Powers of the golden ratio

Question

Let $\phi$ be the golden ratio. I'm tasked to prove by other means than induction that $x$ in the next equation $$\phi^n =\phi F_n +x,$$ is actually a Fibonacci number. I have tried to apply Binet's formula to $\phi^n -\phi F_n$: \begin{align} \phi^n -\phi F_n &= \phi^n -\phi \left(\dfrac{\phi^n -\left(-\frac{1}{\phi}\right)^n}{\sqrt{5}}\right)\\ &=\phi^n -\frac{\phi^{n+1} -\left(-\frac{1}{\phi}\right)^n \phi}{\sqrt{5}}\\ &=\frac{\sqrt{5} \phi^n -\phi^{n+1} +\left(-\frac{1}{\phi}\right)^n \phi}{\sqrt{5}}. \end{align} But then I got stuck. Could you please help me?

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Answer 1

Starting from $$\phi^n -\phi F_n,$$ then plug $$F_n \equiv \frac{1}{\sqrt{5}} (\phi^n -\varphi^n)$$ where $$\phi =\frac{1+\sqrt{5}}{2}, \qquad \varphi =\frac{1-\sqrt{5}}{2},$$ and use properties of $\phi$ and $\varphi$, namely, that $$1-\frac{1}{\sqrt{5}}\phi = -\frac{1}{\sqrt{5}}\varphi, \qquad \phi \varphi=-1,$$ to compute another Fibonacci number.

$$\phi^n -\phi F_n =\phi^n -\phi \cdot \tfrac{1}{\sqrt{5}}(\phi^n -\varphi^n) =\phi^n -\tfrac{1}{\sqrt{5}} \phi^{n+1} +\tfrac{1}{\sqrt{5}} \phi \varphi^n \\ =\left(1-\tfrac{1}{\sqrt{5}} \phi\right)\phi^n +\tfrac{1}{\sqrt{5}}\phi \varphi^n =-\tfrac{1}{\sqrt{5}} \varphi \phi^n +\tfrac{1}{\sqrt{5}} \phi \varphi^n \\ =-\varphi \phi \cdot \tfrac{1}{\sqrt{5}} \left(\phi^{n-1} -\varphi^{n-1} \right) =\tfrac{1}{\sqrt{5}} \left( \phi^{n-1} -\varphi^{n-1} \right) =F_{n-1}.$$

Answer 2

$$ \phi F_n = \phi\frac{\phi^n - \bar{\phi}^n}{\phi - \bar{\phi}} =\frac{\phi^{n+1} - \phi\bar{\phi}^n}{\phi - \bar{\phi}} =\frac{\phi^{n+1} + \phi\bar{\phi}^{n-1}}{\phi - \bar{\phi}} $$ $$ \phi F_n +F_k =\frac{\phi^{n+1} + \phi\bar{\phi}^{n-1}+\phi^k - \bar{\phi}^k}{\phi - \bar{\phi}} $$ and $ \phi F_n +F_k = \phi^n$ if and only if $$ =\frac{\phi^{n+1} + \phi\bar{\phi}^{n-1}+\phi^k - \bar{\phi}^k}{\phi - \bar{\phi}} = \phi^n $$ or $$ \phi^{n+1} + \phi\bar{\phi}^{n-1}+\phi^k - \bar{\phi}^k = \phi^{n+1} - \phi^n\bar{\phi}= \phi^{n+1}+\phi^{n-1} $$

Setting $k=n-1$ gives the result you desire.