My solution: After experimenting with small cases we see that $x^{2k}|P_{2k}(x)$.So let's induct on $k$. We have that: $P_{2k+2}(x)=P_{2k+1}(1-x)P_{2k+1}(1+x)-1=\\ =(P_{2k}(2-x)P_{2k}(x)-1)(P_{2k}(x+2)P_{2k}(-x)-1)-1=-P_{2k}(2-x)P_{2k}(x)-P_{2k}(x+2)P_{2k}(-x)+ P_{2k}(x+2)P_{2k}(-x)P_{2k}(x)P_{2k}(2-x)$ but by induction hypothesis we have $x^{2k+2}|P_{2k}(x+2)P_{2k}(-x)P_{2k}(x)P_{2k}(2-x)$ so we want to show that $x^{2k+2}|Q(x)+Q(-x)$ where $Q(x)=P_{2k}(x)P_{2k}(2-x)$. Note that $Q(x)+Q(-x)$ has only even degrees so we just have to look at coefficients at $x^{2k}$. We have $Q(x)=P_{2k}(x)P_{2k}(2-x)=(a_nx^n+...+a_{2k}x^{2k})(a_n2^n+...+a_{2k}2^{2k}+xT(x))$ for some $T(x)$ that isn't zero. So we have that coefficient with $x^{2k}$ in $Q(x)+Q(-x)$ is $2a_{2k}P_{2k}(2)$. Lemma : $P_n(2)=0$ for $n$ even. For $n=0$ is true. We have $P_{n+2}=P_{n+1}(3)P_{n+1}(-1)-1=(P_n(4)P_n(2)-1)(P_n(2)P_n(0)-1)-1=1-1=0$ so we have proven it. Back to the main problem: $Q(x)+Q(-x)=a_0x^{2k}+a_1x^{2k+2}+a_2x^{2k+4}+...$ where $a_0=0$ by Lemma. Now we have proven that $x^{2k}|P_{2k}(x)\implies x^{2k+2}|P_{2k+2}(x)$. So specially for $k=1008$ we v get the desired result.