The greatest integer less than or equal to the number $R=(8+3\sqrt{7})^{20}$ 
Given $$R=(8+3\sqrt{7})^{20}, $$ if $\lfloor R \rfloor$ is Greatest integer less than or equal to $R$, then which of the following option(s) is/are true?

*

*$\lfloor R \rfloor$ is an even number

*$\lfloor R \rfloor$ is an odd number

*$R-\lfloor R \rfloor=1-\frac{1}{R}$

*$R(R-\lfloor R \rfloor-1)=-1$

My try:  I wrote $R$ as $$R=8^{20}\left(1+\sqrt{\frac{63}{64}}\right)^{20} \approx8^{20}\left(1+\sqrt{0.98}\right)^{20} \approx8^{20}\left(1.989\right)^{20} .$$
Now, $8^{20}\left(1.989\right)^{20}$ is slightly less than $8^{20} \times 2^{20}=2^{80}$,
$$\lfloor 2^{80}\rfloor=2^{80}$$
hence
$$\lfloor R \rfloor=2^{80}-1,$$
so option $2$ is correct.
How does one figure out whether options $3$ and $4$ are correct or wrong?

Building on Anomaly's answer, let $a_n = (8 + 3\sqrt{7})^n + (8 - 3 \sqrt{7})^n$. Prove that $a_n$ can be defined by induction by $a_0 = 2$, $a_1 = 16$, $a_n = 16a_{n-1}  - a_{n-2}$. This will help in obtaining information about $a_{20}$.
