19. Let $f$ and $g$ be functions such that for all real numbers $x$ and $y$,
$$
g(f(x+y))=f(x)+(x+y) g(y) .
$$

Find the value of $g(0)+g(1)+\cdots+g(2013)$.
19. Answer: 0.
Let $y=-x$. Then $g(f(0))=f(x)$ for all $x$. This shows that $f$ is a constant function; namely $f(x)=c$ for some c. So that $g(c)=g(f(0))=f(x)=c$. For all $x, y$, we have
$$
(x+y) g(y)=g(f(x+y))-f(x)=g(c)-c=0 .
$$

Since $x+y$ is arbitrary, we must have $g(y)=0$ for all $y$. Hence,
$$
g(0)+g(1)+\cdots+g(2013)=0 .
$$