How to prove the sequence of $p_n(x)$ are all integer polynomials on x? for $x \in \mathbb R ,|x|< \frac12$
define:
$A_n(x)$
solution of the relation:
$A_{n+1}(x) = \frac{x^2}{1-A_n(x)}$
$A_0(x) = 0$
(I found that $A_n(x) = 2x^2 \cdot \frac{(1+\sqrt{1-4x^2})^n - (1 - \sqrt{1-4x^2})^n}{(1+\sqrt{1-4x^2})^{n+1} - (1 - \sqrt{1-4x^2})^{n+1}}$)
prove that the sequence:
$p_0(x)=1$
$p_{n+1}(x) = p_n(x) \cdot (1-x^2 - A_n(x))$
consists of integer polynomials in $x$

I will sketch a solution assuming the relation given by the OP for $A_n$ which looks correct. First some notations to simplify things: let $c=\sqrt{1-4x^2}$ and $Y_{1,2}=\frac{1 \pm c}{2}$. We will first prove that:
$Y_1^n-Y_2^n=cQ_n(x)$ where $Q_n(x), n \ge 0$ is an integral polynomial in $x$
Since $Y_{1,2}$ are the roots of the equation $Y^2-Y+x^2=0$ it follows by Newton's relations that any symmetric polynomial $\sigma(Y_1,Y_2)$ is an integral polynomial in $x$ and since $Y_1-Y_2=c$ the claim above follows by factoring $Y_1^n-Y_2^n$
But now $1-A_{n+1}(x) = \frac{1-A_n(x)-x^2}{1-A_n(x)}$ implies $(1-A_{n+1}(x))(1-A_{n}(x))=\frac{p_{n+1}(x)}{p_n(x)}$. Noting that also $1-A_{n}(x)=\frac{x^2}{A_{n+1}(x)}$, we get:
$\frac{p_{n+1}(x)}{p_n(x)}=\frac{x^2}{A_{n+2}} \frac{x^2}{A_{n+1}}$ so multiplying these relations gives us:
$p_{n+1}(x)=\Pi_{k=2}^{k=n+2}\frac{x^2}{A_k(x)}\Pi_{k=1}^{k=n+1}\frac{x^2}{A_k(x)}$
By the OP computation $\frac{x^2}{A_k(x)}=\frac{Y_1^{k+1}-Y_2^{k+1}}{Y_1^k-Y_2^k}$,
which immediately gives:
$p_{n+1}(x)=\frac{Y_1^{n+3}-Y_2^{n+3}}{Y_1^2-Y_2^2}\frac{Y_1^{n+2}-Y_2^{n+2}}{Y_1-Y_2}$
Now $Y_1-Y_2=c, Y_1^2-Y_2^2=c$ so:
$p_{n+1}(x)=Q_{n+3}(x)Q_{n+2}(x)$ integral so we are done!
