than $1$. Therefore, there integers $q$ and $r$ with $0<r<n$ such that $n=mq+r$. Then by Proposition \[divisity:Hogat:property1\] part (1) $$\begin{aligned}
  G_{n}^{'}(x) &=& G_{mq+r}^{'}(x) \\
   &=& G_{mq+1}^{'}(x)G_{r}^{'}(x)+g(x)G_{mq}^{'}(x)G_{r-1}^{'}(x)\\
   &=& \left(d(x)G_{mq}^{'}(x)+g(x)G_{mq-1}^{'}(x)\right)G_{r}^{'}(x)+g(x)G_{mq}^{'}(x)G_{r-1}^{'}(x)\\
   &=& d(x)(x) G_{mq}^{'}(x) G_{r}^{'}(x)+g(x)G_{mq-1}^{'}(x)G_{r}^{'}(x)+g(x)G_{mq-1}^{'}(x)G_{r}^{'}(x).\end{aligned}$$ Grouping terms and simplifying we obtain, $$G_{n}^{'}(x)=G_{mq}^{'}(x)G_{r+1}^{'}(x)+g(x)G_{mq-1}^{'}(x)G_{r}^{'}(x).$$ This and the fact that $G_{m}^{'}(x)\mid G_{n}^{'}(x)$ and $G_{m}^{'}(x) \mid G_{mq}^{'}(x)$ imply that $$G_{m}^{'}(x)\mid g(x)G_{mq-1}^{'}(x)G_{r}^{'}(x).$$ From Lemma \[gcdlemmas\] part (3) and Proposition \[gcddistance1;2\] we know that $\gcd(G_{mq}^{'}(x),g(x))=1$ and $\gcd(G_{mq-1}^{'}(x), G_{mq}^{'}(x))=1$, respectively. This implies that $G_{m}^{'}(x)\mid G_{r}^{'}(x).$ That is a contradiction since degree ($G_{r-1}^{'}(x))<$ degree $(G_{m-1}^{'}(x))$. This completes the proof.

\[divisibity:property:first:type\] Let $m$ be a positive integer that is not a power of two. If $G_{m}^{*}(x)$ is a GFP of Lucas type, then for all odd divisors $q$ of $m$, it holds that $G_{m/q}^{*}(x)$ divides $G_{m}^{*}(x)$. More over $G_{m/q}^{*}(x)$ is of the Lucas type.

Let $q$ be an odd divisor of $m$. If $q=1$ the result is obvious. Let us suppose that $q\not = 1$. Therefore, there is an integer $d>1$ such that $m=dq$. Using the Binet formula (\[bineformulados\]), where $a:=a(x)$ and $b:=b(x)$, we have $G_{m}^{*}(x)=G_{dq}^{*}(x)=(a^{dq}+b^{dq})/\alpha$. Let $X=a^{d}$ and $Y=b^{d}$. Using induction it is possible to prove that $X+Y$ divides $X^q+Y^q$. This implies that there is a polynomial $Q(x)$ such that $(X^q+Y^q)/\alpha=Q(x)(X+Y)/\alpha$. 