Tricky probability problem I am having trouble with proving the following assertion:

$X,Y$ are i.i.d. with mean $0$ and variance $1$. If $X+Y$ and $X-Y$
  are independent then $X,Y$ are normally distributed.

Should I be looking at derivatives of the m.g.f of $X$ or something? I've proven that it is even and that $M_X(\theta)=1+\frac{1}{2}\theta^2+o(\theta^3)$ but I cannot proceed.
 A: First of all, observe that:
$$
\theta_1 X+\theta_2Y=\frac{\theta_1+\theta_2}2 (X+Y)+\frac{\theta_1-\theta_2}2 (X-Y).
$$
Note that $M_X(\theta)=M_Y(\theta)=g(\theta)$ because $X$ and $Y$ are coming from the same distribution. Using the independence of $X+Y$ and $X-Y$, and also $X$ and $Y$, we get:
$$
\mathbb E\left(e^{\theta_1 X+\theta_2Y}\right)=\mathbb E\left(e^{\frac{\theta_1+\theta_2}2 (X+Y)}\right)\mathbb E\left(e^{\frac{\theta_1-\theta_2}2 (X-Y)}\right)\implies\\
g(\theta_1)g(\theta_2)=g(\frac{\theta_1+\theta_2}2)^2g(\frac{\theta_1-\theta_2}2)g(\frac{\theta_2-\theta_1}2).
$$
Now choose $f(\theta)=\log(g(\theta))$. We arrive at following functional equation:
$$
f(\theta_1)+f(\theta_2)=2f(\frac{\theta_1+\theta_2}2)+f(\frac{\theta_1-\theta_2}2)+f(\frac{\theta_2-\theta_1}2).
$$

Solving the functional equation:
With change of variables, we get:
$$
f(x+y)+f(x-y)=2f(x)+f(y)+f(-y).
$$
We want to prove that there are constants $\alpha$ and $\beta$ such that:
$$
f(x)=\beta x^2+\alpha x.
$$
In order to do that we follow this method. We first prove the claim for integers. Then we prove it for inverse of integers $\frac 1n$($n\neq 0$), and then for all rational numbers. Finally by continuity of $f$, we derive that for all $x\in\mathbb R$.
Note that $f(0)=0$ and by strong induction we can prove that for all $n\in\mathbb N$ and all $x\in\mathbb R$:
$$
f(nx)=nf(x)+\frac{n(n-1)}2 (f(x)+f(-x))=\frac{n(n+1)}2 f(x)+\frac{n(n-1)}{2}f(-x) (\star).
$$
The equation $(\star)$, is the core of the proof. 
To prove the claim for integers, choose $x=1$, $\alpha=\frac{f(1)-f(-1)}2$ $\beta=\frac{f(1)+f(-1)}2$ and we can see that:
$$
f(n)=\frac{n(n+1)}2 f(1)+\frac{n(n-1)}{2}f(-1)=\beta n^2+\alpha n.
$$
Note that for negative $n$, we have:
$$
f(n(-x))=\frac{n(n+1)}2 f(-x)+\frac{n(n-1)}{2}f(x)\\
=\frac{-n(-n+1)}2 f(x)+\frac{-n(-n-1)}{2}f(-x)=f((-n)x)
$$
which means that the claim is also true for negative $n$.
On the other hand, if we choose $x=\frac 1n$ and $x=-\frac 1n$ in $(\star)$, we can see that:
$$
f(1)=\frac{n(n+1)}2 f(\frac 1n)+\frac{n(n-1)}{2}f(-\frac 1n)\\
f(-1)=\frac{n(n+1)}2 f(-\frac 1n)+\frac{n(n-1)}{2}f(\frac 1n).
$$
By solving these two equations in terms of $f(1)$ and $f(-1)$ we have:
$$
f(\frac 1n)=\frac{\frac 1n(\frac 1n+1)}2 f(1)+\frac{\frac 1n(\frac 1n-1)}{2}f(-1)\\
f(-\frac 1n)=\frac{\frac 1n(\frac 1n+1)}2 f(-1)+\frac{\frac 1n(\frac 1n-1)}{2}f(1)
$$
Now we have proved that for all $\frac 1n$:
$$
f(\frac 1n)=\beta\frac 1{n^2}+\alpha\frac{1}n.
$$
For rational numbers, we can see from $\star$:
$$
f(\frac mn)=\frac{m(m+1)}2 f(\frac 1n)+\frac{m(m-1)}{2}f(-\frac 1n)\\
=\frac{\frac mn(\frac mn+1)}2 f(1)+\frac{\frac mn(\frac mn-1)}{2}f(-1)
$$
So for all rational numbers $q\in\mathbb Q$ we have proved that:
$$
f(q)=\frac{q(q+1)}2 f(1)+\frac{q(q-1)}{2}f(-1)=\beta q^2+\alpha q
$$
By continuity of characteristic function, we have for all $x$:
$$
f(x)=\frac{x(x+1)}2 f(1)+\frac{x(x-1)}{2}f(-1)=\beta x^2+\alpha x.
$$

From the solution of preceding functional equation, we have:
$$
f(\theta)=\beta\theta^2+\alpha \theta\implies g(\theta)=\exp(\beta\theta^2+\alpha \theta)
$$
where by considering the variance of $X$ and its mean value,  we can see $\beta=\frac 12$ and $\alpha=0$. 
Remark: Here we have proved a more general statement regardless of variance and mean value. 
