Prove $f_{n}^{2} = f_{n-2}f_{n+2}+(-1)^{n}, n\ge 3$
(For the sake of space, I'm going to skip the basis step and move straight to the inductive step.)
Inductive Step: Assume P(n) is true, prove P(n+1), that is: $f_{n+1}^{2}=f_{n+1}f_{n+3}+(-1)^{n+1}$
This is the only thing I can muster up.
$f_{n-1}f_{n+3}+(-1)^{n+1}=f_{n-1}\left(f_{n+2}+f_{n+1}\right)+(-1)^{n+1}$
I'm unclear on what to do after this step? (Is it more beneficial to write $f_{n-1} = f_{n+1}-f_{n}$ versus using $f_{3}$?
Assume P(n) is true for all $k\leq n$, prove it holds for all $k\leq n+1$, in particular P(n+1), i.e.$f_{n+1}^{2}=f_{n+1}f_{n+3}+(-1)^{n+1}$
$$f_{n+1}^{2}-f_{n-1}f_{n+3}=(f_{n+2}-f_{n})^2-(f_{n}-f_{n-2})(f_{n+2}+f_{n+1})\\=(f_{n+2}-f_{n})^2-(f_{n}-f_{n-2})(2f_{n+2}-f_{n})\\=f_{n+2}^2-4f_{n+2}f_{n}+2f_{n}^2+2f_{n-2}f_{n+2}-f_{n-2}f_{n}\\=f_{n+2}^2-4f_{n+2}f_{n}+2f_{n}^2+2f_{n}^2-f_{n-2}f_{n}\\=(f_{n+2}-2f_{n})^2-f_{n-2}f_{n}=f_{n-1}^{2}-f_{n-2}f_{n}=(-1)^{n-1}=(-1)^{n++1}$$