The Fibonacci polynomials are defined recursively as functions of one variable: $$ F_0(z) = 0 \\ F_1(z) = 1 \\ \forall n > 1: F_n(z) = zF_{n-1}(z) + F_{n-2}(z) $$ Thus for example $F_5(z) = 1+3z^2 + z^4$, and $$F_6(z) = 3z+4z^3+z^5 =z(1+z^2)(3+z^2) = F_2(z) F_3(z) (3+z^2) $$
It appears from several hundred examples that if a prime $p$ divides $n$ then $F_p(z)$ is a factor of $F_n(z)$. If this conjecture is true, then $F_n(z)$ is always reducible in $\Bbb{Z}(z)$ (that is, $F_n(z)$ is factorable into more than one polynomial with degree at least one and integer coefficients) whenever $n$ is composite. I suspect this is not too hard to prove.
It also appears, again from a large number of examples, that for prime $p$, $F_p(z)$ is always irreducible in $\Bbb{Z}(z)$. But examples do not constitute a proof, and I don't see a good way of attacking the problem of proving irreducibility.
My question is, does anybody know of a proof that for prime $p$, $F_p(z)$ is always irreducible in $\Bbb{Z}(z)$?