<s>[INST] $F(2n-1) = F(n-1)^2 + F(n)^2$, where $F(i) $ is the $i$'th Fibonacci number, for all natural numbers greater than $1$ I'm proceeding by induction, but I have no idea how to do the inductive step. The Fibonacci sequence takes on the definition:
$F(0)=0$
$F(1) = 1$
$F(2) = 1$
So far, my proof is as follows:
Base Case:
$F(2) = 1 = F(1)^2 + F(0)^2$. This case holds
Inductive Step: We assume the claim holds for $n$ and show it holds for $n+1$...
$F(2n+1) = F(2n)+F(2n-1)=F(2n) + F(n-1)^2+F(n)^2$
I am stuck on how to reduce this expression to $F(n)^2 + F(n+1)^2$...
Edit: Thank you for the responses. I realize I made an egregious error in defining the fibonacci sequence. I have corrected it to the definition we use in my class.
Edit by DS : Now changed the indexing to the standard convention of the Fibonacci numbers.
 [/INST] Yet another approach. By setting $M=\small\begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix}$ we have
$$ \begin{pmatrix} F_{n+2} \\ F_{n+1}\end{pmatrix} = M \begin{pmatrix} F_{n+1} \\ F_{n}\end{pmatrix},\qquad M^n=\begin{pmatrix}F_{n+1} & F_n\\ F_{n} & F_{n-1}\end{pmatrix} $$
and $F_{2n+1}$ is the top-left element of $M^{2n}=(M^n)^2$, where
$$ \begin{pmatrix}F_{n+1} & F_n\\ F_{n} & F_{n-1}\end{pmatrix}^2 = \begin{pmatrix}\color{blue}{F_{n+1}^2+F_n^2} & F_n(F_{n-1}+F_{n+1})\\ F_n(F_{n-1}+F_{n+1}) & F_n^2+F_{n-1}^2\end{pmatrix} $$
leads to
$$ F_{2n+1}=F_{n+1}^2+F_n^2,\qquad \boxed{F_{2n\color{red}{-1}}=F_n^2+F_{n-1}^2.}$$

By Binet's explicit formula we have $F_{2n}=F_n L_n$, where Lucas numbers $L_n$ are defined through $L_0=2,L_1=1$ and $L_{n+2}=L_{n+1}+L_n$.
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