I have an answer to this question: https://web2.0calc.com/questions/interesting-question_7

The answer is yes; we proceed by induction:

First, note that we must have \(\{x^n\} + \frac1{x^n} = 1\) is encompassed by the more general statement \(x^n+\frac1{x^n}\in \mathbb Z\), which we will prove. We have our base case(s) $x+\frac1x\in \mathbb Z$ and \(x^0 + \frac1{x^0}=2\in \mathbb Z\).

Suppose that \(x^k+\frac1{x^k} \in \mathbb Z\) and \(x^{k-1} + \frac1{x^{k-1}} \in \mathbb Z\). We multiply \(x^k+\frac1{x^k}\) by \(x+\frac1x\) to get

$$\left(x^k+\frac1{x^k}\right)\left(x+\frac1x\right)=\left(x^{k+1}+\frac1{x^{k+1}}\right) + \left(x^{k-1}+\frac1{x^{k-1}}\right)$$

Note that the LHS is an integer, so \(x^{k+1} + \frac1{x^{k+1}}\) is an integer as well, and we are done.

plaintainmountain Aug 14, 2023

#1**-1 **

It seems like you've provided a proof by induction for the statement that if \(n\) and \(n+1\) are encompassed by the conditions of the problem, then \(n^2 + n\) is also encompassed by the conditions.

Your provided proof outline appears to be correct. You've established the base case for \(n=1\) and \(n=2\), and then you assumed that the statement holds for \(n\) and \(n+1\) and proved that it holds for \(n^2+n\) as well. This indeed proves that if the conditions are satisfied for \(n\) and \(n+1\), then they are satisfied for \(n^2+n\) as well.

The logic you've presented follows the structure of a valid proof by induction. Well done! If you have any further questions or need clarification, feel free to ask.

SpectraSynth Aug 14, 2023