A factorization problem involving Fibonacci and Lucas Polynomials Consider a sequence of polynomial $\{w_n(x)|\, n\geq 0\}$ which are defined recursively by $w_n(x)=xw_{n-1}(x)+w_{n-2}(x)$. With $w_0(x)=0$ and $w_1(x)=1$, one gets the so-called Fibonacci polynomials $w_n(x)=F_n(x)$. With $w_0(x)=2$ and $w_1(x)=x$, one gets the Lucas polynomials $w_n(x)=L_n(x)$. While working on another problem, I noticed that for any odd $n$ the polynomial $F_n(z)^2-L_{2n}(z)+2$ appears to factor as $\phi_n(z)\phi_n(-z)$ for some polynomial $\phi_n(z)$ with integer coefficients. For example, when $n=1$ we have 
\begin{align*}
F_1(z)^2-L_{2}(z)+2&=1-(z^2+2)+2\\&=(1-z)(1+z)\end{align*}
and for $n=3$ we have
\begin{align*}F_3(z)^2-L_{6}(z)+2&=(z^2+1)^2-(z^6+6z^4+ 9 z^2 +2 )+2\\
&=(1+3z+z^2+z^3)(1-3z+z^2-z^3)
\end{align*}
Does anyone see how to prove this factorization property?
I have know idea how to tag this question. Suggestions are appreciated.

The key observation is that, for $n$ odd, $L_{2n}(x) = L_{n}(x)^2 + 2$; this is straightforward to show by induction, as a subcase of the more general identity $L_{m+n}(x) = L_{m}(x)L_{n}(x) + (-1)^{m-n}L_{m-n}(x)$. 
Now, your polynomial $F_{n}(z)^2 - L_{2n}(z) + 2$ reduces to $F_{n}(z)^2 - L_{n}(z)^2 = (F_{n}(z) - L_{n}(z))(F_{n}(z) + L_{n}(z))$. Finally, since, for odd $n$, $F_{n}(z)$ is even and $L_n(z)$ is odd (again this is easy to show via induction), it follows that if you let $G_{n}(z) = F_{n}(z) - L_{n}(z)$, then your polynomial is simply $G_{n}(z)G_{n}(-z)$. 
