Assuming that $X_0=a \in [0,1]$ where $X_n$ is a martingale and $$P(X_{n+1}=\frac{X_n}{2}|\textit F_n)=1-X_n,\>\>\>P(X_{n+1}=\frac{1+X_n}{2}|F_n)=X_n$$
How may we show that $\bf{E[(X_{n+1}-X_n)^2]=\frac{1}{4}E[X_n(1-X_n)}$
My first attempt was to rearrage the probabilities of each term so that for example
$$P(X_{n+1}=\frac{X_n}{2}|\textit F_n)=\frac{E[X_{n+1}|[\frac{X_n}{2}|\textit F_n]]}{X_{n+1}}$$ and then where $\frac{1}{2}X_n|\textit{F_n}=\frac{1}{2}X_{n-1}$ so that $$P(X_{n+1}=\frac{X_n}{2}|\textit F_n)=\frac{E[X_{n+1}|\frac{1}{2}X_{n-1}]}{X_{n+1}}=1-X_n$$
But frankly I'm just getting more and more confused on how to proceed whilst being slightly overwhelmed by each tidbit of intuition that arrives with an additional variable.
It's easy demonstrate that $\mathbb{E}((X_{n+1}-X_n)^2) = \mathbb{E}(X_{n+1}^2 - X_{n}^2)$ (because $X_n$ is martingale). You can see that \begin{eqnarray} \mathbb{E}(X_{n+1}^2) &=& \mathbb{E}(\mathbb{E}(X_{n+1}^2|\mathcal{F}_n))\\ &=& \mathbb{E}(\mathbb{E}(X_{n+1}^2 1_{\{X_{n+1} = \frac{X_n}{2}\}}|\mathcal{F}_n)) + \mathbb{E}(\mathbb{E}(X_{n+1}^2 1_{\{X_{n+1} = \frac{1 + X_n}{2}\}} |\mathcal{F}_n)) \\ &=& \mathbb{E}(\frac{X_n^2}{4}\mathbb{P}(X_{n+1} = \frac{X_n}{2}|\mathcal{F}_n)) + \mathbb{E}((\frac{1+X_n}{2})^2\mathbb{P}(X_{n+1} = \frac{1+X_n}{2}|\mathcal{F}_n))\\ & =& \mathbb{E}(\frac{X_n^2}{4}(1-X_n) + (\frac{1+X_n}{2})^2X_n)\\ &=& \mathbb{E}(\frac{1}{4}X_n + \frac{3}{4}X_n^2) \end{eqnarray} then $\mathbb{E}(X_{n+1}^2 - X_{n}^2) = \mathbb{E}(\frac{1}{4}X_n + \frac{3}{4}X_n^2 - X_n^2) = \frac{1}{4}\mathbb{E}(X_n(1-X_n))$