Prove that all elements of sequence $a_{n}=\frac{ \left(1+\sqrt{n^4-n^2+1}\right)^{n} + \left(1-\sqrt{n^4-n^2+1}\right)^{n}}{2^{n}}$ are integers. $\textbf{PROBLEM}$:
Prove or disprove that all elements of sequence $a_{n}=\frac{ \left(1+\sqrt{n^4-n^2+1}\right)^{n} + \left(1-\sqrt{n^4-n^2+1}\right)^{n}}{2^{n}}$ are integers.
$\textbf{MY THOUGHTS}$:
First thing to note is that $${\forall n \in \mathbb{N}~~~~~ \exists k \in \mathbb{Z}:~~~~~  \left( n^4-n^2 \right) = 4k}$$
Secondly, according to Binomial theorem, $$ a_{n}=\frac{\sum\limits_{i=0}^{\lfloor \frac n2 \rfloor} {n \choose 2i} \cdot \left(n^4-n^2+1 \right)^i}{2^{n-1}}$$
I've tried to prove statement by induction on $n$, using facts above, but didn't manage to do it.
I also thought about using the fact that $$\begin{cases} \left(1+\sqrt{n^4-n^2+1}\right) + \left(1-\sqrt{n^4-n^2+1}\right) = 2\\ \left(1+\sqrt{n^4-n^2+1}\right) \cdot \left(1-\sqrt{n^4-n^2+1}\right)= n^2 - n^4\end{cases}  $$
So, we could think about $\left(1+\sqrt{n^4-n^2+1}\right)$ and $\left(1-\sqrt{n^4-n^2+1}\right)$ as roots of $x^2 -2x + n^2 -n^4$.
In that case we can try to solve equivalent problem: Prove that $ \frac{x_1^n+x_2^n}{2^n}$ is integer if $x_1$ and $x_2$ are roots of $x^2 -2x + n^2 -n^4$.
 A: Let $x_1$ and $x_2$ denote $\left(1 + \sqrt{n^4 - n^2 + 1} \right)$ and $\left(1 - \sqrt{n^4 - n^2 + 1} \right)$. We want to prove that $\frac{x_1^n+x_2^n}{2^n}$ is integer.
Proof can be obtained by induction on n.
Firstly, let's fix $x_1$ and $x_2$. By that I mean: choose any fixed $n = n_0$ and let $x_1 =  \left(1 + \sqrt{n_0^4 - n_0^2 + 1} \right)$ and $x_2 =  \left(1 - \sqrt{n_0^4 - n_0^2 + 1} \right)$. Now we can think of $x_1$ and $x_2$ as of constants.
No matter how one will chose $n_0$ above, for $n=0$ and $n=1$ statement is true.
Let it be true $\forall n \le k$.
As mentioned in comments, $\forall k \ge 2 \in \mathbb{N},~~~~~ x_1^{k+1} + x_2^{k+1} = \left(x_1+x_2\right)\left(x_1^k+x_2^k\right) - x_1x_2\left(x_1^{k-1} + x_2^{k-1}\right)$. Using facts from section "My thoughts" one can easily see that this decomposition immediately proves induction step (no matter how we choose $n_0$).
So, for any chosen $n_0$ we completed proof by induction. So, to prove original statement for any $n$ put $n_0=n$.
In this way we obtained proof.
