Romanian Mathematical Olympiad

a) Let $x$ be an invertible element and let $n$ be a positive integer such that $a^n = 0$. Since $a + x = x(x^{-1}a + 1)$, it is enough to show that $x^{-1}a + 1$ is invertible. Set $b = x^{-1}a$. Then $b^n = x^{-n}a^n = 0$, (because $A$ is commutative), hence $b^{2n+1} = 0$. Consequently,
$$1 = b^{2n+1} + 1 = (b+1)(b^{2n} - b^{2n-1} + \dots - b + 1),$$
i.e., $b+1$ is invertible.

b) Induct on $n$, $n \ge 1$, to prove that $a^n - 1$ is invertible. For $x = -1$ we get that $a - 1$ is invertible. Assume that $b = a^n - 1$ is invertible. From the hypothesis it follows that $a - b^{-1}$ is also invertible, and so is $ab - 1 = (a - b^{-1})b$. Moreover,
$$a^{n+1} - 1 = a + (a(a^n - 1) - 1) = a + (ab - 1)$$
is invertible. Since $A$ is finite, there exist two integers $q > p \ge 1$ such that $a^p = a^q$, i.e. $a^p(a^{q-p} - 1) = 0$. Since $a^{q-p} - 1$ is invertible, it follows that $a^p = 0$.