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
 A: Use the binomial theorem. $(-\sqrt{3})^{2k}=3^k$ and the odd-indexed terms cancel because
$(-\sqrt{3})^{2k+1}=-(\sqrt{3})^{2k+1}$
A: Expand the sums using binomial expansion and all of the root terms disappear:
$$ (2+\sqrt 3)^n + (2-\sqrt 3)^n = 2^n\sum_{r=0}^{n} \binom {n}{r}(\frac {\sqrt 3}{2})^r+ 2^n\sum_{r=0}^{n}\binom {n}{r}(-1)^r(\frac {\sqrt 3}{2})^r$$
All of the terms corresponding to odd values of r are cancelled, leaving only two copies of the terms corresponding to even values of r.
$$(2+\sqrt 3)^n + (2-\sqrt 3)^n =2^{n+1}\sum_{r=0}^{n}\binom {n}{2r}(\frac {\sqrt 3}{2})^{2r}$$
Pulling down the factor of two in the power and multiplying the 2's inside the sum shows that the number must be an integer:
$$(2+\sqrt 3)^n + (2-\sqrt 3)^n =2^{n+1}\sum_{r=0}^{n}\binom {n}{2r}(\frac 34)^{r} = \sum_{r=0}^{n}\binom {n}{2r}3^{r}2^{n-r+1}$$
Since r is at most n, n-r+1 is at minimum 1. Thus there are no roots or negative exponents so it must be an integer.
This works for any (a+root(b))^n 
A: Using the Newton identity,
$$
(2+\sqrt{3})^n + (2-\sqrt{3})^n = \sum_{i=0}^n \left(\binom{n}{i} 2^i\sqrt{3}^{n-i} + 2^i(-\sqrt{3})^{n-i}\right).
$$
The terms where $n-i$ is odd get elided, leaving you with terms where $2m = n-i$, which make $(\pm\sqrt{3})^{2m}$ an integer.
A: $2-\sqrt3$ and $2+\sqrt3$ are the roots of $a^2-4a+1=0$, hence your expression is the solution of the recurrence
$$a_{n+2}=4a_{n+1}-a_n$$ with $a_0=2,a_1=4$.
Obviously, all other terms are integer.
A: Several answers. I add the mine because I want to say something about your Update. 
If you apply the only (non trivial) automorphism of the quadratic field $\Bbb Q(\sqrt d)$ defined by $\sigma(a+b\sqrt d)=a-b\sqrt d$ you have
$$(a+b\sqrt d)^n=A_n+B_n\sqrt d \iff (a-b\sqrt d)^n=A_n-B_n\sqrt d$$ and you have finished taking the sum equal to $2A_n$.
This can not be verified if the irrational $\theta$ is not quadratic because the conjugate of $\theta$ is not unique (there are $k$ in total where $k$ is the degree of the concerned field). You can verified this for example with $\sqrt[3] 2$ whose conjugates are $\sqrt[3] 2$, $\sqrt[3] 2 j$ and $\sqrt[3] 2 j^2$ where $j^2+j+1=0$ (the non real cubic root of $1$).
Without using automorphisms you can see that even for $n=2$ you cannot have for $b\ne 0$
$$(a+b\sqrt[3] 2)^2+(a-b\sqrt[3] 2)^2\in \Bbb Z$$
A: Pell equations are relevant to certain sequences, recurrence relations. E.g., see this answer. I'll use a formula from the answer with $x_1,y_1,D$.
Sometimes a Pell equation doesn't seem to exist for a certain sequence, as seen in this case and this case (see the ends of the answers).
In this case, a Pell equation does exist.
Let $a_n=\frac{(2+\sqrt{3})^n+(2-\sqrt{3})^n}{2}$, $n\ge 0$, $n\in\mathbb Z$.
Then $b_n=2a_n$ ($n\ge 0$, $n\in\mathbb Z$) is the sequence of integers (you wanted to prove they're integers) you want.
$a_0=1$, $a_1=2$.
We could continue for curiosity, but it's not needed:
$a_2=7$, $a_3=26$, $a_4=97$, etc.
$x_1=2$, $y_1=1$, $D=3$.
$2^2-3\cdot 1^2=1$. The equality holds!
Therefore we can expect a Pell equation.
And indeed, $\frac{a_n^2-1}{3}$ with $n\ge 0$, $n\in\mathbb Z$ is always a perfect square.
A: 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.
A: 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.
A: 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.
A: Let $x \in \Bbb Q(\sqrt 3)$ be your number. Since $x$ is fixed by any $\sigma \in \text{Gal}(\Bbb Q(\sqrt 3)/\Bbb Q)$, we know by Galois theory that $x$ is rational. As it is an algebraic integer, it is an integer.

This can be used to show that
$$y=(3-2\cos(2\pi/7))^4+(3-2\cos(4\pi/7))^4+(3-2\cos(6\pi/7))^4$$ is also an integer ($y=682$).
(The minimal polynomial of $2\cos(2\pi/7)$ over $\Bbb Q$ is $x^3+x^2-2 x-1$; the $3$ conjugates of
$2\cos(2\pi/7)=\zeta_7+\overline{\zeta_7}=\zeta_7+\zeta_7^{-1},$
 with $\zeta_7=e^{2\pi i /7}$, are of the form $\zeta_7^k+\zeta_7^{-k}=2\cos(2k\pi/7)$).
