Saudi Arabian Mathematical Competitions

(a) Let $(1+\sqrt{5})^{n}=x_{n}+y_{n} \sqrt{5}$, where $x_{n}, y_{n}$ are positive integers, $n=1,2, \ldots$ Then
$$(1-\sqrt{5})^{n}=x_{n}-y_{n} \sqrt{5}, \quad n=1,2, \ldots$$
hence
$$\begin{equation*} x_{n}^{2}-5 y_{n}^{2}=(-4)^{n}, \quad n=1,2, \ldots \tag{1} \end{equation*}$$
If $n$ is even, consider $a_{n}=x_{n}^{2}-4^{n}$ and we have
$$\begin{gathered} \sqrt{a_{n}}+\sqrt{a_{n}+4^{n}}=\sqrt{x_{n}^{2}-4^{n}}+\sqrt{x_{n}^{2}}=\sqrt{5 y_{n}^{2}}+\sqrt{x_{n}^{2}} \\ =y_{n} \sqrt{5}+x_{n}=(1+\sqrt{5})^{n} \end{gathered}$$
If $n$ is odd, consider $a_{n}=5 y_{n}^{2}-4^{n}$ and we have
$$\begin{gathered} \sqrt{a_{n}}+\sqrt{a_{n}+4^{n}}=\sqrt{5 y_{n}^{2}-4^{n}}+\sqrt{5 y_{n}^{2}}=\sqrt{x_{n}^{2}}+\sqrt{5 y_{n}^{2}} \\ =x_{n}+y_{n} \sqrt{5}=(1+\sqrt{5})^{n} \end{gathered}$$

(b) If $n$ is even, then we have $a_{n}=x_{n}^{2}-4^{n}=5 y_{n}^{2}$, where
$$\begin{gathered} y_{n}=\frac{1}{2 \sqrt{5}}\left[(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}\right] \\ =\frac{2^{n}}{2 \sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]=2^{n-1} F_{n} \end{gathered}$$
where $F_{n}$ is the $n^{\text{th}}$ Fibonacci number. In this case we get $a_{n}= 5 \cdot 4^{n-1} F_{n}^{2}$, hence $5 \cdot 4^{n-1} \mid a_{n}$ and the quotient is $F_{n}^{2}$.