If $a+1/a$ is an integer, then so is $a^t+1/a^t$ for $t\in\mathbb N$ I need to show if $a$ is in $\mathbb{R}$ but not equal to $0$, and $a+\dfrac{1}{a}$ is integer, $a^t+\dfrac{1}{a^t}$ is also an integer for all $t\in\mathbb N$.
Can you provide me some hints please?
 A: Hint: If $\displaystyle a+\frac1a$ is an integer then $\displaystyle \left(a+\frac1a\right)^2,\left(a+\frac1a\right)^3, \ldots $ are integers.
Multiply the powers out and you should be able to see why  $a^t+\dfrac1{a^t}$ is going to be an integer for positive integer $t$, using a combination of symmetry and induction.
A: Use induction on $$a^{t+1} + \frac1{a^{t+1}} = \left(a^t+\frac1{a^t}\right)\left(a+\frac1a\right) - \left(a^{t-1}+\frac1{a^{t-1}}\right)$$
A: Let $T_n = r^n + \dfrac{1}{r^n}$, $T_0 = 2$, and $T_1 = \alpha$
\begin{align}
   r + \dfrac 1r &= \alpha \\
   r &= \alpha - \dfrac 1r \\
   \dfrac 1r &= \alpha - r\\
\hline
   r^{n+2} &= \alpha r^{n+1} - r^n \\
  \dfrac{1}{r^{n+2}} &=  \dfrac{\alpha}{r^{n+1}} - \dfrac{1}{r^n} \\
   r^{n+2} + \dfrac{1}{r^{n+2}} 
      &= \alpha\left( r^{n+1} + \dfrac{1}{r^{n+1}} \right)
         - \left( r^n + \dfrac{1}{r^n} \right) \\
   T_{n+2} &= \alpha T_{n+1} - T_n \\
\hline
\end{align}
Clearly, if $\alpha$ is an integer, then $T_n$ is an integer for all non negative integers $n$.
