that $r'\mid \gcd(d(x),G_{2k-1}(x))$. By the inductive hypothesis we know that $\gcd(d(x), G_{2k-1}(x)) = \rho$. Thus, $r'\mid \rho$. It is easy to see that $\gcd(d(x),G_{2k+1}(x))$ divides $r'$. So, $r'=\gcd(G_1(x), d(x))$.

We now show that depending on the type of the sequence it holds that $\gcd(d(x),G_1(x))=G_1$. If $\{G_n(x)\}$ is a GFP of Fibonacci type, by definition of $p(x)$ we that $G_1(x)=1$ (see comments after Binet formula ). Suppose that $\{G_n(x)\}$ is a GFP of Lucas type. Recall that $2p_{1}(x)=p_{0}(x)d(x)$ and that $|p_{0}(x)|=1 \text { or } 2$. The conclusion is straightforward since $G_1(x)=(a(x)+b(x))/\alpha=d(x)/\alpha$.

Proof of part (2). Let $S(n)$ be the statement $$\rho=\gcd(d(x), G_{2n}(x))\text{  for  }n\ge 1.$$ To prove $S(2)$ we suppose that $\gcd(d(x), G_{4}(x))=r$. Thus, $r$ divides any linear combination of $d(x)$ and $G_4(x)$. Therefore, $r$ divides $G_4(x)-d(x) G_3(x)$. This and given that $G_4(x)= d(x) G_{3}(x) + g(x) G_{2}(x)$, imply that $r\mid g(x)G_{2}(x)$. Therefore, $r\mid \gcd(d(x),g(x)G_{2}(x))$. Since $\gcd(d(x), g(x))=1$, we have that $r\mid \rho$. It is easy to see that $\rho \mid r$. Thus, $r=\gcd(d(x),G_2(x))$. This proves $S(2)$.

We suppose that $S(n)$ is true for $n=k-1$. That is, suppose that $\gcd(d(x), G_{2k-2}(x)) = \rho$. To prove $S(k)$ we suppose that $\gcd(d(x), G_{2k}(x))=r'$. Thus, $r'$ divides any linear combination of $d(x)$ and 