The number $(3+\sqrt{5})^n+(3-\sqrt{5})^n$ is an integer Prove by induction that this number is an integer:
$$u_n=(3+\sqrt{5})^n+(3-\sqrt{5})^n$$
Progress
I assumed that it holds for $n$ and I tried to do it for $n+1$ but the algebra gets quite messy and I'm unable to prove that the following term is an integer: $\sqrt{5}((3+\sqrt{5})^n-(3-\sqrt{5})^n)$

Let $v_n = (3+\sqrt{5})^n - (3-\sqrt{5})^n$, then
$$
u_1 = 6, v_1 = 2\sqrt{5}
$$
And
$$
u_{n+1} = (3+\sqrt{5})(3+\sqrt{5})^n + (3-\sqrt{5})(3-\sqrt{5})^n = 6u_n + \sqrt{5}v_n
$$
$$
v_{n+1} = (3+\sqrt{5})(3+\sqrt{5})^n - (3-\sqrt{5})(3-\sqrt{5})^n = 6v_n +\sqrt{5}u_n
$$
If $v_n$ is an integer multiple of $\sqrt{5}$ and $u_n$ is an integer, then $v_{n+1}$ is an integer multiple of $\sqrt{5}$ and $u_{n+1}$ is an integer.
