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

Solution 1:

Hint: Let $a=(2+\sqrt{3})^n$ and $b=(2-\sqrt{3})^n$. Then $ab=(4-3)^n=1$. Also $a^{-1}=b$. Now conclude that $a+b=a+a^{-1}$ is integral.

Solution 2:

Consider the matrix

$$ A = \begin{bmatrix} 2 & 3\\ 1 & 2 \end{bmatrix} $$

Clearly $ A^n $ is a matrix with integer entries, therefore the trace $ \operatorname{tr} A^n $ is an integer. On the other hand, $ A $ is a diagonalizable matrix whose eigenvalues are $ \lambda_1 = 2 + \sqrt{3} $, $ \lambda_2 = 2 - \sqrt{3} $; therefore the eigenvalues of $ A^n $ are $ \lambda_1^n $ and $ \lambda_2^n $. Since the trace of a matrix is the sum of its eigenvalues, we conclude that $ \operatorname{tr} A^n = \lambda_1^n + \lambda_2^n $, and the quantity on the right hand side is an integer.