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**DIOGYK** Yes it makes lots of sense, thanks!

I did this:

Since $\displaystyle x+\frac{1}{x}=a, a\in\mathbb{Z}$, and $\displaystyle x^k+\frac{1}{x^k}=b, b\in\mathbb{Z}$, $\displaystyle ab=(x+\frac{1}{x})(x^k+\frac{1}{x^k})=(x^{k+1}+ \frac{1}{x^{k+1}})+(x^{k-1}+\frac{1}{x^{k-1}})$.

So since we assumed that $\displaystyle x^k+\frac{1}{x^k}=b, b\in\mathbb{Z}$, this also assumes that $\displaystyle x^{k-1}+\frac{1}{x^{k-1}}=d, d\in\mathbb{Z}$, right? if so..

Since $\displaystyle a\in\mathbb{Z}$ & $\displaystyle b\in\mathbb{Z}$, $\displaystyle ab\in\mathbb{Z}$, and we assumed that $\displaystyle d\in\mathbb{Z}$, so $\displaystyle x^{k+1}+ \frac{1}{x^{k+1}}=ab-(x^{k-1}+\frac{1}{x^{k-1}})=ab-d$, which means that $\displaystyle x^{k+1}+ \frac{1}{x^{k+1}}\in\mathbb{Z}$, which means that $\displaystyle x^n+\frac{1}{x^n}\in\mathbb{Z}$ right? Is this proof accurate, if so, Big thanks to you Prove It, johnsomeone and emakarov : )