Fibonacci identity: $f_{n+1}f_{n-1} = f_n^2 + (-1)^n$ How do I see that
$$f_{n+1}f_{n-1} = f_n^2 + (-1)^n$$
for $n \ge 2$, where $f_1 = 1$, $f_2 = 1$, and $f_{n+2} = f_{n+1} + f_n$ for $n \in \mathbb{N}$?
 A: The identity you have given actually has a name (Cassini's Identity). Now, for any $n\geq 1$, let $S(n)$ denote the statement
$$
S(n) : F_{n-1}F_{n+1} = F_n^2+(-1)^n.
$$
Base step ($n=1$): $S(1)$ says that $F_0F_2=F_1-1$ and this is true since both sides equal zero. 
Inductive step: For some fixed $k\geq 1$, suppose that
$$
S(k) : F_{k-1}F_{k+1} = F_k^2+(-1)^k
$$
holds. To be proved is that
$$
S(k+1) : F_{k}F_{k+2} = F_{k+1}^2+(-1)^{k+1}.
$$
Beginning with the left-hand side of $S(k+1)$,
\begin{align}
F_kF_{k+2}&= F_k(F_k+F_{k+1})\\[0.5em]
&= F_k^2+F_kF_{k+1}\\[0.5em]
&= (F_{k-1}F_{k+1}-(1)^k)+F_kF_{k+1}\tag{by $S(k)$}\\[0.5em]
&= (F_{k-1}+F_k)F_{k+1}-(-1)^k\\[0.5em]
&= F_{k+1}F_{k+1}-(-1)^k\\[0.5em]
&= F_{k+1}^2+(-1)^{k+1}.
\end{align}
we end up with the right-hand side of $S(k+1)$, completing the inductive step.
Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$
A: Here's a slick proof using a determinant (turns out this is the same as given in the link by Aretino).
First we prove 
$$\left[\begin{matrix}1&1\\1&0\end{matrix}\right]^n=\left[\begin{matrix}f_{n+1}&f_n\\f_n&f_{n-1}\end{matrix}\right]$$
This is easy enough via induction. The base case $n=1$ holds (when we also add the definition $f_0=0$). Now assuming it holds for $n=k$, we have
$$\left[\begin{matrix}1&1\\1&0\end{matrix}\right]^{k+1}=\left[\begin{matrix}f_{k+1}&f_k\\f_k&f_{k-1}\end{matrix}\right]\left[\begin{matrix}1&1\\1&0\end{matrix}\right]=\left[\begin{matrix}f_{k+1}+f_k&f_{k+1}\\f_{k}+f_{k-1}&f_k\end{matrix}\right]=\left[\begin{matrix}f_{k+2}&f_{k+1}\\f_{k+1}&f_k\end{matrix}\right]$$
as desired.
Now we just take a determinant!
$$\begin{align}&\det\left(\left[\begin{matrix}1&1\\1&0\end{matrix}\right]^n\right)=\det\left( \left[\begin{matrix}f_{n+1}&f_n\\f_n&f_{n-1}\end{matrix}\right]\right)
\\\implies&\det\left(\left[\begin{matrix}1&1\\1&0\end{matrix}\right]\right)^n=f_{n+1}f_{n-1}-f_n^2
\\\implies&(-1)^n=f_{n+1}f_{n-1}-f_n^2
\end{align}$$
Rearranging gives us the result
$$f_{n+1}f_{n-1}=f_n^2+(-1)^n$$
A: This is a special case of Catalan's Identity where r=1 in
$f_{n+1}\,f_{n-1}=f_n^2 - \left(-1\right)^{n-r}f_r^2$
Proof can be found here.
A: Here is a proof by induction (since you tagged it).
First of all, the statement doesn't make sense for $n=1$, so I'm assuming you are tasked with proving it for $n\geq 2$.  To establish the base case note that when $n=2$ we have that
$$f_3f_1=2\cdot1=f_2^2+(-1)^2.$$
Now assume (for induction) that the result it true for $n-1$.  Then in particular we have
$$f_{n-1}^2=f_{n}f_{n-2}-(-1)^{n-1}=f_{n}f_{n-2}+(-1)^{n}.$$  Using this and the fact that $f_{n+2}=f_{n+1}+f_{n-1}$ we have
$$f_{n+1}f_{n-1}=(f_{n}+f_{n-1})f_{n-1}=f_nf_{n-1}+f_{n-1}^2=f_nf_{n-1}+f_{n}f_{n-2}+(-1)^n$$ $$=f_{n}(f_{n-1}+f_{n-2})+(-1)^n=f_n^2+(-1)^n,$$
which is the result.
A: This is not something that you "see", but rather something that you discover while fiddling with Fibonacci numbers and asking yourself "I wonder if there is a relationship between ... and ...?".
The way that you see it is just by creating a table of some quantities that you have been "fiddling" with
$$\begin{array}{cccc}
k&f_k&f_{k-1}f_{k+1}&f_{k}^2\\
0&0&&\\
1&1&0&1\\
2&1&2&1\\
3&2&3&4\\
4&3&10&9\\
5&5&24&25\\
6&8&65&64\\
7&13&168&169\\
8&21&442&441\\
9&34&1155&1156\\
10&55&&\\
\end{array}
$$
It's not difficult to see the pattern.
