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Assume $x \in \mathbb{R}$ and $(x + \frac{1}{x}) \in \mathbb{Z}$.

Use Induction to prove that for all $n \in \mathbb{N},~ (x^n + \frac{1}{x^n}) \in \mathbb{Z}$.

I'm not sure how to use the information given to prove the induction step. What I have so far is.

$P(n) : (x^n + \frac{1}{x^n}) \in \mathbb{Z}$

Base Case Then $(x^0 + \frac{1}{x^0}) = 2 \in \mathbb{Z}$

Induction Hypothesis: $x^k + \frac{1}{x^k}\in \mathbb{Z}$

Want to prove: $x^{k+1} + \frac{1}{x^{k+1}} \in \mathbb{Z}$

My first step was

$x^{k+1} + \frac{1}{x^{k+1}} = x \cdot x^k + \frac{1}{x} \cdot \frac{1}{x^k}$

Winther
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rm120
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1 Answers1

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Use $$ \left( x + \frac{1}{x} \right) \left( x^n + \frac{1}{x^n} \right) = \left( x^{n+1}+\frac{1}{x^{n+1}} \right)+ \left( x^{n-1} + \frac{1}{x^{n-1}} \right). $$ What do you know about the factors on the left and the second term on the right?

Chappers
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  • So for the left hand side, both terms are integers (given and induction hypothesis). Then for the second term on the right, I'm not sure. Would it follow that it's also an integer based on the principle of complete induction? – rm120 Jan 19 '16 at 17:40
  • Yes, provided you check two basis cases. – Chappers Jan 19 '16 at 17:49
  • Got it, thanks. – rm120 Jan 19 '16 at 18:04