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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!

Gizmo
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  • Your question is not well formed. What is it that you must prove? – String Feb 19 '17 at 20:44
  • Hint: expand $(r+1/r)(r^k + 1/r^k)$ and express in terms of the form $r^n + 1/r^n$ – Henry Feb 19 '17 at 20:44
  • String: Sorry, missed a phrase: we want to prove that $r^n + 1/r^n$ is also and integer. – Gizmo Feb 19 '17 at 20:52
  • Hint $ $ Exploit innate symmetry. 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 for expressing power sums in terms of elementary summetric polynomials.

     – Bill Dubuque Feb 20 '17 at 01:45

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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.

Donald Splutterwit
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