I'm not looking for an explicit solution, but some suggestions.
Let $(X_n)$ be a martingale adapted to $(\mathcal F_n)$ where, $\forall n,X_{n+1}X_n^{-1}\in L^1$. Show that $\mathbb E(X_{n+1}X_n^{-1})=1$ and $\mathrm{Cov}(X_{n+1}X_n^{-1},X_n X_{n-1}^{-1})=0$.
My thoughts so far: So $X_n$ is $\mathcal F_n$ measurable meaning that, coupled together with $X_{n+1}$ and $\mathcal F_{n+1}$, this is sort of "previsible". Since $X_n$ is both a martingale and "previsible", we have $X_n=X_0$ (almost everywhere). In particular, $\forall n,\mathbb E(X_{n+1}X_0^{-1})=\mathbb E(X_n X_0^{-1})$, and taking $n=0$ for the base case the RHS is just $1$ and I can induct on the relation to show that this is always $1$ for any $n$.
I'm like 90% sure this is wrong and I'm misunderstanding something, but I'm very lost at the moment as for an approach. A hint would be appreciated.