> Considering fibonacci - golden ratio relation, is it possible to devise a explict connection/translation between the puzzle 104's graph and the x²=1+x equation?

Here's my attempt -

Assume you want to compute the number of paths of length \\(n\\), starting from \\(x\\). You can start counting by taking one of two paths:

- Go along \\(f\\), and end up at \\(x\\). Then count the number paths of length \\(n - 1\\)

- Go along \\(h \circ g\\), and end up at \\(x\\) again. Then count the number of paths of length \\(n - 2\\).

Let \\(f_n\\) be the number of paths. Then

\\[

f_n = f_{n-1} + f_{n-2} \tag{★}

\\]

Now consider

\\[

\phi := \underset{n \to \infty}{\mathrm{lim}} \frac{f_{n}}{f_{n-1}}

\\]

With (★) we have

\\[

\begin{align}

\phi & = \underset{n \to \infty}{\mathrm{lim}} \frac{f_{n}}{f_{n-1}} \\\\

& = \underset{n \to \infty}{\mathrm{lim}} \frac{f_{n-1} + f_{n-2}}{f_{n-1}} \\\\

& = \underset{n \to \infty}{\mathrm{lim}} \left(1 + \frac{f_{n-2}}{f_{n-1}} \right) \\\\

& = 1 + \underset{n \to \infty}{\mathrm{lim}} \frac{f_{n-2}}{f_{n-1}} \\\\

& = 1 + \underset{n \to \infty}{\mathrm{lim}} \frac{f_{n-1}}{f_{n}} \\\\

& = 1 + \underset{n \to \infty}{\mathrm{lim}} \left(\frac{f_{n}}{f_{n-1}}\right)^{-1} \\\\

& = 1 + \left(\underset{n \to \infty}{\mathrm{lim}} \frac{f_{n}}{f_{n-1}}\right)^{-1} \\\\

& = 1 + \phi^{-1} \\\\

\end{align}

\\]

Hence \\(\phi = 1 + \phi^{-1}\\). Now multiply both sides by \\(\phi\\). This gives:

\\[

\phi^2 = \phi + 1

\\]

...as you wanted to see...

Here's my attempt -

Assume you want to compute the number of paths of length \\(n\\), starting from \\(x\\). You can start counting by taking one of two paths:

- Go along \\(f\\), and end up at \\(x\\). Then count the number paths of length \\(n - 1\\)

- Go along \\(h \circ g\\), and end up at \\(x\\) again. Then count the number of paths of length \\(n - 2\\).

Let \\(f_n\\) be the number of paths. Then

\\[

f_n = f_{n-1} + f_{n-2} \tag{★}

\\]

Now consider

\\[

\phi := \underset{n \to \infty}{\mathrm{lim}} \frac{f_{n}}{f_{n-1}}

\\]

With (★) we have

\\[

\begin{align}

\phi & = \underset{n \to \infty}{\mathrm{lim}} \frac{f_{n}}{f_{n-1}} \\\\

& = \underset{n \to \infty}{\mathrm{lim}} \frac{f_{n-1} + f_{n-2}}{f_{n-1}} \\\\

& = \underset{n \to \infty}{\mathrm{lim}} \left(1 + \frac{f_{n-2}}{f_{n-1}} \right) \\\\

& = 1 + \underset{n \to \infty}{\mathrm{lim}} \frac{f_{n-2}}{f_{n-1}} \\\\

& = 1 + \underset{n \to \infty}{\mathrm{lim}} \frac{f_{n-1}}{f_{n}} \\\\

& = 1 + \underset{n \to \infty}{\mathrm{lim}} \left(\frac{f_{n}}{f_{n-1}}\right)^{-1} \\\\

& = 1 + \left(\underset{n \to \infty}{\mathrm{lim}} \frac{f_{n}}{f_{n-1}}\right)^{-1} \\\\

& = 1 + \phi^{-1} \\\\

\end{align}

\\]

Hence \\(\phi = 1 + \phi^{-1}\\). Now multiply both sides by \\(\phi\\). This gives:

\\[

\phi^2 = \phi + 1

\\]

...as you wanted to see...