THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

First solution. An element $x$ of $R$ such that $x^2=x$ is called idempotent. We show that $1$ is the only unit in $R$. The proof is based on the remark below: () Let $x$ and $y$ be elements of a ring $R$. If $x\ne 1$ is idempotent, then $xyz\ne 1$ and $zyx\ne 1$ for all $z$ in $R$; in particular, $xy$ and $yx$ are both non-invertible. Indeed, if $xyz=1$, then $x=xxyz=x^{2}yz=xyz=1$, a contradiction. Hence $xyz\ne 1$; similarly, $zyx\ne 1$. Back to the problem, fix a non-invertible $x$ in $R\setminus \{0\}$; since $R$ is not a skew-field, there exists at least one such. Since $2x$ is not invertible, it is idempotent, so $2x=4x^2=4x$, that is, $2x=0$. We now show that $xy=xyx=yx$ for all $y$ in $R$. To prove the first equality, refer to () to infer that $xy-xyx=xy(1-x)$ is non-invertible, hence idempotent, so $xy-xyx=xy(1-x)xy(1-x)=xy(x-x^2)y(1-x)=0$, since $x^2=x$. Similarly, $yx-xyx=(1-x)yx$ is non-invertible, hence idempotent, so $yx-xyx=(1-x)yx(1-x)yx=(1-x)y(x-x^2)yx=0$, since $x^2=x$.

---

Alternative solution.

Second solution. As in the previous solution, we show that $1$ is the only unit of $R$. We first prove that if $u$ is a unit, and $x\ne 0$ is not, then $u+x$ is not a unit. Let $D$ be the set of all non-invertible elements of $R\setminus \{0\}$; since $R$ is not a skew-field, $D$ is non-empty. If $x$ is a member of $D$, then so is $-x$, and $x=x^2=(-x{)}^2=-x$, so $2x=0$. Then $(1+x{)}^2=1+2x+x^2=1+x$, so $1+x$ is not a unit (otherwise, $1+x=1$, so $x=0$, a contradiction). If $u$ is a unit, and $x$ is a member of $D$, then $ux$ and $1+ux$ are both in $D$, and so is $u+x=u(1+u^{-1}x)$. Let $x$ be a member of $D$, let $y$ be a member of $R$, and write $x(xy)=x^{2}y=xy$ and $(yx)x=y{x}^2=yx$, to infer that $xy$ and $yx$ are both non-units. We are now in a position to prove that $1$ is the only unit of $R$. Let $u$ be a unit, and let $x$ be a member of $D$. Then $x+ux$ and $x+xu$ are both non-units, so $(x+ux{)}^2=x+ux$ and $(x+xu{)}^2=x+xu$. Expand both squares to write $x^2+xux+u{x}^2+(ux{)}^2=x+ux$ and $x^2+x^{2}u+xux+(xu{)}^2=x+xu$, and infer that $ux=xux=xu$. Finally, since $u+x$ is not a unit, $(u+x{)}^2=u+x$, so $u^2=u$; that is, $u=1$.