TITLE: Integral part of sum of huge powers
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: It is the Lucas number $L_{6048}$.
Explicitly:
$$\eqalign{&9062504429942830612498674723940636856073005180754102851103202470654400480862004\cr&470945569990318549786600962729358391458409742980121786528345516555311724482526\cr&469267214426875327939356480512886766136191123865898185909060509390911530211575\cr&250068296151787836515687295491384367878337029779507202772473247457611189758829\cr&765936634113604899116555301784760851901747848131875756210602568727367710580171\cr&895805654908212158642925592215357660076677734346857341254456623431865984405026\cr&892929174068925635521071005404413310101081812578653451740631688019712993398028\cr&457721524261115727978848468969621355164549034177997064126859568812631509397511\cr&016984199127347171026602653375690100545552519714738487571315942718492813756706\cr&082204335257998697204024192225222110769079285160314467805807823101456293333013\cr&846744313345219198803135032572074342248040447170428652954313635095193406457345\cr&312384800612815239265977428989281668436611671607920070840254359108721833323841\cr&188078628899091481176464496313126989415994801739336416785720301217234956031027\cr&978552035253675749937689863092976010498341281125560768668240052298434396165671\cr&537289666399741274176671027424411077085703975089732293442712775771653296648523\cr&231441100434648095729915791965725056427249879681535562607728243485574370058117\cr&109750914437122}$$

REPLY [2 votes]: 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.