Given the recursive function: $$x_{n+1}=\sqrt[k]{x_n^k + q}$$ where $x_0 \in \mathbb{R}$ and $q, k \in \mathbb{Z}$. Such that: $$x_{Fn(1)}=x_{Fn(2)}=x_1$$ $$x_{Fn(n)}-x_{Fn(n-1)}=q$$ where $Fn(n)$ is the $n^{th}$ Fibonnaci number.
Is there any possible solution for $x_0, k$ & $q$?
There would be infinite solutions, with conditions based on the values of $k$ and $x_0$.
The value $q$ can be easily found by solving the second and third equations using $n=2$.
Using $n=1$, and knowing the Fibonacci numbers are $F_n(0)=0$ and $F_n(1)=1$, we can eliminate $x_1$ as part of $x_0$, which leaves $x_0$ and $k$ with just one (first) equation reformulated.
We then find that there are zero to infinite solutions for $x_0$ depending on the value of $k$, and zero to infinite solutions for $k$ depending on the value of $x_0$, with finite solutions in-between.