If $E(X_n^2)<\infty$, then for a Martingale $E(X_n^2)<M$ iff $\sum_{n=1}^\infty E[(X_n-X_{n-1})^2]<\infty$

Let $$\{X_n\}_{n\geq0}$$ be a martingale with $$E(X_n^2)<\infty$$ for all $$n$$. How to prove that:

$$E(X_n^2) for all $$n$$, if and only if $$\sum_{n=1}^\infty E[(X_n-X_{n-1})^2]<\infty$$.

The hunch is to use the optional stopping time theorem, but it's hard to see how to apply it to the case. A tip is very much appreciated.

It is in fact related to the quadratic variation process of $$X_n$$. Note that $$X_n^2 - \sum_{k=0}^n E[(X_k-X_{k-1})^2\;|\mathcal{F}_{k-1}],\quad n\geq0,$$ is a $$\{\mathcal{F}_n\}$$-martingale.
By the martingale property you can show $$E[(X_n - X_{n-1})^2] = E[X_n^2] - E[X_{n-1}^2].$$