Deduce that the next integer greater than $(3+\sqrt{5})^n$ is divisible by $2^n$

Question

I can prove it for base case $n=1$. I have assumed it true for $n=k$ But I cant prove it for $n=k+1$.

Answer 1

This integer is clearly $a_n:=(3+\sqrt{5})^n+(3-\sqrt{5})^n$. This sequence satisfies the recurrence relation $a_{n+2}=6a_{n+1}-4a_n$, from which we can prove $2^n|a_n$ by induction (the base case will use two consecutive cases of $n$). Indeed $2|a_1=6,\,4|a_2=28$, and $k_n:=2^{-n} a_n\implies k_{n+2}=3k_{n+1}-k_n$, giving what is clearly a sequence of integers.

Answer 2

Hint: Let $a=(3+5^{0.5})^n$ and $b=(3-5^{0.5})^n$. Calculate $a+b$ and try to prove $b<0,5$.

Answer 3

Hint

$a=\dfrac{3+\sqrt5}2,b=\dfrac{3-\sqrt5}2=\dfrac1a<1$

So, $a,b$ are the roots of $t^2-3t+1=0$

So, if $x_n=a^n+b^n,$

$x_{n+2}=3x_{n+1}-x_n$

By induction, $x_n$ is an integer and is the next integer of $a^n$