Linear Recurrences and Sequence

Question

$$a_n = \lfloor\left(3+\sqrt5\right)^n\rfloor+1$$

Prove that for each n the number $a_n$ is divisible by $2^n$ without a remainder.

Hints

1. try to use the following sequence $b_n = (3+\sqrt{5}\,)^n+(3-\sqrt{5}\,)^n$ and find a recurrence formula for $b_n$.

2. Use the fact that $0 < 3-\sqrt{5} < 1$

Edit: The minus sign in the sequence.

Answer 1

Hint: the recurrent sequence for $b_n$ is: $$b_n=6b_{n-1}-4b_{n-2}, b_0=2, b_1=6.$$ Note that $b_n$ is an increasing sequence and its terms are always positive even numbers. Also, note that since $0<3-\sqrt{5}<1 \Rightarrow 0<(3-\sqrt{5})^n<1$, then: $$a_n=b_n.$$