Fibonacci Square Identity On a real analysis worksheet i was given, the following question is posited.

Let $f_n$ be the Fibonacci sequence defined by $f_1=1, f_2=1$ and for $n>2, f_{n+1}=f_n+f_{n-1}$.  Show that for all $n>2,$ $$f_{n+1}f_{n-1}=(-1)^nf_n^2$$

I can't help but think that this is a typo.  All Fibonacci numbers are positive, and a basic substitution of $n=3$ would give
$$f_{4}f_2=3\neq-4=(-1)^3f_3^2$$
If it is a typo, does anyone know the actual identity that is being questioned here?
 A: It's supposed to be $f_{n + 1} f_{n - 1} = f_n^2 + (-1)^n$, which is called Cassini's identity.
A: A simple proof for the Cassini's identity is in the following form . Consider the 
following matrix that is well-know to $Q_p$ matrix
$$
Q_2=\left[
\begin{array}{cc}
0 & 1 \\
1 & 1
\end{array}
\right] \, .
$$
with the induction on $n$, we can prove that the $n$th power of matrix $Q_2$, is in the following form
$$
Q_2^n= \left[
\begin{array}{cc}
F_{n-1} & F_{n} \\
F_{n} & F_{n+1}
\end{array}
\right]\, .
$$
In addition, the determinant of matrix $Q_2$ is $(-1)$ and because of this we conclude that
$$
det(Q_2^{n})={(-1)}^{n}\, .
$$
the above relation, results that
$$
det(
\left[
\begin{array}{cc}
F_{n-1} & F_{n} \\
F_{n} & F_{n+1}
\end{array}
\right]
)={(-1)}^{n}
\Longrightarrow
F_{n-1}F_{n+1}-F_{n}^2={(-1)}^{n}\, .
$$
Another proof for the Cassini's identity is as follows
\begin{eqnarray}
F_{n+1}F_{n-1}-F_{n}^2&=&(F_{n-1}+F_{n})F_{n-1}-F_{n}^2 \\
\\
&=& F^2_{n-1}+F_n(F_{n-1}-F_n) \\
\\
&=&F^2_{n-1}-F_nF_{n-2} \\
\\
&=&-(F_nF_{n-2}-F^2_{n-1})
\end{eqnarray} 
With repetition the mentioned process, we get
\begin{eqnarray}
-(F_nF_{n-2}-F^2_{n-1})&=& {(-1)}^2(F_{n-1}F_{n-3}-F^2_{n-2}) \\
\\
&=&{(-1)}^3(F_{n-2}F_{n-4}-F^2_{n-3})  \\
\\
&& \vdots \\
&=& {(-1)}^n(F_{1}F_{-1}-F^2_{0})  \\
\\
&=& {(-1)}^n
\end{eqnarray}
The last row is true, because of the following relation between Fibonacci numbers
$$
F_{-n}={(-1)}^{n+1}F_n \, .
$$
