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?

• It should be $F_{n+1}F_{n-1}-F_n^2=(-1)^n$. – carmichael561 Nov 12 '16 at 19:45
• Thank you. I knew something was wrong. – Iceman Nov 12 '16 at 19:46

It's supposed to be $f_{n + 1} f_{n - 1} = f_n^2 + (-1)^n$, which is called Cassini's identity.
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
The last row is true, because of the following relation between Fibonacci numbers $$F_{-n}={(-1)}^{n+1}F_n \, .$$