If n is a natural number, show that the following expression is also a natural number
$$\frac{(1+\sqrt 5)^n-(1-\sqrt 5)^n}{2^n\sqrt 5}$$ (I tried to write $a+\sqrt 5b=(1+\sqrt 5)^n$, a and b are integers,then the expression is equal to $\frac{b}{2^{n-1}}$. But I don't know how to go on... Hoping someone could see this
n=1: $b=1\quad$ $2^{n-1}=1\quad$ quotient=1
n=2: $b=2\quad$ $2^{n-1}=2\quad$ quotient=1
n=3: $b=8\quad$ $2^{n-1}=4\quad$ quotient=2
n=4: $b=24\quad$ $2^{n-1}=8\quad$ quotient=3
Thus I guess $b=(n-1)2^{n-1}$ for $n>1$, but I'm at a loss how to prove it
Hint: Consider the quadratic equation $$x^2-x-1=0$$ If it has roots $\alpha, \beta$, it can be shown quite easily using induction that $$\alpha ^n=F_{n+1}\alpha+F_n$$ where $F_n$ is the $n^{th}$ term of the Fibonacci sequence, viz., $0,1,1,2,3,5,...$. The same can also be said for powers of $\beta$. Can you now proceed?