Solving functional equation $f(x+y)=f(x)+f(y)+xy$ We are given $f(0)=0$. Then when $x+y=0$:
$$0=f(-y)+f(-x)+xy$$
Can I now use $x=0$ and obtain:
$$0=f(-y)?$$ Is this correct? Is there a better way to solve this equation? 

I show that the only solution is $f = \frac{1}{2}x^2 + bx$ (assuming $f$ continuous).
My strategy was to transform $f(x+y) = f(x) + g(y) + xy$ into a linear equation.
Write $f(x+y) = f(x) + g(y) + xy$ as $f(x+y) - \frac{1}{2}(x+y)^2 = (f(x) - \frac{1}{2}x^2) + (g(y) - \frac{1}{2}y^2)$. Substitute $g(z) = f(z) - \frac{1}{2}z^2$. Then we have:
$g(x+y) = x+y$, which is linear, so $g(x) = bx$ (assuming continuity).
Thus $f(z) - \frac{1}{2}z^2 = bz$, so $f(z) = \frac{1}{2}z^2 + bz$.
