$X,Y \sim N(\mu,\sigma^2)$ Suppose $\sigma_X^2=\sigma_Y^2 \neq 0 \implies X+Y,X-Y$ are independent. $X,Y \sim N(\mu,\sigma^2)$
Suppose $\sigma_X^2=\sigma_Y^2 \neq 0 \implies X+Y,X-Y$ are independent.
I know that $\sigma_{X+Y}^2=\sigma_X^2+\sigma_Y^2 , \sigma_{X-Y}^2=\sigma_X^2+\sigma_Y^2$.
If I'll show that $\phi_{X+Y}(t)\cdot\phi_{X-Y}(t)=\phi_{(X+Y)-(X-Y})(t)$ , it's enough ?
$\phi:=$characteristic function
There is another way ?
Thanks !

Without additional assumptions on the joint distribution of $(X,Y)$ (similar questions in MSE assume for example that $X$ and $Y$ are i.i.d) the claim in the OP may not hold.

*

*Consider for example the following situation: $X\sim N(0,1)$, $\epsilon\sim Be(\pm1,1/2)$, $\epsilon$ and $X$ independent, and $Y=\epsilon X$. Since
$$E[e^{itY}]=\frac12\big(E[e^{itX}]+E[e^{-itX}]\Big)=e^{-t^2/2}$$
we have that $Y\sim N(0,1)$. On the other hand,
\begin{align}
E[e^{it(X+Y)}]&=\frac12E[e^{i2tX}]+\frac12=\frac12e^{-2t^2}+\frac12\\
E[e^{it(X-Y)}]&=\frac12E[e^{i2tX}]+\frac12=\frac12e^{-2t^2}+\frac12
\end{align}
but
\begin{align}
E[e^{it(X+Y)+is(X-Y)}]&=\frac12(E[e^{2itX}]+E[e^{2isX}])\\
&=\frac12(e^{-2t^2} + e^{-2s^2})\\
&\neq
E[e^{it(X+Y)}]E[e^{is(X-Y)}]
\end{align}


*If the joint distribution of $X$ and $Y$ is a binormal, then any linear combination $aX+bY$ is normally distributed; furthermore, for any $a,b,c,d$, $(aX+bY,cX+dY)$ is binormal (possibly degenerated). If in addition, $E[X]=\mu=E[Y]$, $\sigma_X=\sigma=\sigma_Y$, and $\operatorname{cov}(X,Y)=0$, then $X+Y\sim N(2\mu,2\sigma^2)$, $X-Y\sim N(0,2\sigma^2)$
and so,
\begin{align}
E[e^{it(X+Y)+is(X-Y)}]&=E[e^{i(t+s)X}e^{i(t-s)Y}]\\
&= e^{i\mu (t+s) -\sigma^2(t+s)^2/2}e^{i\mu (t-s) -\sigma^2(t-s)^2/2}\\
&=e^{2i\mu t -\sigma^2(t^2+s^2)}\\
&=E[e^{it(X+Y)}]E[e^{is(X-Y)}]
\end{align}
