Exercise 3.2.6

Show that the golden ratio $\phi$ and its conjugate $\hat \phi$ both satisfy the equation $x^2 = x + 1$.

This is so obvious, that it is painful:

$$ \phi^2 - \phi - 1 = \bigg(\frac{1 + \sqrt5}{2}\bigg)^2 - \frac{1 + \sqrt5}{2} - 1 = \frac{1 + 2\sqrt{5} + 5 - 2 - 2\sqrt{5} - 4}{4} = 0$$

And now the conjugate:

$$ \hat\phi^2 - \hat\phi - 1 = \bigg(\frac{1 - \sqrt5}{2}\bigg)^2 - \frac{1 - \sqrt5}{2} - 1 = \frac{1 - 2\sqrt{5} + 5 - 2 + 2\sqrt{5} - 4}{4} = 0$$