$R=(3+\sqrt 5)^{2n}$ and $f=R-[R]$, where $[.]$ is an integer part function. Then find $R(1-f)$

Question

$f$ basically stands for the fractional part of $R$, but since value of $n$ isnt know, I don’t understand how it’s possible to derive it, because it is tricky to write in general terms. How do I solve it?

Answer 1

I am adding it as an answer because it was way too clumsy for a comment.

Let $(3+\sqrt{5})^{2n} + (3 - \sqrt{5})^{2n} = I$ where $I$ is an integer

$[(3+\sqrt{5})^{2n}] = I-1$(because $(3-\sqrt{5})^{2n}$ is a small fraction)

$f = (3 + \sqrt{5})^{2n} - [(3 + \sqrt{5})^{2n}] = (3+\sqrt{5})^{2n}-I+1=(3+\sqrt{5})^{2n}-(3+\sqrt{5})^{2n} - (3 - \sqrt{5})^{2n}+1$

$f=-(3 - \sqrt{5})^{2n}+1$

$1-f = (3-\sqrt{5})^{2n}$