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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?

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  • $\begingroup$ binomial theorem on (a + 1/a)^t and induction? $\endgroup$
    – abel
    Nov 27, 2014 at 15:20

3 Answers 3

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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.

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Use induction on $$a^{t+1} + \frac1{a^{t+1}} = \left(a^t+\frac1{a^t}\right)\left(a+\frac1a\right) - \left(a^{t-1}+\frac1{a^{t-1}}\right)$$

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  • $\begingroup$ It is worth mentioning that the above approach comes to mind because of the useful $a^n+b^n=(a^{n-1}+b^{n-1})(a+b)-(a^{n-2}+b^{n-2})ab$ $\endgroup$
    – rah4927
    Nov 28, 2014 at 4:36
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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 $n$.

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