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.
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.
$$ (2+\sqrt{5})^{2n}+(2-\sqrt{5})^{2n}= (2+\sqrt{5})^{2n}+\frac{1}{(2+\sqrt{5})^{2n}}$$
$$ K=\lfloor(2+\sqrt{5})^{2n}\rfloor+1$$