Harvard-MIT Mathematics Tournament

Solution:

Note that
$$\begin{aligned} (x+y)^{k} & = \sum_{i=0}^{k} \binom{k}{i} x^{i} y^{k-i} \\ & = \frac{1}{2} \sum_{i=0}^{k} \binom{k}{i} \left(x^{i} y^{k-i} + x^{k-i} y^{i}\right) \end{aligned}$$
by pairing the $i$th and $(k-i)$th terms. But $x^{k-i} y^{i} = \left(x^{i} y^{k-i}\right)^{-1}$ since $x$ and $y$ are $k$th roots of unity. Moreover, since $x$ and $y$ have absolute value $1$, so does $x^{i} y^{k-i}$, so $x^{k-i} y^{i}$ is in fact its complex conjugate. It follows that their sum is real, thus so is $(x+y)^{k}$.

This can also be shown geometrically. The argument of $x$ (the angle between the vector $x$ and the positive $x$-axis) is an integer multiple of $\frac{2 \pi}{k}$, as is the argument of $y$. Since $x+y$ bisects the angle between $x$ and $y$, its argument is an integer multiple of $\frac{\pi}{k}$. Multiplying this angle by $k$ gives a multiple of $\pi$, so $(x+y)^{k}$ is real.