Show that these non-linear recursions produce integers only The recurrence is of third order:

Start with
  \begin{align*}a_0(x)&=1\\ a_1(x)&=1\\ a_2(x)&=x \end{align*} and then
  \begin{align*}a_{n+3}(x)&=\frac{a_{n+2}^2(x)-a_{n+1}^2(x)}{a_{n}(x)}.\end{align*}

The calculation and factorization of $a_n(x)$ for $3\le n \le 6$ gives the following results: 
\begin{align*} a_3(x)&= (x+1)(x-1)\\
a_4(x)&=(x^2 + x -1)(x^2-x-1)\\
a_5(x)&=x(x^2 -2)(x^4-4x^2+2)\\
a_6(x)&=(x^6 +x^5-5x^4-4x^3+6x^2+3x-1)(x^6 -x^5-5x^4+4x^3+6x^2-3x-1)\\
\end{align*}
Questions: 


*

*Any reference for this ?  

*Show that $a_n(x)$ is an integer coefficient polynomial.  It would be of
degree $F_{n+1} -1$, where  $F_n$ is a Fibonacci number.

*Is it true that $a_n(x)$ divides $a_{2n+1}(x)$?

*For $k \in \mathbb N, k>1$, is there a general close form for $a_n(k)$?


For $k=2$, we obviously have $a_n(2)=F_{n+1}$.
For $k=3$, we have $a_n(3)=F_{2F_{n+1}}$. 
What about $a_n(4)$?

Nice recursion!! However, let us change the name and indexing of your sequence. Notice that if $\,U_n(x)$ is the Chebyshev polynomial of the second kind, then $\, U_n(x/2)$ is a polynomial in $x$ with integer coefficients. Define $\, P_n(x) := U_{F_n - 1}(x/2).\,$ Your sequence $\, a_n(x) = P_{n+1}(x).\,$ Please note the different indexing.
Check the following equations hold true for all integer $\,n$
$$ P_1(x) = P_2(x) = 1, \,\, P_3(x) = x,\,\,
 P_n(x)P_{n+3}(x) = P_{n+2}^2(x) - P_{n+1}^2(x).$$
Both $\,U_{n-1}(\cos\theta) = \sin(n\theta)/\sin(\theta)\,$ and $\,F_n = U_{n-1}(i/2)/i^{n-1}\,$ are divisibility sequences,
hence $\,P_n(x)\,$ divides $\,P_{kn}(x)\,$ for all integer $\,n,k\,$ where $\,P_n(x)\ne 0.\,$ We also have
$\,P_n(x) = -(-1)^n P_{-n}(x)\,$ for all integer $\,n.\,$
