Let's consider the sequence $b_n=\left(\frac{\sqrt{5}+1}{2}\right)^n-\left(\frac{\sqrt{5}-1}{2}\right)^n$. Note that $$b_n^2 = \left(\frac{\sqrt{5}+3}{2}\right)^n+\left(\frac{3-\sqrt{5}}{2}\right) -2 = a_n$$

Therefore, the to-be-proved claim is equivalent to showing that $b_n=k$ when $n$ is odd and $b_n=\sqrt{5}k$ when $n$ is even, where $k$ is an integer. This can be done using mathematical induction.

Note that $b_{n+2} = \sqrt{5}b_{n+1} - b_n$ holds for any $n\ge 1$. Meanwhile, we have $b_1=1$ and $b_2=\sqrt{5}$. Assuming $b_{2l-1}=u$ and $b_{2l}=\sqrt{5}v$ where $l$, $u$, and $v$ are all positive integers. Then we have $$\begin{array}{rl} b_{2l+1} =& \sqrt{5}b_{2l} - b_{2l-1} = 5v - u \\ b_{2l+2} = &\sqrt{5}b_{2l+1} - b_{2l} = \sqrt{5}(5v-u) - \sqrt{5}v =\sqrt{5}(4v-u) \end{array}$$

Hence, by the principle of mathematical induction, we assert the claim holds.