Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.7 AIME Prove it Romania

Let $S$ be the sum of all invertible elements of a finite ring.
Prove that $S^2 = S$ or $S^2 = 0$ .

Solution

If $1 + 1 \ne 0$ , then $x \ne -x$ , for every invertible $x$ , whence $S = 0$ (we can group the invertible elements into pairs $(x, -x)$ ).

If this is not the case, notice that $xS = S$ , for every invertible $x$ . Adding all these relations, $S^2 = kS$ , where $k$ is the number of invertible elements. Then $S^2 = S$ for odd $k$ and $S^2 = 0$ for even $k$ (since $1 + 1 = 0$ ).

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