Product of integral and fractional part of binomial expansion

Question

Let $R = (5\sqrt{5} + 11)^{2n+1} = [R] + f$, where $[.]$ denotes the greatest integer function. Prove that $Rf = 4^{2n+1}$.

I need help with the above question. When I try expanding, I will get some integral part and some terms of √5. But those terms being greater than 1, how to determine integral and fractional part of the given expression?

Answer 1

As $0<5\sqrt5-11=\dfrac{125-121}{5\sqrt5+11}<1$

and $(5\sqrt5+11)^{2n+1}-(5\sqrt5-11)^{2n+1}$ is rational

$\implies f=(5\sqrt5-11)^{2n+1}$

Can you take it from here?