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Let $r$ be a real number such that $r + 1/r$ is an integer. Prove that for every natural number $n$, $r^n + 1/r^n$ is also an integer. (In addition, I have to use induction, strong induction, or a minimum counterexample).

I initially tried minimum counterexample, assuming that for a fixed $r$, that $k$ is the smallest natural number for which the statement is false. I then had: $$r^{k-1} + (1/r)^{k-1} = n \in \mathbb{N}$$ with the intent of showing a contradiction for the case $k$, but couldn't get anywhere after combining the terms into a single fraction. Similar attempts with induction and strong induction went nowhere.

All and any help is appreciated. Thank you kindly!

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    Your question is not well formed. What is it that you must prove? – 2017-02-19
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    Hint: expand $(r+1/r)(r^k + 1/r^k)$ and express in terms of the form $r^n + 1/r^n$ – 2017-02-19
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    String: Sorry, missed a phrase: we want to prove that $r^n + 1/r^n$ is also and integer. – 2017-02-19
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    **Hint** $ $ Exploit [innate symmetry.](http://goo.gl/b0qqsb) For $\rm\:y=r,\,z = r^{-1}\:$ we know $\rm\:\color{#c00}{yz,\ y+z\in\Bbb Z}.\,$ Now use the recurrence $$\rm\quad\ \: y^{n+1}+z^{n+1}\ =\ (\color{#c00}{y+z})\ (y^n+z^n) -\ \color{#c00}{yz}\: (y^{n-1}+z^{n-1})\quad for\ \ all\ \ \ n \ge 0\qquad\quad $$ to deduce by induction that $\rm\,y^n+z^n\in\Bbb Z\,$ for all $\rm\,n\ge 0.\ $ **Remark** $ $ Above is a special case of [Newton's identities](http://goo.gl/eFRW6Z) for expressing power sums in terms of elementary summetric polynomials. – 2017-02-20

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First note that that the result is true for $k=0$. It is also true for $k=1$ (by hypothesis)

For $k \geq 2$ \begin{eqnarray*} r^{k}+\frac{1}{r^{k}}=(r+\frac{1}{r})(r^{k-1}+\frac{1}{r^{k-1}})-(r^{k-2}+\frac{1}{r^{k-2}}) \end{eqnarray*} & so the result follows by induction.