If $\frac{X+Y}{\sqrt{2}}\stackrel{d}{=}X\stackrel{d}{=}Y$, use the CLT to show that $X,Y\in N(0,1)$ 
(a) Suppose that $X$ and $Y$ are i.i.d. $N(0,1)$. Show $\frac{X+Y}{\sqrt{2}}\stackrel{d}{=}X\stackrel{d}{=}Y$.


(b) Conversely: Suppose $X$ and $Y$ are i.i.d. with mean $0$ and variance $1$, and suppose further that
\begin{equation*}
\frac{X+Y}{\sqrt{2}}\stackrel{d}{=}X\stackrel{d}{=}Y
\end{equation*}
Show that both $X$ and $Y$ have a $N(0,1)$ distribution. (Use the Central Limit Theorem)

I have solved part (a) as follows:
(a) Since $X,Y\in N(0,1)$, we have that: $\phi_X=\phi_Y=\exp\big(\frac{-t^2}{2}\big)$, thus:
\begin{align*}
\phi_{\frac{X+Y}{\sqrt{2}}}&=\phi_{\frac{X}{\sqrt{2}}+\frac{Y}{\sqrt{2}}}\\
&=\phi_{\frac{X}{\sqrt{2}}}\phi_{\frac{Y}{\sqrt{2}}} \quad\text{by independence}\\
&=\phi_X\big(\frac{t}{\sqrt{2}}\big)\phi_Y\big(\frac{t}{\sqrt{2}}\big)\\
&=\exp\big(\frac{-(t/\sqrt{2})^2}{2}\big)\exp\big(\frac{-(t/\sqrt{2})^2}{2}\big)\\
&=\exp\big(\frac{-t^2}{4}+\frac{-t^2}{4}\big)\\
&=\exp\big(\frac{-t^2}{2}\big)\\
&=\phi_X=\phi_Y\\
&\iff \frac{X+Y}{\sqrt{2}}\stackrel{d}{=}X\stackrel{d}{=}Y
\end{align*}
but I am stuck on part (b). In particular, I am not sure how to bring the CLT into it, any help would be greatly appreciated.
 A: Hint: For part b), think of the following steps: 1) If $n=2^k$ for some $k$, $X_1,X_2, X_1',X_1'',X_2',X_2'',X_i$ iid copies of $X$, then
$
X \stackrel{D}{=} \frac{X_1+X_2}{\sqrt{2}} \stackrel{D}{=} \frac{(1/\sqrt{2})[X_1'+X_1''+X_2'+X_2'']}{\sqrt{2}} \stackrel{D}{=} \frac{X_1'+X_1''+X_2'+X_2''}{\sqrt{4}} \stackrel{D}{=}\cdots \stackrel{D}{=} \frac{1}{\sqrt{n}} \sum_{i=1}^n X_i.
$


*What does the CLT have to say about the asymptotic distribution of the RHS


*Since the LHS distribution does not depend on $n$, it must be equal to the limiting distribution of the RHS.
A: With the help of @LostStatistician18's hint I have a solution that goes into more detail than the solution on the linked page in the comments, so I will post it here in hopes that it might help someone in the future.
Let $\{X_n\}_{n\ge1}$ be independent observations of $X$ and $\{Y_n\}_{n\ge1}$ be independent observations of $Y$. Then $X_n\stackrel{d}{=}X$ and $Y_n\stackrel{d}{=}Y$ for all $n$. Thus, we have that:
\begin{align*}
X\stackrel{d}{=}\frac{X_n+Y_n}{\sqrt{2}} \quad\text{for all $n$}
\end{align*}
So in particular, $X\stackrel{d}{=}\frac{X_1+Y_1}{\sqrt{2}}$ but we also have that: $X_1\stackrel{d}{=}X\stackrel{d}{=}\frac{X_2+Y_2}{\sqrt{2}}$ and $Y_1\stackrel{d}{=}Y\stackrel{d}{=}\frac{X_3+Y_3}{\sqrt{2}}$, and so:
\begin{align*}
X&\stackrel{d}{=}\frac{\frac{1}{\sqrt{2}}(X_2+Y_2+X_3+Y_3)}{\sqrt{2}}\\
&=\frac{X_2+Y_2+X_3+Y_3}{\sqrt{4}}
\end{align*}
But now $X_2\stackrel{d}{=}X\stackrel{d}{=}\frac{X_4+Y_4}{\sqrt{2}}$, $X_3\stackrel{d}{=}X\stackrel{d}{=}\frac{X_5+Y_5}{\sqrt{2}}$, $Y_2\stackrel{d}{=}Y\stackrel{d}{=}\frac{X_6+Y_6}{\sqrt{2}}$ and $Y_3\stackrel{d}{=}Y\stackrel{d}{=}\frac{X_7+Y_7}{\sqrt{2}}$, so iterating the process above gives:
\begin{align*}
X&\stackrel{d}{=}\frac{\frac{1}{\sqrt{2}}(X_4+Y_4+X_5+Y_5+X_6+Y_6+X_7+Y_7)}{\sqrt{4}}\\
&=\frac{X_4+Y_4+X_5+Y_5+X_6+Y_6+X_7+Y_7}{\sqrt{8}}\\
&\stackrel{d}{=}\frac{X_1+X_2+X_3+X_4+X_5+X_6+X_7+X_8}{\sqrt{8}} \quad\text{since $X_n\stackrel{d}{=}X\stackrel{d}{=}Y\stackrel{d}{=}Y_m$ for all $n,m$}
\end{align*}
Continuing in this fashion, we see that:
\begin{align*}
X&\stackrel{d}{=}\frac{\sum\limits_{i=1}^{2^n}X_i}{\sqrt{2^n}}\\
&=\frac{S_{2^n}-0\cdot n}{1\cdot\sqrt{2^n}}\xrightarrow{d}Z\in N(0,1)\,\,\text{by the CLT}
\end{align*}
And so indeed: $X\in N(0,1)$ as we wished to show and an identical argument shows that $Y\in N(0,1)$ as well.
