6
$\begingroup$

I am trying to solve the following exercise. If $x+\frac{1}{x}$ is a natural number, then $x^{n}+\frac{1}{x^{n}}$ is a natural number for all $n\in \mathbb{N}.$

Here what I've done. If $x+\frac{1}{x}$ and $x^{n}+\frac{1}{x^{n}}$ are natural numbers, then $(x+\frac{1}{x})(x^{n}+\frac{1}{x^{n}})\in \mathbb{N}$. But $(x+\frac{1}{x})(x^{n}+\frac{1}{x^{n}})=x^{n+1}+\frac{1}{x^{n-1}}+x^{n-1}+\frac{1}{x^{n+1}}$ and I got that $x^{n+1}+\frac{1}{x^{n+1}}\in \mathbb{N}$. But there is a problem here: I am not supposed to use the second principle of mathematical induction.

My question is: How does one prove that by using the first principle of mathematical induction?

I would appreciate your help.

  • 0
    Counter-example: $x+\frac1x=1$ $\Big(x+\frac1x\Big)^2=x^2+2+\frac1{x^2}=1^2=1$ $x^2+\frac1{x^2}=1-2=-1$ which is not a natural number.2018-10-17

2 Answers 2

3

Hint $\ $ Put $\rm\ y = \frac{1}x\ $ in: $\rm\ x+y,\: xy\in \mathbb Z\ \Rightarrow\ x^n+y^n\in \mathbb Z\:$ by induction, since

$\rm x^{n+1}+y^{n+1}\ =\ (x+y)\: (x^n+y^n) - xy\: (x^{n-1} + y^{n-1})$

Look up Lucas sequences to learn more about such quadratic number theory.

3

By expanding $\big(x+\frac{1}{x}\big)^n$ using the binomial theorem and collecting terms of the form one gets $x^{k}+\frac{1}{x^{k}}$, one gets a proof using strong induction, that is, supposing that $x^{k}+\frac{1}{x^{k}}$ is an integer for all $k.

  • 0
    @Rankeya, you're right.2012-03-14