A proof that $\frac{(2\phi)^n-(-1)^n}{\phi^{2n}-(-1)^n}\cdot\left(2^n-\phi^n\right)\cdot\sqrt5\in\mathbb Q$ for all $n\in\mathbb Z$

Question

During computation of some series (with help of a CAS), at an intermediate step I encountered an expression, that after dropping non-essential parts looks like this:$$\mathcal E_n=\frac{(2\phi)^n-(-1)^n}{\phi^{2n}-(-1)^n}\cdot\left(2^n-\phi^n\right)\cdot\sqrt5$$ where $\phi=\frac{1+\sqrt5}2$ is the golden ratio (for $n=0$ take a limit and let $\mathcal E_0=0$).

Although it is not immediately obvious from its form, it appears that $\mathcal E_n$ takes only rational values for all integer $n$. So far it is only an observation, and I am asking you to help me to find a proof. For example, it could be a recurrence relation with only rational coefficients.

Answer 1

Let $\psi = -1/\phi$. Then the Fibonacci numbers $(F_n)$ and the Lucas numbers $(L_n)$ are given by

$$ F_n = \frac{\phi^n - \psi^n}{\phi - \psi}, \qquad L_n = \phi^n + \psi^n. $$

Using this, your number $\mathcal{E}_n$ can be written as

$$ \mathcal{E}_n = \frac{4^n + (-1)^n + 2^n L_n}{F_n} \in \Bbb{Q}. $$