Proving that if $tX_1 + sX_2 \stackrel{D}{=} \sqrt{t^2 + s^2}X$ then $X \stackrel{D}{=} N(0, \delta^2)$ Assume that for all $s, t \in \mathbb{R}$ the following property
$$tX_1 + sX_2 \stackrel{D}{=} \sqrt{t^2 + s^2}X \tag{1}$$
is true.
Moreover $X_1, X_2, X$ are i. i. d.
My task is to prove that if $(1)$ stands then $X \stackrel{D}{=}N(0, \sigma^2)$. So $X$ is symmetrically normally distributed. 
How can it be proved? I suppose that characteristic functions should be used but I have no idea how. I will appreciate any hints or tips.
 A: There are 2 ingredients.  As you guessed, characteristic functions are involved.  Here is a hint outline.
Step 1: derive the identity $\phi(s)\phi(t)=\phi(\sqrt{s^2+t^2})$ obeyed by the characteristic function $\phi$. Then, somehow manipulate that to obtain a condition like $f(s)f(t)=f(s+t)$ or $g(s)+g(t)=g(s+t)$, valid for all real $s,t$.
Step 2: realize that this puts a severe limitation of what $f$ or $g$ (and ultimately $\phi$) can be.
A: Hints: Assume that $X_1$ is not $0$ with positive probability. let $\phi$ be the characteristic function of $X_1$. Then $\phi (t) \phi (s)=\phi (\sqrt {t^{2}+s^{2}})$. If $\phi (t)=0$ for some $t$ then we get $\phi (\sqrt {t^{2}+s^{2}})=0$ for all $s$ and hence $\phi (u)=0$ for all $u >|t|$. S how that this leads to a contradiction to the given property.  Otherwise, by  a well known fact there is a continuous function $\psi$ such that $e^{\psi (t)}=\phi (t)$ and $\psi (0)=0$. Let $g(t)=\psi (\sqrt t)$ for $t >0$. Show that $g(t+s)=g(t)+g(s)$. I will leave the rest of the argument to you. 
