Integral part of sum of huge powers 
Question: What is the integral part of the following expression?$$(a+\sqrt{b})^{2n}+(a-\sqrt{b})^{2n}$$

The question has specific values of $a=2,b=5$ and $2n=2016$.
I was able to simplify (or complexify) it to:$$\sum_{i=0}^n\binom{2n}{2i}a^{2i}b^{n-i}.$$
I think I need to use $$\binom{2n}{2i}=\binom{2n-1}{2i-1}+\binom{2n-1}{2i},$$
but because the powers of $a$ and $\sqrt{b}$ dont change I cant make a closed form.
Please help.
 A: Define $c_0 = 2$, $c_1 = 4$, and $c_{n + 2} = 4c_{n + 1} + c_n$ for $n \geqslant 0$, then$$
c_n = (2 + \sqrt{5})^n + (2 - \sqrt{5})^n. \quad \forall n \in \mathbb{N}
$$
Note that$$
\begin{pmatrix}c_{n + 2}\\c_{n + 1}\end{pmatrix} = \begin{pmatrix}4 & 1\\ 1 & 0\end{pmatrix} \begin{pmatrix}c_{n + 1}\\c_n\end{pmatrix}.
$$
Denote $\displaystyle A = \begin{pmatrix}4 & 1 \\ 1 & 0\end{pmatrix}$, then$$
\begin{pmatrix}c_n\\c_{n - 1}\end{pmatrix} = A^{n - 1} \begin{pmatrix}c_1\\c_0\end{pmatrix}.
$$
(Tedious computation starts.) Because$$
2016 = 2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^5,
$$
and$$
A^2 = \begin{pmatrix}17 & 4 \\ 4 & 1\end{pmatrix}, A^{2^2} = \begin{pmatrix}305 & 72 \\ 72 & 17\end{pmatrix}, A^{2^3} = \begin{pmatrix}98209 & 23184 \\ 23184 & 5473\end{pmatrix}, \cdots
$$
then the explicit answer can be computed, but it is too hard to continue computing manually.
A: Note that $\displaystyle 2-\sqrt{5}=\frac{-1}{2+\sqrt{5}}$.
So, $\displaystyle (2-\sqrt{5})^{2n}=\frac{1}{(2+\sqrt{5})^{2n}}\in(0,1)$.
Let $K=(2+\sqrt{5})^{2n}+(2-\sqrt{5})^{2n}$ which is an integer. 
$$ (2+\sqrt{5})^{2n}+(2-\sqrt{5})^{2n}= (2+\sqrt{5})^{2n}+\frac{1}{(2+\sqrt{5})^{2n}}$$
$$ K=\lfloor(2+\sqrt{5})^{2n}\rfloor+1$$
