Let $G_k(m,n)=m\,F_k+n\,F_{k-1}$, where $k,m,n$ are any integers and $(F_k)_{k\in\mathbb{Z}}$ is the extended Fibonacci sequence defined by $F_0=0,F_1=1,F_{k+2} = F_{k+1}+F_k$ for all $k\in\mathbb{Z}.$
Conjecture:
There exist nonzero $(m,n)$ for which $G_k(m,n)$ and $G_{k+1}(m,n)-1$ are coprime for all $k$.
Some small candidate examples are $(m,n) = (6, 12), (12, 84), (18, 6), (18, 36), (24, 18), (30, 90).$ E.g., computations show that $G_k(6,12)$ and $G_{k+1}(6,12)-1$ are coprime for all $k\in [-10^6,10^6]$.
(I suspect that there are infinitely many such pairs $(m,n)$. It would be very interesting to know how to determine them, other than as candidates obtained by testing a large range of $k$-values.)
The conjecture might be proved by somehow using the known fact that any three consecutive Fibonacci numbers $F_{k+1},F_k,F_{k-1}$ are pairwise coprime, but I don't see how to proceed with this.
Question: Is the above conjecture correct? (Proof? Disproof? References?) If so, how can the pairs $(m,n)$ be determined?
Motivation: The conjecture implies a negative answer to a recently asked question; viz., it implies that there exist rational $x$ such that iterating $f:x\mapsto{a+b\over a+1}$(with $x={a\over b}$ in least terms) yields a sequence of iterates $(x,f(x),f(f(x)),\ldots)$ converging to $\varphi={1+\sqrt{5}\over 2}$ (the Golden Mean). This is because it can be shown that if $(m,n)$ is any one of the conjectured pairs, then for $x={m-1\over n}$ the $k$th iterate is $f^k({m-1\over n})={G_{k+1}(m,n)-1\over G_k(m,n)}$, which converges to $\varphi$ due to the fact that ${F_{k+1}\over F_k}\to \varphi.$
More generally, for the parametric family of maps $f_c:x\mapsto{a+b\over a+c}$(with $x={a\over b}$ in least terms), $c\in\mathbb{Z},$ we find ${f_c}^k({m-c\over n})={G_{k+1}(m,n)-c\over G_k(m,n)}\to\varphi\ $ if $(m,n)$ is any one of the pairs in the following conjecture:
Conjecture:
For any integer $c$, there exist nonzero $(m,n)$ for which $G_k(m,n)$ and $G_{k+1}(m,n)-c$ are coprime for all $k$.
I couldn’t prove your generalized conjecture, but I have an algorithm to determine if a given triple $(m,n,c)$ satisfies $$\gcd\left(G_k(m,n),G_{k+1}(m,n)-c\right)=1$$ for all $k$. I’d like to give huge thanks to user @aman. Were it not for their answer in my spin-off question Congruences of consecutive Fibonacci numbers, I wouldn’t have been able to give this answer.
Consider the expression $$\gcd\left(G_{k-r}(m,n)-cF_r,G_{k-r+1}(m,n)-cF_{r+1}\right).\label{1}\tag{1}$$ By using that $\gcd(a,b)=\gcd(a,a+b)$, we can deduce that this equals $$\gcd\left(G_{k-r+1}(m,n)-cF_{r+1},G_{k-r+2}(m,n)-cF_{r+2}\right).$$ By a trivial two-sided induction, $\eqref{1}$ attains the same value for every integer $r$. In particular, setting $r=0$, $r=k$, we get $$\gcd\left(G_k(m,n),G_{k+1}(m,n)-c\right)=\gcd\left(G_0(m,n)-cF_{k},G_1(m,n)-cF_{k+1}\right)=\gcd\left(cF_k-n,cF_{k+1}-m\right).$$ In other words, we just want to prove whether there exist integers $m$, $n$, such that for no prime $p$, there exists a solution to $$\label{2}\tag{2}cF_{k+1}\equiv m\pmod{p},\\cF_k\equiv n\pmod{p}.$$
This next part is due to @aman (although heavily adapted). We consider the equation $$c^2F_{k-r}\equiv(-1)^r\left(cF_{r+1}n-cF_rm\right)\pmod{p}.\label{3}\tag{3}$$ As we’ve already shown, this holds for $r=-1$, $r=0$. Again, by a trivial two-sided induction, using only that $$a\equiv b\pmod{p},\quad c\equiv d\pmod{p}\Rightarrow a\pm c\equiv b\pm d\pmod{p},$$ we can prove that $\eqref{3}$ holds for every integer $r$. In particular, for $r=k-1$, $$c^2\equiv (-1)^{k-1}\left(cF_kn-cF_{k-1}m\right)\equiv(-1)^{k-1}\left(n^2-(m-n)m\right)\equiv(-1)^{k-1}\left(n^2+mn-m^2\right)\pmod{p}.$$ The only candidate primes for $\eqref{2}$ are therefore those that satisfy either $$p\mid m^2-mn-n^2-c^2\text{ or }p\mid m^2-mn-n^2+c^2.$$ Therefore, to check a triple $(m,n,c)$, it suffices to just check the Pisano periods modulo each of these primes.
Here are some examples of triples for $1\leq c\leq100$.