if $X,Y$ i.i.d $\mathcal{N}(0,1)$, then $X+Y$ is independent of $X-Y$

Question

I found on another thread* that if $(X+Y)$ is independent $(X-Y)$, and if $X,Y$ are i.i.d., then $X,Y$ are $\mathcal{N}(0,1)$ distributed.

Is also the opposite true? Being $X,Y$ i.i.d $\mathcal{N}(0,1)$, then $X+Y$ is independent of $X-Y$?

here * $X$ and $Y$ i.i.d., $X+Y$ and $X-Y$ independent, $\mathbb{E}(X)=0 $and $\mathbb{E}(X^2)=1$. Show $X \sim N(0,1)$

Answer 1

\begin{align*} \operatorname{cov}(X+Y,X-Y)\equiv&\,\mathbb E[(X+Y)(X-Y)]-[\mathbb E(X+Y)][\mathbb E(X-Y)]=\mathbb E(X^2)-\mathbb E(Y^2)-0\times0\\ =&\,1-1=0.\end{align*}

Hence, $X+Y$ and $X-Y$ are uncorrelated. It is not difficult to see that $X+Y$ and $X-Y$ are jointly normal. Now recall that two jointly random normal variables are independent if and only if they are uncorrelated.