75th Romanian Mathematical Olympiad

Denote $U(A)$ the set of invertible elements in $A$. For $k\in \mathbb{N}$, $k\ge 2$, and $x\in A$, define $kx=\underset{k\text{ terms}}{\underbrace{x+x+\cdots +x}}$. In particular $k\cdot 1=k$. By the given conditions $2\in U(A)$. Denote by (1) the equality $x+x^4=x^2+x^3$ for all $x\in A$.

Changing $x$ by $-x$ in (1) we get $-x+x^4=x^2-x^3$, for any $x\in A$.

By subtraction, the last two relations give $2x=2x^3$, so, as $2\in U(A)$, we obtain $x=x^3$ for all $x\in A$.

For $x\in U(A)$, multiplying by $x^{-1}$ we get $x^2=1$ for any $x\in U(A)$. As the set $U(A)$, of the invertible elements of the monoid $(A,\cdot )$, is a group with $x^2=1$, for any $x\in U(A)$, it results that the group $(U(A),\cdot )$ is commutative.

For $x=2\in U(A)$ from the previous relations, we get $4=1$, so $3=0$, meaning that the ring $A$ is of characteristic $3$.

Let $a,b\in A$, such that $a\ne b$ and $a,a+b\in U(A)$. If we suppose $b\in U(A)$, then $$2ab=(a+b{)}^2-a^2-b^2=1-1-1=-1=2,$$ implying $ab=1=a^2$. This gives $a=b$, a contradiction. That is $b\notin U(A)$.

We also have $$1+ab=a^2+ab=a(a+b)\in U(A),$$ a product of invertible elements.