(Moscow 2000, exercise A1) Let $x$ and $y$ be two different real numbers such that $x^{2}-2000 x=y^{2}-2000 y$. What is the value of $x+y$?
Let $c=-x^{2}+2000 x$, then $x$ and $y$ are the two roots of $X^{2}-2000 X+c$, so $x+y=2000$.

The technique of solving quadratic equations allows solving certain types of higher-degree equations.