Romanian Mathematical Olympiad

We show that (a) implies (b). Let $x \in R \setminus U(R)$, where $U(R)$ is the set of invertible elements of $R$, and let $y \in R \setminus U(R)$ be such that $xy = x + y$, i.e., $(x - 1)(y - 1) = 1$. Similarly, $(y - 1)(z - 1) = 1$ for some $z \in R \setminus U(R)$. Write
$$x - 1 = (x - 1) \cdot 1 = (x - 1)(y - 1)(z - 1) = 1 \cdot (z - 1) = z - 1,$$
to infer that $x = z$. Hence $x - 1 \in U(R)$, showing that $\{t - 1 : t \in R \setminus U(R)\} \subseteq U(R)$.
Since $x \in R \setminus U(R)$, its powers are all different from $1$, and finiteness of the multiplicative group $U(R)$ then forces them all in $R \setminus U(R)$. Hence $x^n - 1 \in U(R)$ for all positive integers $n$, by the set inclusion in the previous paragraph. With reference again to finiteness of $U(R)$, $x^p - 1 = x^q - 1$ for some distinct positive integers $p$ and $q$, say, $p < q$. Then $x^p(x^{q-p} - 1) = 0$, so $x^p = 0$, on account of $x^{q-p} - 1 \in U(R)$. Consequently, $x$ is indeed nilpotent.

We now show that (b) implies (a). Let $x \in R \setminus U(R)$ and choose an integer $p \ge 3$ such that $x^p = 0$. Write $1 = 1 - x^p = (1-x)(1+x+x^2+\cdots+x^{p-1})$ and let $y = -(x+x^2+\cdots+x^{p-1}) = -x(1+x+\cdots+x^{p-2})$. Then $y^p = 0$, so $y \in R \setminus U(R)$. Clearly, $(1-x)(1-y) = 1$, i.e., $xy = x+y$. This completes the proof.