SAMC

(a) Let $(1+\sqrt{5}{)}^n=x_n+y_n\sqrt{5}$, where $x_n,y_n$ are positive integers, $n=1,2,\dots$ Then $$(1-\sqrt{5}{)}^n=x_n-y_n\sqrt{5},n=1,2,\dots$$ hence $$\begin{array}{c}{x}_n^2-5y_n^2=(-4{)}^n,n=1,2,\dots \end{array}\text{\hspace{1em}(1)}$$ If $n$ is even, consider $a_n=x_n^2-4^n$ and we have $$\begin{array}{c}\sqrt{a_n}+\sqrt{a_n+4^n}=\sqrt{x_n^2-4^n}+\sqrt{x_n^2}=\sqrt{5y_n^2}+\sqrt{x_n^2}\\ =y_n\sqrt{5}+x_n=(1+\sqrt{5}{)}^n\end{array}$$ If $n$ is odd, consider $a_n=5y_n^2-4^n$ and we have $$\begin{array}{c}\sqrt{a_n}+\sqrt{a_n+4^n}=\sqrt{5y_n^2-4^n}+\sqrt{5y_n^2}=\sqrt{x_n^2}+\sqrt{5y_n^2}\\ =x_n+y_n\sqrt{5}=(1+\sqrt{5}{)}^n\end{array}$$

(b) If $n$ is even, then we have $a_n=x_n^2-4^n=5y_n^2$, where $$\begin{array}{c}{y}_n=\frac{1}{2\sqrt{5}}\left[(1+\sqrt{5}{)}^n-(1-\sqrt{5}{)}^n\right]\\ =\frac{2^n}{2\sqrt{5}}\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n\right]=2^{n-1}{F}_n\end{array}$$ 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 $F_n^2$.