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 $kx = \underbrace{x+x+\cdots+x}_{k \text{ terms}}$. 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 $a \neq b$ 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.