$x+1/x$ an integer implies $x^n+1/x^n$ an integer

The base case of $n=1$ is true, and suppose it holds for all $k<n$ in order to do the induction step. Then $$(x^{n-1}+1/x^{n-1})(x+1/x)=x^n+1/x^{n-2}+x^{n-2}+1/x^n=(x^n+1/x^n)+(x^{n-2}+1/x^{n-2})$$ so $$x^n+1/x^n=(x^{n-1}+1/x^{n-1})(x+1/x)-(x^{n-2}+1/x^{n-2})$$ which is an integer, so the result follows by induction.

Hint: Expand $(x+\frac{1}{x})^n$ using the binomial theorem. Show that this is a linear combination of elements of the form $x^m+\frac{1}{x^m}$ for $m \leq n$ and of $1$. Then use induction.

Hint let $$a_{n}=x^n+\dfrac{1}{x^n}$$ then we have $$a_{n+2}=(x+\dfrac{1}{x})a_{n+1}-a_{n}=ka_{n+1}-a_{n},x+\dfrac{1}{x}=k\in Z$$ since $a_{1}=k\in Z,a_{2}=k^2-2\in Z$ and it is by use Mathematical induction