TITLE: A functional equation: $f(x, y + z) = f(x, y) + f(x + y, z)$
QUESTION [5 upvotes]: Can anything be said about the solutions of the following functional equation?
$$
f(x, y + z) = f(x, y) + f(x + y, z)
$$
I don't seem to be able to find much in what I think are the standard references in these cases.

REPLY [5 votes]: Following the beautiful idea of Robert Israel, we will show that the solutions of the equation are precisely functions of the form $f(x,y)=g(x+y)-g(x)$, where $g$ is an arbitrary function.
First, we plug $z = -y$ into the equation. This yields $$f(x, 0) = f(x, y) + f(x + y, -y).\tag{1}$$
In the case $y=0$, this tells us that $f(x,0)=0$ for all $x$, as Robert Israel already noticed. Using this fact in $(1)$, we have that $$f(x,y)=-f(x+y,-y)\tag{2}$$ must hold for all $x,y$ in order for $f$ to be a solution. We will now use this fact in the original equation. The original equation says that $$f(x, y) = f(x, y + z)  - f(x + y, z)$$ holds for all $x,y,z$. Using $(2)$ twice we may rewrite this as $$f(x, y) = - f(x+y+z, - y - z)  + f(x+y+z, -z)$$ Now, plug in $z=-x-y$ and get $$f(x, y) = - f(0, x)  + f(0, x+y).$$ This means that if $f$ solves the original equation, we may define the function $g$ by $g(w)=f(0,w)$ and then $f(x,y)=g(x+y)-g(x)$ will hold. This shows that indeed the solutions of the equation are precisely of the form suggested by Robert Israel.