(\[formula1:for:gcddistance1;2\]) and Lemma \[gcdlemmas\] part (3) imply that $\gcd(G_{1}^{*}(x),G_{2}^{*}(x))=1$.

We suppose that $S(m)$ is true for $m=k-1$. Thus, suppose that $\gcd(G_{k-1}^{*}(x),G_{k}^{*}(x))=1$. We prove that $S(k)$ is true. From Lemma \[gcdforproposition4\] part (1), we know that $$\label{formula2:for:gcddistance1;2}
\gcd(G_{k}^{*}(x),G_{k+1}^{*}(x)) \text{ divides } \gcd(G_{k}^{*}(x), g(x)G_{k-1}^{*}(x)).$$ From Lemma \[gcdlemmas\] part (3) we know that $\gcd(G_{k}^{*}(x),g(x))=1$. So, $$\gcd(G_{k}^{*}(x), g(x)G_{k-1}^{*}(x))= \gcd(G_{k}^{*}(x), G_{k-1}^{*}(x)).$$ This, (\[formula2:for:gcddistance1;2\]) and the inductive hypothesis imply that $\gcd(G_{k}^{*}(x),G_{k+1}^{*}(x)=1$.

We now prove the proposition for consecutive even integers (this proof is actually a direct consequence of the previous proof). From Lemma \[gcdforproposition4\] part (2), we have $\gcd(G_{2k}^{*}(x), G_{2k+2}^{*}(x))$ divides $\gcd(G_{2k}^{*}(x),d(x)G_{2k+1}^{*}(x))$. From Lemma \[gcdlemmas\] part (2) we know that $\gcd(d(x),G_{2k}^{*}(x))=1$. This implies that $\gcd(G_{2k}^{*}(x),d(x)G_{2k+1}^{*}(x))= \gcd(G_{2k}^{*}(x),G_{2k+1}^{*}(x))$. From the previous part –that is, the case when $m$ and $n$ are consecutive integers– of this proof we conclude that $\gcd(G_{2k}^{*}(x),G_{2k+1}^{*}(x))=1$. This proves that $\gcd(G_{2k}^{*}(x), G_{2k+2}^{*}(x))=1$.

Finally we prove the proposition for consecutive odd integers. From the recursive definition of GFP we have that $\gcd(G_{2k+1}^{*}(x),G_{2k-1}^{*}(x))$ equals $$\gcd(d(x)G_{2k}^{*}(x)+g(x)G_{2k-1}^{*}(x),G_{2k-1}^{*}(x))=\gcd(d(x)G_{2k}^{*}(x),G_{2k-1}^{*}(x))$$ From the first case in this proof we know that $\gcd(G_{2k}^{*}(x),G_{2k-1}^{*}(x))=1$. This implies that $\gcd(G_{2k+1}^{*}(x),G_{2k-1}^{*}(x))=\gcd(d(x),G_{2k-1}^{*}(x))$. This and Lemma \[gcdlemmas\] imply that $$\gcd(G_{2k+1}^{*}(x),G_{2k-1}^{*}(x))=\gcd(d(x),G_{2k-1}^{*}(x))=G_1^{*}(x).$$ This completes the proof of part (1).

Identities and other properties of Generalized Fibonacci Polynomials {#mainresults:section}
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In this section 