Given $(R_n)$ a sequence of independent random variables such that $P(R_n = -1) = P(R_n = 1) = \frac{1}{2}$, we want to show that $X_n := (R_1 + \ldots + R_n)^2 - n$ is indeed a martingale.
Now in the solution of this example, things go like
$$X_{n+1} = (R_1 + \ldots R_{n+1})^2 - (n+1)$$ hence we start with
$$ \begin{align} X_{n+1} - X_n & = (R_1 + \ldots R_{n+1})^2 - (n+1) - (R_1 + \ldots + R_n)^2 - n \\\\ & = (R_1 + \ldots R_n)^2 + R_{n+1}^2 - n - 1 - 2R_{n+1}(R_1 + \ldots + R_n) - (R_1 + \ldots + R_n)^2 - n \\\\ & = 2R_{n+1}(R_1 + \ldots + R_n) \end{align} $$
Where he says $R_{n+1}^2 = 1$
I don't get why $R_{n+1}^2 = 1$ and also I don't get why the term $-n-n = -2n$ has vanished.
Then it goes on like ($H$ belongs to the filtration $\mathcal{F}_n$ to which the martingales are measurable) $$\int_H (X_{n+1} - X_n)d\mathbb{P} = \int_H R_{n+1}\cdot 2\mathbb{1}_H(R_1 + \ldots R_n)d\mathbb{P}$$
Since $R_{n+1}$ is independent from $\mathcal{F}_n$ we get
$$ = \underbrace{\mathbb{E}[R_{n+1}]}_{=0}\int2\mathbb{1}_HR_{n+1} = \mathbb{E}[R_{n+1}]\cdot 2\mathbb{E}[H]\mathbb{E}[R_1 + \ldots R_n] = 0$$
I think there are some mistakes in this solution but I'm still a beginner in these topics so I'm asking you for clarifications about the whole exercise.
Thank you!