$A$ and $B$ are i.i.d RVs, assume $\mathbb{E}[A^2]$ exists, $\mathrm{var}{A}=1$, and that MGF of $A$ exists near zero. $C=(A+B)/\sqrt{2}$ and $D=(A-B)/\sqrt{2}$. Now prove that if. $C$ and $D$ are i.i.d., $A, B, C, D$ are $N(0,1)$ random variables.
Basic computation got me to figure out that the expectation and variance of all four RVs are 0 and 1 respectively, and I also know that RVs that have an MGF such that $M(2s)=(M(s))^4$ have a normal distribution, but I am not sure how to proceed from there.
Put $\phi(t) = Ee^{itA} = Ee^{itB}$. We have
\begin{gather} \phi(\frac{t+s}{\sqrt{2}}) \phi(\frac{t-s}{\sqrt{2}}) = Ee^{i A ( \frac{t+s}{\sqrt{2}})} Ee^{i B (\frac{t-s}{\sqrt{2}})} = Ee^{i( A ( \frac{t+s}{\sqrt{2}}) + B (\frac{t-s}{\sqrt{2}}))}= Ee^{i(t \frac{A+B}{\sqrt{2}} + s \frac{A-B}{\sqrt{2}} )} = Ee^{i(t \frac{A+B}{\sqrt{2}} )} Ee^{i(s \frac{A-B}{\sqrt{2}} )} =\\ = \phi(\frac{t}{\sqrt{2}}) \phi(\frac{t}{\sqrt{2}}) \phi(\frac{s}{\sqrt{2}}) \phi(-\frac{s}{\sqrt{2}}) \text{ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ (1)} \end{gather} Moreover, $Ee^{i(t \frac{A+B}{\sqrt{2}} )} = Ee^{i(t \frac{A-B}{\sqrt{2}} )} $ and thus \begin{gather} \phi(\frac{t}{\sqrt{2}}) \phi(\frac{t}{\sqrt{2}}) = \phi(\frac{t}{\sqrt{2}}) \phi(-\frac{t}{\sqrt{2}}) \text{ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ (2)} \end{gather} Taking $\sqrt{2}t$ and $\sqrt{2}s$ instead of $t$ and $s$ we get from (1) and (2) that $$\phi(t+s) \phi(t-s) = \phi^2(t)\phi(s)\phi(-s), \text{ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ (3)}$$ $$\phi^2(t) = \phi(t) \phi(-t). \text{ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ (4)}$$ Hence $$\phi(t+s) \phi(t-s) = \phi^2(t)\phi^2(s), \text{ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ (5)}.$$ Taking $-t$ instead of $t$ in (4) we get $$\phi^2(-t) = \phi(t) \phi(-t) \text{ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ (6)}$$ It follows from (4) that either $\phi(t) = 0$ or $\phi(t) = \phi(-t)$. It follows from (6) that either $\phi(-t) = 0$ or $\phi(t) = \phi(-t)$. Hence we have $\phi(t) = \phi(-t)$ for all $t$. Thus $\overline{\phi(t)} = \overline{Ee^{itA}} = Ee^{-itA} = \phi(-t) = \phi(t)$ . We got that $\phi(t) \in \mathbb{R}$.
Further, MGF exists near $0$, so $\exists EA^n$ for all $n \ge 1$ and hence $\phi(t) \in C^{\infty}(\mathbb{R})$. But $\phi(0)=1$, thus $\phi(t) \ne 0$ near zero. Put $T = \inf(t>0: \phi(t) = 0)$. If $T < \infty$, then let $t \to T-$ and $s \in (0, \frac{T}2)$. We have $\phi(t), \phi(s) \ne 0$ and it follows from (5) that $\phi(t+s) \ne 0$. We got a contradiction. Hence, $T = \infty$ and $\phi(t) \ne 0$ for all $t \ge 0$. Similarly we get that $\phi(t) \ne 0$ for all $t$. We also know that $\phi(t) \in \mathbb{R}$ is continuous and $\phi(0)=1$. Thus $\phi(t) > 0$.
Put $g(t) = \ln \phi(t)$. Then $g(t) \in C^{\infty}(\mathbb{R})$. It follows from (5) that $$g(t+s) + g(t-s) = 2(g(t) + g(s)).\text{ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ ${}$ (7)}$$
Fix $t \ne 0$. We know that $$g((k+1)t) + g((k-1)t) = 2(g(kt) + g(t))$$ for all integer $k$ and that $\phi(0)=0$. Taking $k=1$ we get that $g(2t) = 4g(t)$ and by induction we get that $g(pt) = p^2g(t)$ for integer $p \ge 0$. Moreover, we may get similarly that $g(pt) = p^2g(t)$ for all integer $p$. Put $u = tq$ for some integer $q$. Thus $$g(\frac{p}{q}u) = g(pt) = p^2g(t) = p^2g(\frac{u}{q}) = \frac{p^2}{q^2} f(u).$$ Hence $g(\frac{p}{q}) = \frac{p^2}{q^2} g(1)$ and by continuity of $g$ we get that $g(x) = cx^2$.
We got that $$Ee^{iAt} = e^{g(t)} = e^{ct^2}$$ for some $c \in \mathbb{R}$. We know that $|Ee^{iAt}| \le E|e^{iAt}| = 1$, hence $c \le 0$. If $c=0$ then $A=0$, $DA = 0$ and we got a contradiction. Hence $c < 0$. Thus $A \sim N(0, \sigma^2)$. But $DA =1$. Hence $A \sim N(0,1)$ and $B \sim N(0,1)$ also. Now it's easy to see that $C \sim N(0,1)$ and $D \sim N(0,1)$ , q.e.d.