We plug in the pairs $(a, x),(a, 2 x),(a+x, x)$ and $(a-x, x)$ to get

$$\begin{aligned} f(a+x)+f(a-x)-2 f(a) & =g(a) x^{2} \\ f(a+2 x)+f(a-2 x)-2 f(a) & =4 g(a) x^{2} \\ f(a+2 x)+f(a)-2 f(a+x) & =g(a+x) x^{2} \\ f(a-2 x)+f(a)-2 f(a-x) & =g(a-x) x^{2} \end{aligned}$$

respectively. Combining these equations in the form $2 E_{1}-E_{2}+E_{3}+E_{4}$ the left hand side vanishes, yielding an equation in $g:(g(a+x)+g(a-x)-2 g(a)) x^{2}=0$, i.e.

$$g(a)=\frac{g(a+x)+g(a-x)}{2}$$

Since $g$ is continuous, it must be linear, i.e. $g(x)=c_{1} x+c_{0}$. However, the original equation for $x=y$ together with the concavity condition now gives us

$$0 \geqslant f(2 x)+f(0)-2 f(x)=\left(x c_{1}+c_{0}\right) x^{2}$$

for all $x$, which is only possible if $c_{1}=0$. Thus $g(x) \equiv c_{0}=2 A$ is constant and

$$f(x+y)+f(x-y)-2 f(x)=2 A y^{2}$$

This suggests that $f$ is a quadratic function, so we can set $f(x)=A x^{2}+f_{1}(x)$. Then $(*)$ becomes $f_{1}(x+y)+f_{1}(x-y)-2 f_{1}(x)=0$, so an easy induction gives us

$$f_{1}(n x)-f_{1}(0)=n\left(f_{1}(x)-f_{1}(0)\right) \quad \text { for all } \quad n \in \mathbb{Z}$$

By setting $f_{1}(0)=C$ and $f_{1}(1)=B+C$ we obtain $f_{1}(x)=B x+C$ and $f(x)=$ $A x^{2}+B x+C$ for all $x \in \mathbb{Q}$. By concavity of $f$ we conclude that $f(x)=A x^{2}+B x+C$ for all real $x$.

## Remark.

In fact, $(*)$ implies that the second derivative of $f$ is constant by taking $y \rightarrow 0$ and the problem is solved. The solution presented here avoids use of derivatives.