Prove the Fibonacci relation $F_{2k} = (F_{2(k-1)}^2 + 1)/F_{2(k-2)}$

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Let $F_k$ be the Fibonacci sequence defined by $F_0=F_1=1$ and $F_k=F_{k-1} + F_{k-2}$

I am trying to show the relation $F_{2k}=(F_{2(k-1)}^2 + 1)/F_{2(k-2)}$ for $k=2,3,4,...$

I first attempted to prove this by induction however I cannot seem to see any simplification. I then used Cassini's identity $F_{n}^2 - F_{n+1}F_{n-1}=(-1)^{n-1}$ to obtain

$(F_{2(k-1)}^2 + 1)/F_{2(k-2)} = (F_{2(k-1)+1}F_{2(k-1)-1} + (-1)^{2(k-1)-1}+1)/F_{2(k-2)}$

$=(F_{2k-1}F_{2k-3}+2)/F_{2k-4}$

However I am similarly stuck with this. Can anyone help with finding a relation which will help me to prove the induction or point me in the right direction.

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Your definition of the Fibonacci numbers is not consistent with the customary definition, which is $$\begin{align*} F_0 &= 0, \\ F_1 &= 1, \\ F_n &= F_{n-1} + F_{n-2}. \end{align*}$$ You instead have $F_0 = 1$, which shifts your sequence by $1$. This is the reason for the discrepancy.

We will use the customary definition, in which case the identity to be proven is $$F_{2n+1} = \frac{F_{2n-1}^2 + 1}{F_{2n-3}},$$ which is obtained by substituting $2k = 2n+1$, and is equivalent to the identity $$1 = F_{2n+1} F_{2n-3} - F_{2n-1}^2.$$ A proof by induction is straightforward. Also possible is an approach via the identity $$F_n = \frac{1}{\sqrt{5}} \left( \varphi^n - \bar \varphi^n \right),$$ where $\varphi = (1 + \sqrt{5})/2$ and $\bar \varphi = (1 - \sqrt{5})/2$.