Induction Proof: Fibonacci Numbers Identity with Sum of Two Squares Using induction, how can I show the following identity about the fibonacci numbers? I'm having trouble with simplification when doing the induction step.
Identity: $$f_n^2 + f_{n+1}^2 = f_{2n+1}$$
I get to:
$$f_{n+1}^2 + f_{n+2}^2$$
Should I replace $f_{n+2}$ using the recursion? When I do that, I end up with the product of terms, and that just doesn't seem right. Any guidance on how to get manipulate during the induction step?
Thanks!

Since fibonacci numbers are a linear recurrence - and the initial conditions are special - we can express them by a matrix $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}$$ this is easy to prove by induction:


*

*$$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^1 = \begin{pmatrix} F_{2} & F_1 \\ F_1 & F_{0} \end{pmatrix}$$

*$$\begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} F_n + F_{n+1} & F_{n+1} \\ F_n + F_{n-1} & F_{n} \end{pmatrix} = \begin{pmatrix} F_{n+2} & F_{n+1} \\ F_{n+1} & F_{n} \end{pmatrix}$$



Now your theorem follows from $$\begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}^2 = \begin{pmatrix} F_{n+1}^2 + F_n^2 & - \\ - & - \end{pmatrix} = \begin{pmatrix} F_{2n+1} & F_{2n} \\ F_{2n} & F_{2n-1} \end{pmatrix}$$

https://en.wikipedia.org/wiki/Fibonacci_number
