Bernstein's Theorem (Probability)

I am doing an exercise that establishes the Bernstein's Theorem (it's called that way in french i don't know if it is known with this name in english). It states the following :

Let $(\Omega,\mathcal{A},\mathbb{P}) $ be a probability space. We suppose that $X$ and $Y$ are two independent random variables with same law of probability. Let $U=X+Y$ and $V=X-Y$. Then $U$ is independant of $V$ iff $X$ and $Y$ are two variables of normal distribution.

The goal of the exercise is to show this with a supplementary assumption that is $E(X^2)=E(Y^2)< \infty$.

I have done the first two parts of the exercise with is establishing the $ \implies$ way and we have shown that the following relation characterize the characteristic function of the standard normal distribution, \begin{align} \phi^2=\phi '^2-\phi \phi '' \end{align}

Where $\forall t \in \mathbb{R}, \phi(t)=\mathbb{E}(e^{itX})$ is the characteristic function of X.

For the direct implication of the theorem i am stuck, the exercise suppose that $\mathbb{E}(X)=0, Var(X)=1$, $X$ independant of $Y$ and they are of same probability law.

What i have shown :

$\forall t \in \mathbb{R}, \phi_U(t)=\phi_{X+Y}(t)=\phi_X(t)^2 $ and $ \phi_V(t)=\phi_X(t)\phi_X(-t)$ using independance and the fact that $X$ and $Y$ have same law. Then i showed that, :

$\forall u,v \in \mathbb{R}, \phi_{(U,V)}(u,v)=\phi_U(u) \phi_V(v)=\phi_X(u)^2\phi_X(v)\phi_X(-v) $ because $U$ and $V$ are supposed independant.

And this is where i'm stuck because the next question is

Which relation $\phi_{U,V}$ has to satisfie for $U$ and $V$ to be independant

But i feel like the relation is the one i used to write $\phi_{(U,V)}(u,v)=\phi_U(u) \phi_V(v)$.

The next question is

Deduce that for all $(u,v) \in \mathbb{R}^2$ we have $\phi_X(u+v)\phi_X(u-v)=\phi_X(u)^2\phi_X(v) \phi_X(-v)$.

Then we use this relation to get the one that characterize the characteristic function of the standard normal distribution and then i would be easy to conclude.

What am i not understanding properly ?

Thank you !

Hint: Your equation $$ \phi_{(U,V)}(u,v)=\phi_U(u) \phi_V(v)=\phi_X(u)^2\phi_X(v)\phi_X(-v) $$ is correct. Therefore, in order to answer the next question, you need to prove $$ \phi_{(U,V)}(u,v)\stackrel{?}=\phi_X(u+v)\phi_X(u-v) $$ To prove this, write $$ \phi_{(U,V)}(u,v)=\mathbb E\left[e^{i(uU+vV)}\right],\tag{$*$} $$ substitute $U=X+Y$ and $V=X-Y$ into $(*)$, then rearrange.