TITLE: Fibonacci sequence: Given $n$ and $\mathrm{Fib}(n)$, is it possible to calculate $\mathrm{Fib}(n-1)$?
QUESTION [6 upvotes]: Given $n$ and $\newcommand{\Fib}{\mathrm{Fib}} \Fib(n)$, is it possible to calculate the previous number in the Fibonacci sequence - $\Fib(n-1)$ using integer math in constant time? In other words, I believe I'm looking for a closed form formula.
For example: Given $n=10$ and $\Fib(10)=55$, I'd like to determine that $\Fib(9)=34$ without using $\Fib(7) + \Fib(8)$.

REPLY [5 votes]: This is a dissenting note on some of these answers.
I am purposly making this an answer,
though it really is an anti-answer.
The problem with any answer that uses $\phi$ is that, 
as $n$ gets larger, 
$\phi$ has to be computed with increasingly great precision. 
This will not take constant time,
either for the computation of $\phi$
or the multiplication by $\phi$.