74th Romanian Mathematical Olympiad

Let $\alpha ,\beta ,\gamma$ be the reduced arguments of the complex numbers $a,b,c$. Since $abc\in \mathbb{R}$, it follows that $\sin (\alpha +\beta +\gamma )=0$, i.e., $\alpha +\beta +\gamma =k\pi$, where $k\in \mathbb{Z}$. Therefore, we can write $\gamma =k\pi -\alpha -\beta$ for some integer $k$. Furthermore, from $a+b+c\in \mathbb{R}$ we have $\sin \alpha +\sin \beta +\sin \gamma =0$, thus $\sin \alpha +\sin \beta =-\sin \gamma =-\sin (k\pi -\alpha -\beta )$, leading to $|\sin \alpha +\sin \beta |=|\sin (\alpha +\beta )|$, which is equivalent to: $$\left|\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}\right|=\left|\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha +\beta }{2}\right|.$$ If $\sin \frac{\alpha +\beta }{2}=0$, then there is an integer $q$ such that $\frac{\alpha +\beta }{2}=q\pi$, and $\gamma =(k-2q)\pi$, hence $\sin \gamma =0$, which leads to $c\in \mathbb{R}$. If $\sin \frac{\alpha +\beta }{2}\ne 0$, then $|\cos \frac{\alpha +\beta }{2}|=|\cos \frac{\alpha -\beta }{2}|$, from which we have ${\cos }^2\frac{\alpha +\beta }{2}={\cos }^2\frac{\alpha -\beta }{2}$, thus $\cos (\alpha +\beta )=\cos (\alpha -\beta )$, leading to $\sin \alpha \sin \beta =0$, i.e., at least one of the numbers $a$ or $b$ is real. Therefore, at least one of the numbers $a,b$, or $c$ is real. Let this number be $a$. From the hypothesis, we obtain that $b+c\in \mathbb{R}$, and since $a$ is nonzero, we also have $bc\in \mathbb{R}$. If $b$ is real, then $c$ will also be real, so $C_n$ is real for any $n\in \mathbb{N}$. If $b\in \mathbb{C}\setminus \mathbb{R}$, then $b$ and $c$ are the roots of a quadratic equation with real coefficients, thus $b=\bar{c}$. Consequently, $C_n=a^n+b^n+c^n=a^n+b^n+{\bar{b}}^n=a^n+b^n+{\bar{c}}^n\in \mathbb{R}$.

Alternative solution. Let $|a|=|b|=|c|=r>0$. For a complex number $z$ with modulus $r>0$, we have $\bar{z}=\frac{r^2}{z}$. Since $A$ is a real number, we obtain: $$a+b+c=\bar{a}+\bar{b}+\bar{c}=\frac{r^2}{a}+\frac{r^2}{b}+\frac{r^2}{c}=r^2\cdot \frac{ab+bc+ca}{abc}\in \mathbb{R},$$ which implies $ab+bc+ca\in \mathbb{R}$. For $n=0$ we have $C_0=3$, and for $n=1$ we have $C_1=A\in \mathbb{R}$. Also, for $n=2$ we have $C_2=A^2-2(ab+bc+ca)\in \mathbb{R}$. For $n\ge 2$, we have: $$C_{n+1}=A\cdot C_n-(ab+bc+ca)\cdot C_{n-1}+B\cdot C_{n-2}.$$ By induction, it now follows that $C_n\in \mathbb{R}$ for any $n\in \mathbb{N}$.