2.1 Find $f(2013)$, if for any real $x$ and $y$ the equality holds

$$
f(x-y)=f(x)+f(y)-2xy
$$
Solution. Substitute $x=y=0$. We get $f(0)=2 f(0)+0$, from which we obtain that $f(0)=0$. Substitute $x=y$. We get $0=f(0)=f(x)+f(x)-2 x^{2}$. Hence, $f(x)=x^{2}$.

Answer: 4052169. (C)

![](https://cdn.mathpix.com/cropped/2024_05_06_997c53c06135e2de9c16g-02.jpg?height=67&width=1157&top_left_y=1003&top_left_x=498)