.t.\;\;ab=0\implies a^2b=a(ab)=a\cdot 0 =0$$
OTOH
$$0=a^2b=a(ab)\implies\;a\;\;\text{is a zero divisor (why?!)}$$
