Problem: Let $(F_n)_{n\geq1}$ denote the Fibonacci sequence with $F_1 = F_2 = 1$. Show that $(1, F_{2k-1}, F_{2k+1})$ is a Markoff triple for each integer $k \geq 1$.
For reference a Markoff triple is a tuple $(x,y,z)$ of positive integers satisfying$$x^2+y^2+z^2 = 3xyz.$$
Now this amounts to showing that $$1+F_{2k-1}^2+F_{2k+1}^2 = 3F_{2k-1}F_{2k+1}.$$ I am not sure how to proceed. I have some disjoint thoughts: First I have the identity that$$F_1^2+\cdots+F_n^2=F_nF_{n+1}$$ by a previous exercise.
Moreover I could try to use induction since the equality is true for $k=1$, but I'm not sure how to use the IH to show its true for any $k \in \Bbb N$
Hints appreciated.
First, for any $k \geqslant 1$, plugging $F_{2k} = F_{2k + 1} - F_{2k - 1}$ and $F_{2k + 2} = F_{2k + 3} - F_{2k + 1}$ into $F_{2k + 2} = F_{2k + 1} + F_{2k}$ yields $F_{2k + 3} = 3F_{2k + 1} - F_{2k - 1}$.
Next comes the technique of Vieta jumping. The identity$$ 1 + F_{2k - 1}^2 + F_{2k + 1}^2 = 3F_{2k - 1} F_{2k + 1} $$ obviously holds for $k = 1$. Now assume that it holds for $k$, then $x = F_{2k - 1}$ is a root of the quadratic equation $x^2 - 3F_{2k + 1} x + (F_{2k + 1}^2 + 1) = 0$. By Vieta's formula, the other root is $x' = 3F_{2k + 1} - x = F_{2k + 3}$, thus$$ 1 + F_{2k + 3}^2 + F_{2k + 1}^2 = 3F_{2k + 3} F_{2k + 1}. $$ By induction, the identity holds for all $k \geqslant 1$.