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.