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

Because \(\phi = \frac{1 + \sqrt{5}}{2}\):

\[\begin{equation} \begin{split} \phi^2 & = \left( \frac{1 + \sqrt{5}}{2} \right)^2 \\ & = \frac{1 + 2 \sqrt{5} + 5}{4} \\ & = \frac{3 + \sqrt{5}}{2} \\ & = \frac{1 + \sqrt{5}}{2} + 1 \\ & = \phi + 1 \end{split} \end{equation}\]

And then because \(\hat{\phi} = \frac{1 - \sqrt{5}}{2}\):

\[\begin{equation} \begin{split} \hat{\phi^2} & = \left( \frac{1 - \sqrt{5}}{2} \right)^2 \\ & = \frac{1 - 2\sqrt{5} + 5}{2} \\ & = \frac{3 - \sqrt{5}}{2} \\ & = \frac{1 - \sqrt{5}}{2} + 1 \\ & = \hat{\phi} + 1 \end{split} \end{equation}\]