Show that $u_{n+1}=6u_n-4u_{n-1}$ if $u_n=(3+\sqrt{5})^n+(3-\sqrt{5})^n$ for $n=1,2,...$ I have already shown that for each $n,u_n$ is an integer.
Now we can show that $u_n=\dfrac{1}{2^n}(\sqrt{5}+1)^{2n}+\dfrac{2^{3n}}{(\sqrt{5}+1)^{2n}}$
But the problem is while showing $u_{n+1}=6u_n-4u_{n-1}$ , I can see that the whole calculation is taking pages. Is there any shortcut, crisp way of proving it?
PS - I am not showing here the gory, intricate calculations here that cost me pages as this might make the users viewing it, uneasy.

Hint Let $a_n=(3+\sqrt{5})^n$, can you show that $a_{n+1}=6a_n-4a_{n-1}$ and the same for $b_n=(3-\sqrt{5})^n$ ?
