Is there something special about $x^2-x-1$? I was working on proofs by induction and had to produce numerous proofs about the properties the Fibonacci sequence as exercises.
It was pointed out to me while proving that $ F_n = \frac 1{\sqrt 5} \left(\phi^n-\left(-\frac 1\phi\right)^n\right)= F_{n-1}+F_{n-2}$, that the roots of the second degree polynomial 
$$y = x^2-x-1$$
are the golden ratio $\phi = \frac {1+\sqrt 5}2 = \lim_{n \to \infty} \frac {F_{n+1}}{F_n}$ and its negative inverse $\left(-\frac 1\phi\right)$. 
I never realized that before and now I wonder whether there is something special about the polynomial $y=x^2-x-1$ appart from the fact that its positive root is the golden ratio $\phi$?

Perhaps what  is more interseting than: $y = x^2 - x - 1$ 
Is this equation: $x = \frac 1x + 1$
Is there some number that equals its reciprocal makes plus 1?
I think the decimal expansion (rounded off at 3 decimals) makes it is easier to follow some of the implications.
$\frac {1}{1.618} = 0.618\\
(1.618)(0.681) = 1\\
(1.618)^2 = 1 + 1.618\\
(0.618)^2 = 1-0.618$
As it related to the Fibonacci sequence.  We start with the conjecture that $n$ gets large $\frac {F_{n+1}}{F_{n}}$ approaches some constant.
$\frac {F_{n+1}}{F_{n}} = \frac {F_{n}}{F_{n-1}}$
And for the Fibonacci sequence $F_{n+1} = F_n + F_{n-1}$
$\frac {F_{n} + F_{n-1}}{F_{n}} = \frac {F_{n}}{F_{n-1}}\\
1 + \frac {F_n}{F_{n-1}} = \frac {F_{n}}{F_{n-1}}$
and we say $\frac {F_{n}}{F_{n-1}}$ equals $x$ and we are back to 
$1 + \frac {1}{x} = x$
One more:
The continued fraction.
$x = \dfrac {1}{1+\frac {1}{1+\frac{1}{1+\cdots}}}\\
x = \frac {1}{1+x}\\
x^2 + x - 1 = 0$
The roots of which are the negative of 
$x^2 - x - 1 = 0$
