Peter wrote: The Fibonacci sequence $\{F_{n}\}$ is defined by \[F_{1}=1, \; F_{2}=1, \; F_{n+2}=F_{n+1}+F_{n}.\] Show that $F_{2n-1}^{2}+F_{2n+1}^{2}+1=3F_{2n-1}F_{2n+1}$ for all $n \ge 1$. We have that $F_{2n-1}^{2}+F_{2n+1}^{2}+1=3F_{2n-1}F_{2n+1}$ $\Leftrightarrow$ $(F_{2n-1}-F_{2n+1})^{2}+1=F_{2n-1}F_{2n+1}$ $\Leftrightarrow$ $F_{2n}^{2}+1=F_{2n-1}F_{2n+1}$ From a known property of Fibonacci sequence we write that $F_{2n-1}F_{2n+1}+F_{2n-2}F_{2n}=F_{4n-1}$ and $F_{2m-1}=F_{m}^{2}+F_{m-1}^{2}$ $\Rightarrow$ $F_{2n-1}F_{2n+1}+F_{2n-2}F_{2n}=F_{4n-1}=F_{2n-1}^{2}+F_{2n}^{2}$ then $F_{2n}^{2}+1=F_{2n-1}F_{2n+1}$ $\Leftrightarrow$ $F_{2n-1}^{2}-1=F_{2n-2}F_{2n}$. On the other hand $F_{2n-2}F_{2n}+F_{2n-3}F_{2n-1}=F_{4n-3}=F_{2n-2}^{2}+F_{2n-1}^{2}$ so $F_{2n-1}^{2}-1=F_{2n-2}F_{2n}$ $\Leftrightarrow$ $F_{2n-2}^{2}+1=F_{2n-3}F_{2n-1}$ therefore,we got that $F_{2n}^{2}+1=F_{2n-1}F_{2n+1}$ $\Leftrightarrow$ $F_{2n-2}^{2}+1=F_{2n-3}F_{2n-1}$ continuing this process we will get that \[F_{2n}^{2}+1=F_{2n-1}F_{2n+1}\Leftrightarrow F_{2}^{2}+1=F_{1}F_{3}\] and the latter is obvious.

Peter wrote: The Fibonacci sequence $ \{F_{n}\}$ is defined by \[ F_{1}=1,\; F_{2}=1,\; F_{n+2}=F_{n+1}+F_{n}.\] Show that $ F_{2n-1}^{2}+F_{2n+1}^{2}+1 = 3F_{2n-1}F_{2n+1}$ for all $ n\ge 1$. Well known: \[ \phi,\phi'=\frac{1\pm\sqrt{5}}{2}\qquad F_{n}=\frac{\phi^{n}-\phi'^{n}}{\sqrt{5}}\] We are trying to prove that $ F_{2n}^{2}+1 = F_{2n-1}F_{2n+1}$ But $ F_{2n}^{2}=\frac{\phi^{4n}+\phi'^{4n}-2}{5}$ and $ F_{2n-1}F_{2n+1}=\frac{(\phi^{2n-1}-\phi'^{2n-1})(\phi^{2n+1}-\phi'^{2n+1})}{5}=\frac{\phi^{4n}+\phi'^{4n}+\phi^{2}+\phi'^{2}}{5}$ And now the result follows, as \[ \frac{-2}{5}+1=\frac{2\frac{1^{2}+\sqrt{5}^{2}}{4}}{5}\]

Induction works too.

Pythagorasauras wrote: Induction works too. Yes, you're right! $$\color{blue}\bf{My \ solution}$$\[F^2_{2n-1}+F^2_{2n+1}+1=3F_{2n-1}F_{2n+1}\]$\iff$ \[F^2_{2n-1}+(F_{2n-1}+F_{2n})^2+1=3F_{2n-1}(F_{2n-1}+F_{2n})\]$\iff$ \[F^2_{2n-1}+F^2_{2n-1}+2F_{2n-1}F_{2n}+F^2_{2n}+1=3F^2_{2n-1}+3F_{2n-1}F_{2n}\]$\iff$ \[F^2_{2n}+1=F^2_{2n-1}+F_{2n-1}F_{2n}\]$\iff$ \[F^2_{2n}+1=F_{2n-1}(F_{2n-1}+F_{2n})\]$\iff$ \begin{align*} &F^2_{2n}+1=F_{2n-1}\cdot F_{2n+1}. \ ({\color{blue}{\clubsuit}}) \end{align*}We will prove $({\color{blue}{\clubsuit}})$ using induction: For $n=1,$ it's obvious. Let's assume that $({\color{blue}{\clubsuit}})$ holds and then prove it for $n+1,$ or just the following equation: \[F^2_{2n+2}+1=F_{2n+1}\cdot F_{2n+3}.\] \[F^2_{2n+2}+1=F_{2n+1}\cdot F_{2n+3}\]$\iff$ \[F^2_{2n+2}+1=F_{2n+1}(F_{2n+1}+F_{2n+2})\]$\iff$ \[F^2_{2n+2}+1=F^2_{2n+1}+F_{2n+1}\cdot F_{2n+2}\]$\iff$ \[(F_{2n}+F_{2n+1})^2+1=F^2_{2n+1}+F_{2n+1}\cdot F_{2n+2}\]$\iff$ \[F^2_{2n}+2F_{2n}F_{2n+1}+F^2_{2n+1}+1=F^2_{2n+1}+F_{2n+1}\cdot F_{2n+2}\]$\iff$ \[F^2_{2n}+2F_{2n}F_{2n+1}+1=F_{2n+1}\cdot F_{2n+2}\]$\iff$ \[F^2_{2n}+1=F_{2n+1}\cdot F_{2n+2}-2F_{2n}F_{2n+1}\]$\iff$ \[F^2_{2n}+1=F_{2n+1}(F_{2n+2}-2F_{2n})\]$\iff$ \[F^2_{2n}+1=F_{2n+1}(F_{2n+2}-F_{2n}-F_{2n})\]$\iff$ \[F^2_{2n}+1=F_{2n+1}(F_{2n+1}-F_{2n})\]$\iff$ \[F^2_{2n}+1=F_{2n+1}\cdot F_{2n-1},\]which is equivalent to $({\color{blue}{\clubsuit}}),$ and we have assumed that it's true. So, the induction is completed. $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \square$