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.

Write $a=3+\sqrt5$, $b=3-\sqrt5$.  Then $u_n=a^n+b^n$. Also
$$u_{n+1}-6u_n+4u_{n-1}=a^{n+1}-6a^n+4a^{n-1}+b^{n+1}-6b^n+4b^{n-1}
=(a^2-6a+4)a^{n-1}+(b^2-6b+4)b^{n-1}$$
etc. You need this to equal zero.
