<s>[INST] Identity on Fibonacci numbers: $F_{2n}^2=F_{2n+2}F_{2n-2}+1$? Let $F_n$ be the Fibonacci Sequence ($F_1=F_2=1, F_{n+2}=F_{n+1}+F_{n}$).
Prove that $F_{2n}^2=F_{2n+2}F_{2n-2}+1$.
I've tried everything from induction to telescoping series but I haven't got close. They seem to make the identity even more messy. The main problem is with the squared term and the 1 at the end.
Does Pell's Equation or continued fractions have anything to do with this?
 [/INST] The difference equation $F_{n+2} = F_{n+1} + F_{n}$ admits the solution
\begin{align}
F_{n} = \frac{\alpha^{n} - \beta^{n}}{\alpha - \beta}
\end{align}
where $2 \alpha = 1 + \sqrt{5}$ and $2 \beta = 1 - \sqrt{5}$. Now,
\begin{align}
F_{2n-2} F_{2n+2} &= \frac{1}{5} \left(\alpha^{2n-2} - \beta^{2n-2} \right) \left( \alpha^{2n+2} - \beta^{2n+2} \right) \\
&= \frac{1}{5} \, \left( \alpha^{4n} - \alpha^{2n-2} \beta^{2n+2} - \alpha^{2n+2} \beta^{2n-2} + \beta^{4n} \right) \\
&= \frac{1}{5} \, \left( \alpha^{4n} - \alpha^{4} - \beta^{4} + \beta^{4n} \right) \\
&= \frac{1}{5} \left( L_{4n} - L_{4} \right) = \frac{L_{4n} - 7}{5} \tag{1}
\end{align}
where $L_{n}$ are the Lucas numbers and have the form $L_{n} = \alpha^{n} + \beta^{n}$. Consider the square of $F_{2n}$.
\begin{align}
F_{2n}^{2} &= \frac{1}{5} \, \left( \alpha^{4n} - 2 \alpha^{2n} \beta^{2n} + \beta^{4n} \right) = \frac{L_{4n} - 2}{5}. \tag{2}
\end{align}
By comparison of equations (1) and (2) it is evident that
\begin{align}\tag{3}
F_{2n}^{2} = F_{2n-2} F_{2n+2} + 1 
\end{align}
which is the desired result. 
</s>