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In an attempt to find the $n^{th}$ Fibonacci number by Binet's formula, the derivation of this formula starts by using the quadratic equation

$$x^2-x-1 = 0$$

If we try to find roots of this equation using quadratic roots formula, we find that we have 2 roots:

$$ \sigma = \frac{1+\sqrt 5}{2}, \gamma= \frac{1-\sqrt 5}{2}$$

Then, we notice that:

\begin{align} x^2 = x+1 \\ x^3 = 2x+1 \\ \vdots\\ x^6 = 8x+5 \\ \vdots\\ \end{align}

Question: why do we start with $x^2-x-1 = 0$ specifically and not with others? Is it because it just fulfill the need?

Oscar Lanzi
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Avv
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    Do you see the similarity between $x^2 = x+1 = x^1+x^0$ and $f_2 = f_1+f_0$? – B. Goddard Jan 05 '22 at 22:42
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    You have to study the little theory about recurrent sequences of the second order and the fact that they constitute a vector space with dimension two with basis $e^{na}$ and $e^{nb}$ where $a,b$ are the roots (assume distinct, as is the case here) of the characteristic quadratic equation you have considered. – Jean Marie Jan 05 '22 at 22:49
  • @JeanMarie. Thanks Prof. – Avv Jan 05 '22 at 23:39
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    @Avv See also my answer here which describes an elementary, self-contained way to solve second order recurrences, without assuming prior knowledge of characteristic polynomials or any of the theory behind. – dxiv Jan 06 '22 at 00:22

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