Let $\{F_n(z)\,|\, n\geq 1\}$ be the Fibonacci polynomials, defined recursively by $F_1=1, F_2=z$ and $F_{n+2}=zF_{n+1}+F_n$. Now consider the polynomial $$\varphi_{n,m}(z)=4-F_n^2F_m^2(z^2+3).$$ I would like to show that $\varphi_{n,m}(z)\neq f(z)f(-z)$ for any $f(z)\in\mathbb Z[z]$ for a large family of odd $n,m$. When $n=1$, $m=1$ or $n=m$, the polynomials do split as $f(z)f(-z)$, but in all other cases they appear to be irreducible. I used Mathematica to check all other pairs of odd $n,m$ less than $100,$ and they all are irreducible.
So my question is:
Question: How can we prove that $\varphi_{n,m}$ is irreducible in $\mathbb Z[z]$ for all odd $n,m$ where $n\neq m$ and, $n,m>1$?
For example, the simplest cases are $$\varphi_{3,5}=1 - 25 z^2 - 80 z^4 - 126 z^6 - 106 z^8 - 48 z^{10} - 11 z^{12} - z^{14}$$ $$\varphi_{3,7}=1 - 43 z^2 - 227 z^4 - 569 z^6 - 787 z^8 - 645 z^{10} - 320 z^{12} - 94 z^{14} - 15 z^{16} - z^{18}$$ $$\varphi_{5,7}=1 - 55 z^2 - 405 z^4 - 1557 z^6 - 3440 z^8 - 4594 z^{10} - 3860 z^{12} - 2086 z^{14} - 724 z^{16} - 156 z^{18} - 19 z^{20} - z^{22}$$ Even if we can't show irreducibility, I would still be very interested in showing that $\varphi_{n,m}(z)$ cannot be factored as $f(z)f(-z)$ for an infinite family of odd pairs $n,m$.
Edit: In case anyone is curious about the context, this question arose in my joint paper with Vajira Manathunga, The Conway polynomial and Amphicheiral knots. See the discussion following Proposition 3.13. Note also that Proposition 3.14 gives a sketch of the proof that the polynomials split in the $n=1$ and $n=m$ cases.