part (2) is similar and we omit it. If $r$ is equal to $ \gcd(G_{n}(x), G_{n+1}(x))$, then $r$ divides any linear combination of $G_{n}(x)$ and $G_{n+1}(x)$. Therefore, $r\mid (G_{n+1}(x)-d(x) G_{n}(x))$. This and the recursive definition of $G_{n+1}(x)$ imply that $r\mid g(x)G_{n-1}(x)$. Therefore, $r\mid \gcd(g(x)G_{n-1}(x),G_{n}(x))$. Since $\gcd(g(x),G_{n}(x))=1$, we have that $\gcd(g(x)G_{n-1}(x),G_{n}(x))=\gcd(G_{n-1}(x),G_{n}(x))$. This completes the proof.

\[gcddistance1;2\] Let $m$ and $n$ be positive integers with $0<|m-n|\le 2$.

(1) If $\{G_{t}^{*}(x)\}$ is a GFP of Lucas type, then $$\gcd(G_m^{*}(x),G_n^{*}(x))=
     \begin{cases}
             G_{1}^{*}(x), & \mbox{if $m$ and $n$ are both odd;} \\
             1, & \mbox{otherwise. }
    \end{cases}$$

(2) If $\{G_{t}^{\prime}(x)\}$ is a GFP of Fibonacci type, then $$\gcd(G_m^{\prime}(x),G_n^{\prime}(x))=
     \begin{cases}
            G_{2}^{\prime}(x) & \mbox{if $m$ and $n$ are both even;} \\
             1, & \mbox{otherwise. }
    \end{cases}$$

We prove part (1) using several cases based on the values of $m$ and $n$. The proof of part (2) is similar and we omit. We first provide the proof for the case when $m$ and $n$ are consecutive integers using induction on $m$. Let $S(m)$ be the statement $$\gcd(G_{m}^{*}(x), G_{m+1}^{*}(x))=1\text{  for } m\ge 1.$$ First we prove $S(1)$. From Lemma \[gcdforproposition4\] part (1), we know that $$\label{formula1:for:gcddistance1;2}
\gcd(G_{1}^{*}(x),G_{2}^{*}(x)) \text{ divides } \gcd(G_{1}^{*}(x), g(x)G_{0}^{*}(x)).$$ Since $$\gcd(G_{0}^{*}(x),G_{1}^{*}(x))=\gcd(p_{0}(x),p_{1}(x))=1,$$ we have that $$\gcd(G_{1}^{*}(x),g(x)G_{0}^{*}(x))=\gcd(G_{1}^{*}(x),g(x)).$$ This, 