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?
hint: First prove the following: $\forall n \geq 1: F_{n+1}\cdot F_{n-1}-F_n^2=(-1)^n$, then use it to prove: $\forall n \geq 1: F_{n+2}\cdot F_{n-2}-F_n^2 = (-1)^{n+1}$, then replace $n$ with $2n$ to get the sought identity.
We can try to derive the recurrence for the sequence $(F_{2n})$. (This was already suggested in the comments.) Then we can get matrix form for it, similar as the well-known matrix form for Fibonacci numbers.
This is a very natural thing to do for anybody who saw derivation of similar identity for Fibonacci numbers based on determinants.
So we first notice that $F_{n+2}=F_{n+1}+F_n=2F_n+F_{n-1}=3F_n-F_{n-2}$. So we have $$F_{2n+2}=3F_{2n}-F_{2n-2}.$$
This gives us $$ \begin{pmatrix} F_{2n+2} & F_{2n} \\ F_{2n} & F_{2n-2} \end{pmatrix}= \begin{pmatrix} 3 &-1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} F_{2n} & F_{2n-2} \\ F_{2n-2} & F_{2n-4} \end{pmatrix}. $$
So by induction we get $$ \begin{pmatrix} F_{2n+2} & F_{2n} \\ F_{2n} & F_{2n-2} \end{pmatrix}= \begin{pmatrix} 3 &-1 \\ 1 & 0 \end{pmatrix}^n \begin{pmatrix} 1 & 0 \\ 0 &-1 \\ \end{pmatrix}. $$
If we take determinants on both sides, we get $$F_{2n+2}F_{2n-2}-F_{2n}^2=-1,$$ which is equivalent to the identity in the question.
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.
