<s>[INST] Prove that $f(x) = ax^2$ For some  function $f \in C(R)$ the following equality holds: $f(x) + f(y) = f(\sqrt{x^2 + y^2})$ for any $x, \ y \in R$. 
Prove that $f(x) = ax^2 \ \forall x\in \mathbb{R}$, where $a = f(1)$.
 [/INST] The functional equation
$$
f(x) + f(y) = f \left( \sqrt{x^2 + y^2} \right)
$$
implies
$$
f(0) + f(0) = f \left( \sqrt{0^2 + 0^2} \right) = f(0) 
$$
so $f(0) = 0$. Further
$$
f(-\lvert x \vert) = f(-\lvert x \vert) + f(0) 
= f \left( \sqrt{(-\lvert x \rvert)^2 + 0^2} \right) = f(\lvert x \rvert) 
$$
which means $f$ is symmetric regarding the $y$-axis:
$$
f(x) = f(-x)
$$
So we can write
\begin{align}
f(x) + f(y) &= f(\lvert x \rvert) + f(\lvert y \rvert) \\
&= f(\sqrt{X}) + f(\sqrt{Y}) \\
&= f(\sqrt{X + Y}) \\
\end{align}
so $g = f \circ \sqrt{.}$ is subject to Cauchy's functional equation
$$
g(x+y) = g(x) + g(y) \quad (x,y \in \mathbb{R}, x,y \ge 0)
$$
which over the reals has many interesting solutions, among them the family
$$
g(x) = a x
$$
with $a \in \mathbb{R}$. If a continuous solution is requested, which is the case here, it is the solution. This means
$$
f(\sqrt{X}) = a X \iff \\
f(x) = a x^2
$$
for non-negative $x$. It is symmetric and of course $f(1) = a$.
</s>