Induction proof $F(n)^2 = F(n-1)F(n+1)+(-1)^{n-1}$ for n $\ge$ 2 where n is the Fibonacci sequence Prove that $F{_n}^2 = F_{n-1}F_{n+1}+(-1)^{n-1}$ for n $\ge$ 2 where n is the Fibonacci sequence
F(2)=1, F(3)=2, F(4)=3, F(5)=5, F(6)=8 and so on.
Initial case n = 2: 
$$F(2)=1*2+-1=1$$  It is true.
Let k = n $\ge$ 2
To show it is true for k+1
How to prove this?
 A: \begin{align}
F_n\cdot F_{n+2} + (-1)^n &= F_n\cdot\left(F_{n+1} + F_n\right) + (-1)^n 
\\&= F_n\cdot F_{n+1} + F^2_n + (-1)^n 
\\&= F_n\cdot\left(F_n + F_{n-1}\right) + F^2_n + (-1)^n 
\\&= F^2_n + F_n\cdot F_{n-1} + F^2_n + (-1)^n 
\\&= F^2_n + F_n\cdot F_{n-1} + \left(F^2_n - (-1)^{n-1}\right) 
\\&= F^2_n + F_n\cdot F_{n-1} + F_{n-1}\cdot F_{n+1} 
\\&= F^2_n + F_n\cdot F_{n-1} + F_{n-1}\cdot \left(F_n + F_{n-1}\right) 
\\&= F^2_n + F_n\cdot F_{n-1} + F_{n-1}\cdot F_n + F^2_{n-1} 
\\&= F^2_n + 2\cdot F_n\cdot F_{n-1} + F^2_{n-1} 
= \left(F_n + F_{n-1}\right)^2 = F^2_{n+1}
\end{align}
A: You can also show via induction the fundamental matrix identity of the Fibonacci sequence
$$
\begin{bmatrix}F(n+1)&F(n)\\F(n)&F(n-1)\end{bmatrix}
=
\begin{bmatrix}1&1\\1&0\end{bmatrix}^n
$$
Then computing the determinants on both sides results in
$$
F(n+1)F(n-1)-F(n)^2=(-1)^n
$$

Another consequence of this matrix identity is the doubling formula
\begin{align}
\begin{bmatrix}F(2n+1)&F(2n)\\F(2n)&F(2n-1)\end{bmatrix}
&=
\begin{bmatrix}F(n+1)&F(n)\\F(n)&F(n-1)\end{bmatrix}^2
\\[0.8em]&=
\begin{bmatrix}F(n+1)^2+F(n)^2&F(n)(F(n+1)+F(n-1))\\F(n)(F(n+1)+F(n-1))&F(n)^2+F(n-1)^2\end{bmatrix}
\\[0.8em]&=
\begin{bmatrix}F(n+1)^2+F(n)^2&F(n)(F(n)+2F(n-1))\\F(n)(2F(n+1)-F(n))&F(n)^2+F(n-1)^2\end{bmatrix}
\end{align}
leading to very fast evaluation algorithms for large Fibonacci numbers.
A: Hint.  Write
$$\eqalign{F(n+1)^2\!\!&{}-F(n)F(n+2)\cr
  &=F(n+1)[F(n)+F(n-1)]-F(n)[F(n+1)+F(n)]\cr}$$
and now simplify this a bit more.
