About the property of a sequence of polynomials 
There is a sequence of polynomials $P_n(x)$ such that $P_0(x) = x^3 - 4x$ and, for $n>0$, 
  $$P_n(x) = P_{n-1}(1+x)P_{n-1}(1-x) - 1.$$ 
  Prove that $x^{2016}$ divides $P_{2016}(x)$.

I tried induction, but then I had to prove that $x^n \vert P_n(x)P_n(2 - x) + P_n(-x)P_n(2+x)$
 A: 
We show by induction that $x^n(x-2)(x+2)$ divides $P_n(x)$ for any even integer $n$.

For $n=0$, $P_0=x(x+2)(x-2)$ and the property holds. Let $n>0$. Note that
$$P_{n+2}(x)=(P_{n}(2+x)P_{n}(-x)-1)(P_{n}(2-x)P_{n}(x)-1)-1.$$
If $x^n(x-2)(x+2)$ divides $P_n(x)$ then, for some polynomial $Q_n(x)$,
$$P_{n}(2+x)P_{n}(-x)=(x+2)^{n+1}x^{n+1}(x-2)Q_n(x)$$
and 
$$P_{n}(2-x)P_{n}(x)=-(x+2)x^{n+1}(x-2)^{n+1}Q_n(-x).$$
Therefore, for some polynomial $R_n(x)$,
\begin{align*}
P_{n+2}(x)&=((x+2)^{n+1}x^{n+1}(x-2)Q_n(x)-1)(-(x+2)x^{n+1}(x-2)^{n+1}Q_n(-x)-1)-1\\
&=x^{n+1}(x+2)(x-2)\left(x^{n+1}R_n(x)-(x+2)^nQ_n(x)+(x-2)^nQ_n(-x)\right).
\end{align*}
Since $n$ is even, 
$$\left.\left(x^{n+1}R_n(x)-(x+2)^nQ_n(x)+(x-2)^nQ_n(-x)\right)\right|_{x=0}=0-2^nQ_n(0)+(-2)^nQ_n(0))=0$$
which implies that we have an extra factor $x$ (or, as remarked by Nick Pavlov, since $P_{n+2}$ is even, if the odd polynomial $x^{n+1}$ divides $P_{n+2}$ then $x^{n+2}$ divides it as well).
Hence $x^{n+2}(x+2)(x-2)$ divides $P_{n+2}$ and the proof of the inductive step is complete.
