algebra precalculus - Prove that the next integer greater than $(3+sqrt{5})^n$ is divisible by $2^n$ where $n$ is a natural number.

Prove that the next integer greater than $(3+\sqrt{5})^n$ is divisible by $2^n$ where $n$ is a natural number. The problem is given in a chapter of induction.
It is actually a part of this whole question:

If $S_n = (3+\sqrt{5})^n + (3-\sqrt{5})^n$, show that $S_n$ is an integer and that $S_{n+1} = 6S_n - 4S_{n-1}$.
Deduce that the next integer greater than $(3+\sqrt{5})^n$ is divisible by $2^n$.

I could do the first two parts. The first part I have done by induction and the second part by simply using the given formula. I cannot proceed at all in the third part. I think it may need induction. Please give any solutions regarding the third one and see whether the second one can be proved using induction.