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 $a = f(1)$.
 A: 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 $f(1) = a$.
A: The crux of this proof is to show that $f$ agrees on $\mathbb{Q}$ with $ax^2$.
Inductively applying the functional equation, we have ($n \in \mathbb{N}$)
$$ \underbrace{f(1) + \dots + f(1)}_{n^2 \text{ times}} = n^2 f(1) = f(\sqrt{n^2}) = f(n)
$$
Hence $f(n) = an^2$ at every positive integer $n$. Observe that $f$ is symmetric about the origin since
$$ f(-x) + f(x) = f(\sqrt{x^2 + x^2}) = f(x) + f(x) \implies f(-x) = f(x)
$$
Hence $f(n) = an^2$ on $\mathbb{Z}$. If $p/q$ is a rational number (without loss of generality let $p,q > 0$), we have
$$ p^2 f(1/q) = \sum_{r=0}^{p^2-1} f(1/q) = f(\sqrt{p^2/q^2}) = f(p/q)
$$
set $p = q$, then $q^2 f(1/q) = f(1)$. Hence $f(x) = ax^2$ on $\mathbb{Q}$. Since continuous functions agreeing on a dense subset of $\mathbb{R}$ must be the same, $f(x) = ax^2$ everywhere.
A bit rough on the details but I think you can work it out yourself.
A: 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 $f(0) = 0$,
$f(x) = cx^2/2$.
