Induction proof $F(n)^2 = F(n-1)F(n+1)+(-1)^{n-1}$ for n $\ge$ 2 where n is the Fibonacci sequence

Question

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?

Answer 1

\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 F_{n+1} + \left(F^2_n - (-1)^{n-1}\right) \\&= F_n\cdot F_{n+1} + F_{n-1}\cdot F_{n+1} \\&= ( F_n+ F_{n-1})\cdot F_{n+1}=F^2_{n+1} \end{align}

Answer 2

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.

Answer 3

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 $$ F(n+1)F(n-1)-F(n)^2=(-1)^n $$

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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.