P2936
We have
\begin{equation}
\begin{split}
\varphi^2 - \varphi - 1
& = \left( \frac{1 + \sqrt{5}}{2} \right)^2 - \left( \frac{1 + \sqrt{5}}{2} \right) - 1 \\
& = \frac{(1 + \sqrt{5})^2}{4} - \frac{1 + \sqrt{5}}{2} - 1 \\
& = \frac{1 + 2 \sqrt{5} + 5}{4} - \frac{1 + \sqrt{5}}{2} - 1 \\
& = \frac{1}{4} + \frac{\sqrt{5}}{2} + \frac{5}{4} - \frac{1}{2} - \frac{\sqrt{5}}{2} - 1 \\
& = \frac{1}{4} + \frac{5}{4} - \frac{1}{2} - 1 \\
& = \frac{1}{4} + \frac{5}{4} - \frac{1}{2} - \frac{4}{4} \\
& = \frac{1}{4} + \frac{1}{4} - \frac{1}{2} \\
& = \frac{2}{4} - \frac{1}{2} \\
& = \frac{1}{2} - \frac{1}{2} \\
& = 0 \\
\end{split}
\end{equation}
The case for $1 - \varphi$ can similarly be directly computed. $\square$