\\
                G_{m}^{*}(x)T(x)+(-1)^{(m+1)t}(g(x))^{mt} G_{r}^{*}(x),             & \hbox{ if }  $q$ \hbox{ is even}
            \end{array}
        \right.$$

    where $t=\left\lceil \frac{q}{2} \right\rceil$,

(3) if $n>1$, then there is a polynomial $T_n(x)$ such that $$G_{2^nr}^{*}(x)= G_{r}^{*}(x)T_n(x)+(2/\alpha)(g(x))^{2^{n-1}r}.$$

We prove part (1). Since $G_{m}^{*}(x)$ and $G_{n}^{*}(x)$ are of the Lucas type, they both satisfy the Binet formula (\[bineformulados\]). Thus, $$G_{m}^{*}(x)G_{n}^{*}(x)=\left(\frac{a^m+b^m}{\alpha}\right)\left(\frac{a^n+b^n}{\alpha}\right)
=\frac{a^{m+n}+b^{m+n}}{\alpha^2}+\frac{(ab)^{m}\left(a^{n-m}+b^{n-m}\right)}{\alpha^2}.$$ So, $G_{m}^{*}(x)G_{n}^{*}(x)=\left[G_{n+m}^{*}(x)+(ab)^mG_{n-m}^{*}(x)\right]/\alpha.$ This and $ab=-g(x)$, imply that $$G_{m+n}^{*}(x)=\alpha G_{m}^{*}(x)G_{n}^{*}(x)-(-g(x))^{m} G_{n-m}^{*}(x).$$ This completes the proof of part (1).

We first prove part (2) when $q$ is odd by induction. Suppose $q=2t-1$, and let $S(t)$ be the following statement. For every positive integer $t$ there is a polynomial $T_t(x)$ such that $$G_{m(2t-1)+r}^{*}(x)=G_{m}^{*}(x)T_t(x)+(-1)^{m(t-1)+t+r}(g(x))^{(t-1)m+r} G_{m-r}^{*}(x).$$ From part (1), taking $T_1(x)=\alpha G_{r}^{*}(x)$, it is easy to see that $S(t)$ is true if $t=1$.

We suppose that $S(k)$ is true. That is, suppose that there is a polynomial $T_{k}(x)$ such that $$\label{formula1:for:Dic2}
G_{m(2k-1)+r}^{*}(x)=G_{m}^{*}(x)T_k(x)+(-1)^{m(k-1)+t+r}(g(x))^{(k-1)m+r} G_{m-r}^{*}(x).$$

We prove that $S(k+1)$ is true. Notice that $G_{m(2k+1)+r}^{*}(x)=G_{(2km+r)+m}^{*}(x)$. Therefore, from part (1) we have $$G_{m(2k+1)+r}^{*}(x)=\alpha G_{m}^{*}(x)G_{2km+r}^{*}(x)+(-1)^{m+1}g^m(x)G_{m(2k-1)+r}^{*}(x).$$ This and $S(k)$ (see (\[formula1:for:Dic2\])), imply that $G_{m(2(k+1)+r}^{*}(x)$ equals $$\alpha G_{m}^{*}(x)G_{2km+r}^{*}(x)+(-1)^{m+1}(g(x))^{m}G_{m}^{*}(x)T_k (x)+(-1)^{km+(t+1)+r}(g(x))^{km+r} G_{m-r}^{*}(x).$$ Therefore, $G_{m(2(k+1)+r}^{*}(x)$ equals $$G_{m}^{*}(x)[\alpha G_{2km+r}^{*}(x)+(-1)^{m+1}(g(x))^{m}T_k (x)]+(-1)^{km+(t+1)+r}(g(x))^{km+r} G_{m-r}^{*}(x).$$ This, with $T_{k+1}(x):=\alpha G_{2km+r}^{*}(x)+(-1)^{m+1}(g(x))^mT_k (x)$, implies $S(k+1)$. This complete the proof when $q$ is odd.

We now prove part (2) when $q$ 