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

We suppose that $H(k)$ is true. That is, suppose that there is a polynomial $T_{k}(x)$ such that $$\label{formula2:for:Dic2}
G_{m(2k)+r}^{*}(x)=G_{m}^{*}(x)T_k(x)+(-1)^{(m+1)k}(g(x))^{mk} G_{r}^{*}(x).$$

We prove that $H(k+1)$ is true. Notice that $G_{2m(k+1)+r}^{*}(x)=G_{((2k+1)m+r)+m}^{*}(x)$. Therefore, from part (1) we have that $$G_{2m(k+1)+r}^{*}(x)=\alpha G_{m}^{*}(x)G_{(2k+1)m+r}^{*}(x)+(-1)^{m+1}(g(x))^m G_{2mk+r}^{*}(x).$$ This and $H(k)$ (see (\[formula2:for:Dic2\])), imply that $G_{m(2(k+1))+r}^{*}(x)$ equals $$\alpha G_{m}^{*}(x)G_{(2k+1)m+r}^{*}(x)+(-1)^{m+1}(g(x))^mG_{m}^{*}(x)T_{k} (x)+(-1)^{(k+1)(m+1)}(g(x))^{m(k+1)}G_{r}^{*}(x).$$ Therefore, $G_{m(2(k+1))+r}^{*}(x)$ equals $$G_{m}^{*}(x) \left[\alpha G_{(2k+1)m+r}^{*}(x)+(-1)^{m+1}(g(x))^mT_{k} (x)\right]+(-1)^{(m+1)(k+1)}(g(x))^{m(k+1)} G_{r}^{*}(x).$$ This with $T_{k+1}(x):=\alpha G_{(2k+2)m+r}^{*}(x)+(-1)^{m}(g(x))^mT_{k} (x)$, implies $H(k+1)$.

We finally prove part (3) by induction. Since $G_{n}^{*}(x)$ is of the Lucas type, by the Binet formula it is easy to see that $G_0(x)=2/\alpha$. Let $S(n)$ be the statement: for every positive integer $n$ there is a polynomial $T_n(x)$ such that $G_{2^nr}^{*}(x)= G_{r}^{*}(x)T_n(x)+(2/\alpha)g^{2^{n-1}r}(x)$.

Proof of $S(2)$. From part (1) we have $G_{2^2r}^{*}(x)= G_{2r+2r}^{*}(x)=\alpha(G_{2r}^{*}(x))^2-(2/\alpha)(g(x))^{2r}$. Applying again the result in part (1) for $G_{2r}^{*}(x)$ (and simplifying) we obtain $$\begin{aligned}
G_{2^2r}^{*}(x)&=&\alpha[\alpha(G_{r}^{*}(x))^2-\frac{2}{\alpha}(-g(x))^{r}]^2-\frac{2}{\alpha}(g(x))^{2r}\\
&=&G_{r}^{*}(x)[\alpha^3(G_{r}^{*}(x))^3+(-1)^{r+1}4\alpha G_r^{*}(x)(g(x))^{r}]+\frac{4}{\alpha}(g(x))^{2r} -\frac{2}{\alpha}(g(x))^{2r}\\
&=&G_{r}^{*}(x)T_2(x)+\frac{2}{\alpha}(g(x))^{2r}.\end{aligned}$$ where $T_2(x)=\alpha^3(G_{r}^{*}(x))^3+(-1)^{r+1}4\alpha G_{r}^{*}(x)(g(x))^{r}$. This proves $S(2)$.

