Let $X$ and $Y$ be i.i.d. Suppose further that $X+Y$ and $X-Y$ are independent. Show that $\varphi_{X}(2u)=(\varphi_{X}(u))^{3}\varphi_{X}(-u)$.
What I tried to do was work backwards, starting with the right side: $(\varphi_{X}(u))^{3}\varphi_{X}(-u)=(\varphi_{X})^{3}\overline{\varphi_{X}(u)} = E\{exp(i\langle u, X \rangle)\}^{3} E\{exp(-i\langle u, X\rangle)\}$. Then, by independence, $= \left[\prod_{j=1}^{n}\varphi_{x _{j}}(u_{j})\right]^{3}\prod_{j=1}^{n}\varphi_{x_{j}}(-u_{j})$.
I'm really not sure if I'm approaching this right, or how I'm going to get the two sides to equal each other. I'm a little lost, so I would REALLY appreciate help on this one! Thank you so much in advance!
Denote $Z = X+Y$ and $W=X-Y$. Since $X$ and $Y$ are independent then, by the properties of characteristic functions
$$\phi_Z(u) = \phi_X(u)\phi_Y(u) = (\phi_X(u))^2$$
and
$$\phi_W(u) = \phi_X(u)\phi_Y(-u) = \phi_X(u)\phi_X(-u)$$
since $X$ and $Y$ are also identical.
So $$(\phi_X(u))^3\phi_X(-u) = \phi_Z(u)\phi_W(u)$$
Can you take it from here?
ADDENDUM (Now that the OP has found the way, I am completing this answer).
Consider the random variable $S = Z+W = 2X$. Since $Z$ and $W$ are independent, then
$$\phi_S(u) = \phi_Z(u)\phi_W(u)$$ But also, since $S=2X$ $$\phi_S(u) = \phi_X(2u)$$
Linking equalities, we obtain
$$(\phi_X(u))^3\phi_X(-u) = \phi_X(2u)$$