Show that $f_{n+1}\cdot f_{n-1}-f^2_n=(-1)^n$ if $f_n $ is the $n$th Fibonacci number

Question

I've reviewed this answer: https://math.stackexchange.com/a/606286/584468 and I'm getting lost on how he did $f_{k+2}f_k−f^2_{k+1}=(f_k−f_{k+1})f_k−f^2_{k+1}$

When I thought that $f_{k+2}=f_{k+1}+f_k$?

But I'm getting:

$$(f_{k+1}+f_k)\cdot f_k-f^2_{k+1} = f_{k+1}\cdot f_k +f^2_k - f^2_{k+1}$$ $$=(f_{k+1})(f_k-f_{k+1}) + f^2_k$$ $$=(f_{k+1})(f_k-(f_k+f_{k-1})) + f^2_k $$ $$=f_{k+1}\cdot(-f_{k-1}) + f^2_k$$ $$=-f_{k+1}\cdot f_{k-1} + f^2_k$$ $$=(-1)(f_{k+1}\cdot f_{k-1} - f^2_k)$$ $$=(-1)(-1)^k$$

Answer 1

You're right. The linked answer is in error; it's using the reverse Fibonacci recurrence $f_{k+1}+f_k=f_{k-1}$ consistently, and not noticing that this isn't quite the same as the original.