Let $R$ be a ring with elements $a,b$ such that $a^2 = 0 = b^2$ and $a+b$ is a unit. How to show that $R$ is a ring of $2\times2$ matrices?
COMMENT: The converse is clearly true, just take $a = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$ and $b = \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}$. So it would be interesting if some one could generalize this to dimension $n$.
Let $c$ denote the inverse of $a+b$, i.e., $ac+bc=ca+cb=1$. Multiplying this by $a$ and $b$ from the left and from the right and using $a^2=b^2=0$, we obtain $abc=a$, $cba=a$, and $cab=b$.
Let us show that $R=aR\oplus bR$. It follows from $ac+bc=1$ that $R=aR+bR$. If $ax=by$, then $bax=0$, implying $0=cbax=ax$ in view of $cba=a$.
The $R$-modules $aR$ and $bR$ are isomorphic. Indeed, the rule $bcx\mapsto ax$, $x\in R$, defines an $R$-module homomorphism $bR=bcR\to aR$ because $bcx=0$ implies $ax=0$ due to $abc=a$. Its kernel vanishes: $ax=0$ implies $abcx=0$ because $abc=a$ and, hence, $0=cabcx=bcx$.
It remains to apply Propositions 5 and 6 on page 52 of "Structure of Rings" by N.Jacobson (1964) and to conclude that $R$ is a ring of $2\times2$ matrices.
Edit. By suggestion of Martin Brandenburg, I reproduce the Propositions quite literally:
Proposition 5 $\dots$ Conversely, if $\frak A$ is a ring with an identity $1$ and ${\frak A}={\frak I}_1\oplus\dots\oplus{\frak I}_n$ is a direct decomposition of $\frak A$ into right ideals which are isomorphic $\frak A$-modules, then there exists a set of matrix units $\{e_{ij}\mid i,j=1,\dots,n\}$ such that ${\frak I}_j=e_{jj}{\frak A}$.
Proposition 6. Let $\{e_{ij}\mid i,j=1,\dots,n\}$ be a set of matrix units in a ring $\frak A$ with identity $1$, and $\frak B$ the subring consisting of the elements which commute with the $e_{ij}$, $i,j=1,\dots,n$. Then every element of $\frak A$ can be written in one and only one way as $\sum b_{ij}e_{ij}$ where $b_{ij}\in{\frak B}$ for all $i$ and $j$. Hence ${\frak A}\cong{\frak B}_n$. The ring $\frak B$ is isomorphic to $e_{11}{\frak A}e_{11}$.
Second thought Edit. Those Propositions are an old stuff. It is probably better to prove them here in two lines.
For any $R$-module $M$, we have $\text{End}_R(\underbrace{M\oplus\dots\oplus M}_n)\cong\text{Matr}_n(\text{End}_RM)$. On the other hand, $\text{End}_RR\cong R$.
Question. The problem looks a bit artificial. Is it a known exercise?
