The functional equation $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$ How to find the all continuous functions $f\colon \mathbb{R}\to\mathbb{R}$ such that $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$

Let $f: \mathbb R \to \mathbb R$ be a continous solution of the functional equation and let $g(x) := ax^2+bx$ where $a$ and $b$ are chosen in such a way that $f(-1) = g(-1)$ and $f(1) = g(1)$. Then $\tilde f := f-g$ is also a solution and we have $\tilde f(-1) = \tilde f(1) = 0$. We will show $\tilde f \equiv 0$, so that $f = g$.
By replacing $f$ with $\tilde f$ way may assume that $f$ is a solution with $f(-1) = f(1) = 0$, and we will show $f \equiv 0$. Let $Z = \{x \in \mathbb R: f(x) = 0 \}$. A priori we have $\{-1,+1\} \subseteq Z$. Plugging $(1,-1,x)$ into the functional equation yields
$$f(x) = \frac{f(x+1) + f(x-1)}{2},$$
which implies $\mathbb Z \subseteq Z$. If we can show
$$2x,x \in Z \Rightarrow \frac{x}{2} \in Z$$
we are finished since then every dyadic rational $a/2^n$ is in $Z$ and these are dense in $\mathbb R$ (here the continuity of $f$ is needed). So let $2x \in Z$ and $x \in Z$. Plug $(\frac{x}{2},\frac{x}{2},x)$ into the functional equation to get 
$$2f\left(\frac{x}{2}\right) + f(x) + f(2x) = f(x) + 2f\left(\frac{3}{2}x\right),$$
hence $f\left(\frac{x}{2}\right) = f\left(\frac{3}{2}x\right)$. Then plugging in $(\frac{x}{2},\frac{x}{2},\frac{x}{2})$ yields 
$$3f\left(\frac{x}{2}\right) + f\left(\frac{3}{2}x\right) = 3f(x),$$
hence $f(\frac{x}{2}) = 0$ and $\frac{x}{2} \in Z$.
