of two, then for all odd divisors $q$ of $m$, it follows that $G_{m/q}^{*}(x)$ divides $G_{m}^{*}(x)$. More over $G_{m/q}^{*}(x)$ is of the Lucas type.

It is easy to see that $E_2(m/q)=E_2(m)$. Therefore, the conclusion follows by Proposition \[propiedadDivision\].

\[propiedadDivisioncase2\] Let $d_{k}=\gcd(G_0 ^*(x),G_{k} ^{*}(x))$ where $G_{k}^{*}(x)$ is GFP of the Lucas type. Suppose that there is an integer $k'>0$ such that $d_{k'}=2$. If $m$ is the minimum positive integer such that $d_{m}=2$, then $m|n$ if and only if $d_{n}=2$.

We suppose $m$ is the minimum positive integer such that $d_{m}=2$. Suppose that $m|n$, by Proposition \[propiedadDivision\] we know that $\gcd(G^* _{m}(x),G_{n}^{*}(x))=G_{m}^{*}(x)$ (we recall that $G^*_{0}(x)=p_0(x)$ and $|p_0(x)|=1$ or $2$). This and the fact that $2|G_{m} ^*(x)$, implies that $\gcd(G_{0} ^*(x),G_{n} ^*(x))=2$. This proves that $d_{n}=2$.

Suppose that there is $n\in \mathbb{N}-\{m\}$ that satisfies that $d_{n}=2$ (note $2|\gcd(G^{*}_{m}(x),G_{n} ^*(x)$). From the division algorithms we have that there are integer $q$ and $r$ such that $n=mq+r$ where $0\le r< m$. This and Proposition \[Dic2\] part (2), imply that $$\gcd(G^{*} _{m}(x),G_{n} ^*(x))=
    \left\{
         \begin{array}{ll}
            \gcd(G^* _{m}(x),(g(x))^{(t-1)m+r} G_{m-r}^{*}(x)) & \hbox{ if } $q$ \hbox{ is odd;} \\
            \gcd(G^* _{m}(x),(g(x))^{mt} G_{r}^{*}(x))             & \hbox{ if }  $q$ \hbox{ is even.}
        \end{array}
    \right.$$ This, Lemma \[gcdlemmas\] part (3), 