we see that the above expression is equal to, $$\left[a^{n+m}(a^2+g(x))+b^{n+m}(b^2+g(x))\right]/(a-b)^{2}.$$ This, with the facts that, $(a^2+g(x))=a(a-b)$ and $(b^2+g(x))=-b(a-b)$, shows that the above expression is equal to $$\left(a^{n+m+1}-b^{n+m+1}\right) /(a-b)=R_{n+m+1}(x).$$

We prove part (2) the proof of part (3) is identical and we omit it. Suppose that $G_{k}^{*}(x)$ is equivalent to $G_{k}^{'}(x)$ and that $G_{k}^{*}(x)$ is of the Lucas type for all $k$. For simplicity let us suppose that $\alpha=1$ (the proof when $\alpha\neq 1$ is similar, so we omit it). Using the Binet formulas (\[bineformulados\]) and (\[bineformulauno\]) we obtain that $G_{n}^{'}(x) G_{m}^{*}(x)-(-g(x))^{m} G_{n-m}^{'}(x)$ equals $$\displaystyle{\frac{(a^{n}-b^{n})(a^{m}+b^{m})-(-g(x))^{m}(a^{n-m}-b^{n-m})}{(a-b)}}.$$ After performing the indicated multiplication and simplifying we obtain that this expression is equal to $$\left[\frac{a^{n+m}-b^{n+m}}{a-b} \right]+ \left[\frac{a^{n}b^{m}-a^{m}b^{n}-(-g(x))^m a^{n-m}+(-g(x))^mb^{n-m}}{a-b}\right].$$ Since $-g(x)=ab$, it is easy to see that the expression in the right bracket is equal to zero. Thus, $(a^{n+m}-b^{n+m})/(a-b)= G_{n+m}^{'}(x)$. This completes the proof of part (2).

\[divisity:Hogat:property2\] Let $\{ G_{n}^{*}(x)\}$ and $\{ G_{n}^{'}(x)\}$ be equivalent GFP sequences. If $m\ge 0$ and $n\ge 0$, then

(1) $(a-b)^2G^{'}_{m+n+1}(x) = \alpha^2 G_{m+1}^{*}(x)G_{n+1}^{*}(x)+\alpha^2 g(x)G_{m}^{*}(x)G_{n}^{*}(x),$

(2) $G_{m+n+2}^{*}(x) = \alpha G_{m+1}^{*}(x)G_{n+1}^{*}(x)+g(x)[\alpha G_{m}^{*}(x)G_{n}^{*}(x)-G_{m+n}^{*}(x)].$

In this proof we use $\alpha=1$, the proof when $\alpha\neq 1$ is similar, so we omit it. (Recall, once again, that we use $a:=a(x)$ and $b:=b(x)$.)

Proof of part 