Solution to Fibonacci Recursion Equations Let the sequence $(a_n)_{n\geq0}$ of the fibonacci numbers: $a_0 = a_1 = 1, a_{n+2} = a_{n+1} + a_n, n \geq 0$ 
Show that:
i) $$a^2_n - a_{n+1}a_{n-1} = (-1)^n \text{ for }n\geq1$$ 
I try to show this with induction:
$n=1:$ 
            \begin{align*}
    a^2_1 - a_{2}a_{0}&=1 - a_2a_0\\
    &= 1- (a_1 + a_0 )a_0\\
    &= 1-(1+1)1 = -1 = (-1)^1\\
   \end{align*}
Assume that: $a^2_n - a_{n+1}a_{n-1} = (-1)^n ,\forall n\geq1$
Inductive step:
            \begin{align*}
    a^2_{n+1} - a_{n+2}a_{n}
    &=(a_n + a_{n-1}) ^2 - (a_{n+1}+a_{n})(a_{n-1}-a_{n-2})\\ 
    &=a^2_{n}+2a_na_{n-1}+a^2_{n-1}-a_{n+1}a_{n-1}+a_{n+1}a_{n-2}-a_{n}a_{n-1}+a_{n}a_{n-2}\\
    &=(a^2_{n}-a_{n+1}a_{n-1})+(a^2_{n-1}-a_{n}a_{n-1})+2a_na_{n-1}+a_{n+1}a_{n-2}+a_{n}a_{n-2}\\
    &=\underbrace{(-1)^n + (-1)^{n-1}}_{=0}+2a_na_{n-1}+a_{n+1}a_{n-2}+a_{n}a_{n-2}\\
    &=2a_na_{n-1}+a_{n+1}a_{n-2}+a_{n}a_{n-2}\\
    &=??\\
   \end{align*}
ii) $$\sum \limits_{i=0}^na_i=a_{n+2}-1 , n\geq0$$
iii) $$a^2_{n-1}+a^2_n=a_{2n} \text{ and } a_{n-1}a_n+a_na_{n+1}=a_{2n+1}, n\geq1$$
Thank you for any help

Another flavor on the inductive step argument:
$$
\begin{split}
a_n^2-a_{n+1}a_{n-1}
      &= a_n^2-(a_n + a_{n-1})a_{n-1} \\
      &= a_n^2 - a_n a_{n-1} - a_{n-1}^2 \\
      &= a_n(a_n-a_{n-1}) - a_{n-1}^2 \\
      &= a_n a_{n-2} - a_{n-1}^2 \\
      &= -\left( a_{n-1}^2 - a_n a_{n-2}\right),
\end{split}
$$
reducing to the inductive hypothesis...
