Filipjack wrote: Let $A$ be a ring and let $D$ be the set of its non-invertible elements. If $a^2=0$ for any $a \in D,$ prove that: a) $axa=0$ for all $a \in D$ and $x \in A$; b) if $D$ is a finite set with at least two elements, then there is $a \in D,$ $a \neq 0,$ such that $ab=ba=0,$ for every $b \in D.$ Ioan Băetu Here is the official solution (in Romanian).

a) Let $G:=(A,+)$ be the additive group structure of $A$. Fix $a\in D$. Consider the group endomorphism $\varphi:G\to G$ given by $x\mapsto axa$. For a unit $x$, since $ax$ is not a unit, it follows that $axax=(ax)^2=0$ so $axa=0$. Hence $A^\times\subseteq\ker\varphi$. Assume FTSOC there exists $x\in A\setminus\ker\varphi$. Then $x$ is not a unit. Since $a\in\ker\varphi$, it follows that $a+x\not\in\ker\varphi$ so $a+x$ is not a unit. Thus $ax+xa=(a+x)^2=0$ so $\varphi(x)=axa=a(ax+xa)=0$, contradicting the choice of $x$. $\square$ b) Let $a=a_1\cdots a_k$ be the longest nonzero product of elements in $D$. It exists since $D$ is finite, and the $a_i$ are distinct by part a). Clearly $a\in D$ since $a_1a=0$. If there exists $b\in D$ such that $ab$ or $ba$ is nonzero, $ab$ or $ba$ would be a longer nonzero product of elements in $D$, contradicting the maximality of $a$. Thus $ab=ba=0$ for all $b\in D$, so this choice of $a$ works. $\square$