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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

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    $\begingroup$ Do you know the binomial formula? $\endgroup$ – Arthur Aug 25 '16 at 10:15
  • $\begingroup$ @Arthur : yes , but is there something more fundemantal that $ (2+\sqrt[m]{3})^n+(2-\sqrt[m]{3})^n $ is integer? $\endgroup$ – Arjang Aug 25 '16 at 10:27
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    $\begingroup$ 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. $\endgroup$ – Watson Aug 25 '16 at 10:34
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    $\begingroup$ Choosing a basis for $ \mathbf Q(\sqrt{3})/\mathbf Q $ and writing the multiplication map by $ 2 + \sqrt{3} $ as a matrix, then interpreting the given quantity as the value of the field trace gives my argument in the answer. $\endgroup$ – Starfall Aug 25 '16 at 10:40
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    $\begingroup$ @Arjang: however working with rational exponents doesn't give the same result, e.g. math.stackexchange.com/questions/89067 $\endgroup$ – Watson Sep 3 '16 at 17:56

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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.

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  • $\begingroup$ It just remind me of the Fibonacci sequence. $\endgroup$ – Mythomorphic Aug 25 '16 at 10:45
  • $\begingroup$ @hkmather802 That's not so strange, seeing that the Fibonacci sequence may be written as $$F_n=\frac{1}{2}\left(\left(\frac{1+\sqrt5}2\right)^n+\left(\frac{1-\sqrt5}2\right)^n\right)$$ $\endgroup$ – Arthur Aug 25 '16 at 12:21
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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.

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  • $\begingroup$ This is a very elegant solution as it doesn't invoke more powerful tools such as the binomial theorem. $\endgroup$ – user1892304 Aug 25 '16 at 10:29
  • $\begingroup$ @user1892304 Thank you :) $\endgroup$ – Dietrich Burde Aug 25 '16 at 10:34
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    $\begingroup$ Sorry if I am missing something obvious, but why is $a+a^{-1}$ an integer? $\endgroup$ – Miguel Aug 25 '16 at 11:11
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    $\begingroup$ @DietrichBurde $a^2 + 1 = (7+4\sqrt3)^n + 1$. So $a+a^{-1} = \frac{(7+4\sqrt3)^n + 1}{(2+\sqrt3)^n}$. How to see that this is an integer? I hope I am not missing something obvious as is suggested by the number of up votes on your answer $\endgroup$ – user290300 Aug 25 '16 at 12:26
  • $\begingroup$ Use induction. For $n=1$, $a+a^{-1}=4$. $\endgroup$ – Dietrich Burde Aug 26 '16 at 12:04
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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.

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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 $\zeta_7^k+\zeta_7^{-k}=2\cos(2k\pi/7)$).

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    $\begingroup$ This is, of course, what's really happening. $\endgroup$ – Starfall Aug 25 '16 at 10:41
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Use the binomial theorem. $(-\sqrt{3})^{2k}=3^k$ and the odd-indexed terms cancel because $(-\sqrt{3})^{2k+1}=-(\sqrt{3})^{2k+1}$

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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

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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.

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$2-\sqrt3$ 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.

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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 $b\ne 0$

$$(a+b\sqrt[3] 2)^2+(a-b\sqrt[3] 2)^2\in \Bbb Z$$

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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.

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