Thai Mathematical Olympiad

Let () be the given functional equation. Setting $x=y=z=0$ in (), we have $f(0)=0$. Setting $y=z=0$ in () and simplifying, we have $$f(\alpha x)={\alpha }^{2}f(x)\text{ for all }x\in \mathbb{R}.$$ Letting $z=0$ in the functional equation, we get $$f(\alpha x+y)+f(\alpha y)+f(x)=\alpha f(x+y)+2f(x)+2f(y).$$ Using $f(\alpha y)={\alpha }^{2}f(y)=(\alpha +1)f(y)$, it follows that $$f(\alpha x+y)=\alpha f(x+y)+f(x)+(1-\alpha )f(y).$$ Then () simplifies to $$f(x+y)+f(y+z)+f(z+x)=f(x+y+z)+f(x)+f(y)+f(z).$$ Putting $z=-y$ in the above equation, we have $$f(x+y)+f(x-y)=2f(x)+f(y)+f(-y).$$ Let $f_c(x)=(f(x)+f(-x))/2$ and $f_o(x)=(f(x)-f(-x))/2$. Then $$\begin{array}{rl}{f}_c(x+y)+f_c(x-y)& =2f_c(x)+2f_c(y)\\ f_o(x+y)+f_o(x-y)& =2f_o(x)\end{array}$$ which respectively are the quadratic and the Jensen functional equations with the continuous solutions $f_c(x)=a{x}^2$ and $f_o(x)=bx$ (note that $f_o(0)=0$). Thus, $f(x)=a{x}^2+bx$. Substituting $f(x)=a{x}^2+bx$ into (*), we can see that $b=0$. Thus $f(x)=a{x}^2$ is the only continuous solution.