TITLE: If $x+\frac{1}{x}$ is a natural number, then $x^{n}+\frac{1}{x^{n}}$ is a natural number for all $n\in \mathbb{N}.$
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.