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 $\phi$?

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    $\begingroup$ Writing it down as $x^2+x+1$ in $\mathbb{Z_2}$ you can say it's the only irriducible polinomyal! $\endgroup$ Mar 24, 2017 at 20:27
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    $\begingroup$ Note that:$$F_{n+2}=F_{n+1}+F_{n+0}\\x^2=x^1+x^0$$Hm... $\endgroup$ Mar 24, 2017 at 20:28

2 Answers 2


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 $x^2 - x - 1 = 0$

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    $\begingroup$ That is quite interesting, the continued fraction $$x = \dfrac {1}{1+\frac {1}{1+\frac{1}{1+\cdots}}}\\$$ looks very nice! $\endgroup$
    – user408202
    Mar 24, 2017 at 21:11

In general, we have, due to characteristic equations, the following:



If $(\star)$ is a polynomial with no repeating roots, and $x_p$ is a root of the above polynomial, then


where the coefficients are determined by your initial value, particularly $F_0=0,F_1=1$ for the Fibonacci sequence.


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