algebra precalculus - Why is $(2+sqrt{3})^n+(2-sqrt{3})^n$ an integer?

Answers to limit $\lim_{n\to ∞}\sin(\pi(2+\sqrt3)^n)$ start by saying that $(2+\sqrt{3})^n+(2-\sqrt{3})^n$ is an integer, but how can one see that is true?

Update: I was hoping there is something more than binomial formula for cases like $(a+\sqrt[m]{b})^n+(a-\sqrt[m]{b})^n$ to be an integer

Answer

As an alternative to applying the binomial theorem (that is a fine way), the sequence given by
$$a_n=(2+\sqrt{3})^n+(2-\sqrt{3})^n \tag{1}$$

fulfills
$$a_0=2,\qquad a_1=4,\qquad a_{n+2}=4a_{n+1}-a_n \tag{2}$$
hence $a_n\in\mathbb{Z}$ is trivial by induction.
In general, if $\eta,\xi$ are roots of a monic second-degree polynomial with integer coefficients,
$$\eta^{n+2}+\xi^{n+2} = (\eta+\xi)(\eta^{n+1}+\xi^{n+1})-(\eta\xi)(\eta^n+\xi^n) \tag{3}$$
proves just the same, since $(\eta+\xi)\in\mathbb{Z}$ and $\eta\xi\in\mathbb{Z}$ are consequences of Viète's formulas.