<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] This assumes that
$f$ is differentiable.
$f(x) + f(y) 
= f(\sqrt{x^2 + y^2})
$.
If $y=0$,
then
$f(x)+f(0) = f(x)$,
so
$f(0) = 0$.
Since
$f(x)+f(-y)
=f(\sqrt{x^2 + y^2})
=f(x)+f(y)
$,
$f(y) = f(-y)$.
Differentiating wrt $x$,
$\begin{array}\\
f'(x)
&=(\sqrt{x^2+y^2})'f'(\sqrt{x^2+y^2})\\
&=\dfrac{x}{\sqrt{x^2+y^2}}
f'(\sqrt{x^2+y^2})\\
\text{so}\\
\dfrac{f'(x)}{x}
&=\dfrac{f'(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}\\
\end{array}
$
Therefore
$\dfrac{f'(x)}{x}
$
is constant.
Setting
$\dfrac{f'(x)}{x}
=c$,
$f'(x) = cx$
so
$f(x) = cx^2/2+d$.
Since $f(0) = 0$,
$f(x) = cx^2/2$.
</s>