Let $x$ and $y$ be two non-zero numbers such that $x^{2} + x y + y^{2} = 0$ ($x$ and $y$ are complex numbers, but that's not too important). Find the value of

$$
\left(\frac{x}{x+y}\right)^{2013} + \left(\frac{y}{x+y}\right)^{2013}
$$
Clearly, $x /(x+y)+y /(x+y)=1$ and

$$
\frac{x}{x+y} \cdot \frac{y}{x+y}=\frac{x y}{x^{2}+2 x y+y^{2}}=1
$$

Thus, $x /(x+y)$ and $y /(x+y)$ are the roots of $t^{2}-t+1=0$, so the sums $S_{k}=(x /(x+y))^{k}+(y /(x+y))^{k}$ satisfy $S_{0}=2, S_{1}=1$ and

$$
S_{k+2}=S_{k+1}-S_{k}
$$

for $k \geq 0$. We deduce that the sequence $\left(S_{k}\right)$ is periodic with period 6, its values being successively $2,1,-1,-2,-1,1,2,1,-1, \ldots$. We deduce that $S_{2013}=$ -2.