$G_{2k}(x)$. Therefore, $r'$ divides $G_{2k}(x)-d(x) G_{2k-1}(x)$. This and given that $G_{2k}(x)= d(x) G_{2k-1}(x) + g(x) G_{2k-2}(x)$, imply that $r'\mid g(x)G_{2k-2}(x)$. Therefore, $r'\mid \gcd(d(x),g(x)G_{2k-2}(x))$. Since $\gcd(d(x), g(x))=1$, we have that $r'\mid \gcd(d(x),G_{2k-2}(x))$. By the inductive hypothesis we know that $\gcd(d(x), G_{2k-2}(x)) = \rho$. Thus, $r'\mid \rho$. It is easy to see that $\gcd(d(x),G_{2k}(x))$ divides $r'$. So, $r'=\gcd(G_2(x), d(x))$.

We observe that for a GFP of Fibonacci type it holds that $G_2^{\prime}(x)=a(x)+b(x)=d(x)$. So, it is clear that $\gcd(G_{2n}^{\prime}(x),d(x))=d(x)$. For a GFP of Lucas type it holds that $G_0^{*}(x)$ is a non-zero constant. Since $G_2^{*}(x)=d(x)G_1^{*}(x)+G_{0}^{*}g(x)$, and $\gcd(d(x), g(x))=1$, we have $$\gcd(d(x),G_2^{*}(x))=\gcd(d(x),d(x)G_1^{*}(x)+G_{0}^{*}g(x))=\gcd(d(x),G_0^{*}(x)g(x))=1.$$

Proof of part (3). We now prove that $\gcd(g(x), G_1(x))=1$ by cases. If $G_1(x)$ is of the Fibonacci type, the conclusion is straightforward. As a second case we suppose that $G_1(x)$ is of the Lucas type. That is $G_1(x)$ satisfies the Binet formula (\[bineformulados\]). Therefore, we have that $$\gcd(g(x), G_1(x))=\gcd(g(x), L_1(x)) = \gcd(g(x), [a+b]/\alpha)=\gcd(g(x), d(x)/\alpha).$$ Since $\gcd(g(x), d(x))=1$, we have that $\gcd(g(x), d(x)/\alpha)=1$. This completes the proof.

\[gcdforproposition4\] If $\{G_{n}(x)\}$ is a GFP polynomial sequence, then for every positive integer $n$ it holds that

(1) $\gcd(G_{n}(x), G_{n+1}(x))$ divides $\gcd(G_{n}(x),g(x)G_{n-1}(x))=\gcd(G_{n}(x),G_{n-1}(x))$,

(2) $\gcd(G_{n}(x), G_{n+2}(x))$ divides $\gcd(G_{n}(x), d(x)G_{n+1}(x))$.

We prove part (1), the proof of 