Is there any short proof of this classical problem? Let $X,Y$ be two i.i.d. r.v.'s with zero mean and unit variance. If $X+Y$ and $X-Y$ are independent, then $X$ and $Y$ are both standard normal distributed.
Is there any short proof for this problem?
 A: This is a weakened form of Bernstein's Theorem (weakened by unnecessarily assuming identical distributions having finite variances), so the proof is shorter. Here's a sketch, adapted from "On three characterizations of the normal distribution" by M. P. Quine:
Define $\quad U = X+Y, \quad V=(X-Y)^2$ 
and characteristic functions
$$\phi(t) = E e^{itX}\\
\gamma(s,t) = E e^{isU + itV}.
$$
Then, using independence,
$${\partial\gamma \over \partial t} = E( iV e^{itV})\ E( e^{isU}),
$$
so, in terms of $X$ and $Y$, it is found that
$$
{\partial\gamma \over \partial t}\bigg|_{t=0} = 2iE(X^2 e^{isX}) E(e^{isY}) - 2i(E(Xe^{isX}))^2 = -2i \phi''(s)\ \phi(s)+2i(\phi'(s))^2.
$$
However, we also have
$${\partial\gamma \over \partial t}\bigg|_{t=0} = E(e^{isU}) {\partial \over \partial t} E(e^{itV})\bigg|_{t=0} = 2i(\phi(s))^2,
$$
hence the equation
$$ - \phi \ \phi'' + (\phi')^2 = \phi^2
$$
whose unique solution is 
$$\phi(s) = e^{-{1 \over 2}s^2}.
$$ 
QED
NB: The cited article proves the stronger version: If $(X,Y)$ are independent and $(X+Y, X-Y$) are independent, then $X$ and $Y$ are normal with the same (finite) variance.
