Prove that $\alpha^{n} + \frac{1}{\alpha^{n}} \in \mathbb{Z}$ for any $n \in \mathbb{Z}$ Let $\alpha$ be any real number such that $ \alpha + \frac{1}{\alpha} \in \mathbb{Z}$. Prove that $\alpha^{n} + \frac{1}{\alpha^{n}} \in \mathbb{Z}$ for any $n \in \mathbb{Z}$
My attempt was to try induction but first I realized that $\alpha$ has to be $ \pm1$ in order for it be be an integer.  
Base case $n=1$ then $ \pm1^{1}+ \frac{1}{\pm1^{1}} \in \mathbb{Z}$
Assume $P(k): \alpha^{k} + \frac{1}{\alpha^{k}} \in \mathbb{Z}$ is true for some $k \geq 1$  I want to show that $P(k+1)$ is true. 
$\alpha^{k+1} + \frac{1}{\alpha^{k+1}}=\alpha^{k}*\alpha+\frac{1}{\alpha^{k}*\alpha}$ so I get an integer multiplied by $\pm1$ so it is an integer. Thus the statement is true for all $n$
Any clarification would be helpful. 
 A: Hint:
$$\left(\alpha^n+\frac{1}{\alpha^n}\right)\left(\alpha+\frac{1}{\alpha}\right)=\left(\alpha^{n+1}+\frac{1}{\alpha^{n+1}}\right)+\left(\alpha^{n-1}+\frac{1}{\alpha^{n-1}}\right)$$
A: Any algebraic number $\alpha$ given by a root of a quadratic, palindromic polynomial
$$ p(x) = x^2 + kx +1,\qquad k\in\mathbb{Z} $$
has such a property. $p(x)$ being equal to $x^2 p\left(\tfrac{1}{x}\right)$ ensures that the roots of $p$ are of the form $\alpha,\tfrac{1}{\alpha}$; Vieta's theorem ensures that $\alpha+\frac{1}{\alpha} = -k \in\mathbb{Z}$. By denoting as
$$ T_n = \alpha^n+\frac{1}{\alpha^n} $$
we have $T_0,T_1\in\mathbb{Z}$ and 
$$ T_n T_1 = T_{n+1} + T_{n-1}, $$
hence $T_n\in\mathbb{Z}$ for any $n\in\mathbb{N}$ is straightforward to prove by induction on $n$.
A: $\alpha+\frac 1\alpha=k\iff \alpha^2-k\alpha+1=0$
Let's define the sequence $a_n$ by this linear induction : $a_{n+2}-ka_{n+1}+a_n=0$.
Then $a_n=\alpha^n+\alpha^{-n}$ considering conditions $a_0=2, a_1=\alpha+\frac 1\alpha$.
Since $a_0,a_1\in\mathbb Z$ then $a_2=ka_1-a_0\in\mathbb Z$ and so on by induction.
