IMC 1997 · Day 1 · P3

Set $S = A + \omega B$, where $\omega = -\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2}$. We have $$\begin{align*} S \overline{S} &= (A + \omega B)(A + \overline{\omega} B) = A^2 + \omega BA + \overline{\omega} AB + B^2 \\ &= AB + \omega BA + \overline{\omega} AB = \omega (BA - AB), \end{align*}$$ because $\overline{\omega} + 1 = -\omega$. Since $\det(S \overline{S}) = \det S \cdot \det \overline{S}$ is a real number and $\det \omega (BA - AB) = \omega^n \det(BA - AB)$ and $\det(BA - AB) \ne 0$, then $\omega^n$ is a real number. This is possible only when $n$ is divisible by 3.