Find all continuous functions over reals such that $f(x)+f(y) = f(x+y)-xy-1$ for all $x,y \in \mathbb{R}$ 
Find all continuous functions over reals such that $f(x)+f(y) = f(x+y)-xy-1$ for all $x,y \in \mathbb{R}$. 

I saw first that $f(0) = -1$ but then I am struggling to see how to get a formula for $f(x)$. If I do $x = 0$ we get $f(0) + f(x) = f(x)-1$ which doesn't really help. Is there a better way to get a formula for $f(x)$ here?
 A: Since you found $f(0)+f(0)=f(0)-0-1$so that $f(0)=-1$, if you hold $y$ constant and take the derivative with respect to $x$, you get
$$f^{\prime}(x)=f^{\prime}(x+y)-y$$
Then rearrange to
$$f^{\prime\prime}(x)=\lim_{y\rightarrow0}\frac{f^{\prime}(x+y)-f^{\prime}(x)}{y}=\lim_{y\rightarrow0}1=1$$
So $f(x)=\frac12x^2+C_1x+C_2$. Since $f(0)=C_2=-1$, we are down to $f(x)=\frac12x^2+C_1x-1$, and since this works in the original equation, we are done.  
EDIT: OK, so here's a solution that doesn't require twice differentiable functions. Rewrite the recurrence relation as
$$\frac{f(x+y)-f(x)}y=\frac{f(y)-f(0)}y+x$$
Given that we know $f(0)=-1$. Then
$$\begin{align}\lim_{y\rightarrow0}\frac{f(x+y)-f(x)}y&=\lim_{y\rightarrow0}\frac{f(y)-f(0)}y+\lim_{y\rightarrow0}x\\
&=f^{\prime}(x)=f^{\prime}(0)+x\end{align}$$
On integration we get
$$f(x)=\frac12x^2+f^{\prime}(0)x+C_3$$
And recall that we knew
$$-1=f(0)=C_3$$
So we are back to verifying that
$$f(x)=\frac12x^2+f^{\prime}(0)x-1$$
works for all values of $f^{\prime}(0)$.
A: Let $g(x) = f(x) + 1-x^2/2$. Then $g$ is continuous and 
$$g(x+y) = g(x) + g(y)$$
The only functions with those properties (continuous and additive) are those of the form $g(x) = ax$, for $a \in \mathbb{R}$. (See Continuous and additive implies linear )
Hence $f(x) = -1+ax + \frac{x^2}{2}$ for some constant $a$.
EDIT: Why that $g$? First, $xy = \frac{1}{2} (x+y)^2 - \frac{1}{2}x^2 - \frac{1}{2} y^2$, so the relation can be written as
$$f(x) - \frac{1}{2}x^2 + f(y) - \frac{1}{2} y^2 = f(x+y) - \frac{1}{2}(x+y)^2 -1. $$
If $h(x) = f(x) -\frac{1}{2} x^2$, then 
$$h(x)+h(y) = h(x+y) -1.$$
There are two $h$ on the left and only one on the right, and the balance is $-1$. So if we define $g(x) = h(x) +1$, then 
$$g(x)-1 + g(y) -1 = g(x+y)-1 -1.$$
