Fibonacci proof question: $f_{n+1}f_{n-1}-f_n^2=(-1)^n$ Show that 
$$f_{n+1}f_{n-1}-f_n^2=(-1)^n$$
when $n$ is a positive integer and $f_n$ is the $n$th Fibonacci number. 

*

*For a basis $n=1$ the equality holds, as 
$$f_{k+1}f_{k-1}-f_k^2=f_2f_0-f_1^2=1 \cdot 0 - 1^2=-1=(-1)^1.$$ 

*Assume the equality holds for $n=k$. Then we may assume that
$$f_{k+1}f_{k-1}-f_k^2=(-1)^k.$$
For the final inductive step, we wish to prove that $$f_{k+2}f_k-f_{k+1}^2=(-1)^{k+1}.$$

*We begin with the left side.


$$
\begin{align*}
f_{k+2}f_k-f_{k+1}^2&= \left( f_k-f_{k+1} \right)f_k-f_{k+1}^2 \\
&=f_k^2-f_{k+1}f_k-f_{k+1}^2 \\
&=-f_{k+1}\left( f_{k+1}+f_k \right)+f_k^2 \\
&=-f_{k+1}f_{k-1}+f_k^2 \\
&= \left( f_{k+1}f_{k-1}-f_k^2 \right)(-1)^1 \\
&=(-1)^k(-1)^1 \\
&=(-1)^{k+1}. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \blacksquare
\end{align*}
$$
Notes: On line 4 by definition $f_{k+1}+f_k=f_{k-1}$. On line 6 we substituted our assumption.
