$x+1/x$ an integer implies $x^n+1/x^n$ an integer Suppose that $0\neq x\in\mathbb{R}$ and $x + \frac1x\in\mathbb{Z}$. Prove that, for all  $n\ge1$, $x^n + \frac1{x^n}\in\mathbb{Z}$.
I can't figure out and understand the question. Can you give me some hints ?
 A: 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.
A: 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.
A: 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
A: The following is an extension to the hint provided by Martin Brandenburg:
expanding $(1+1/x)^n$, which is an integer by hypothesis:
$$nC0 * x^m * 1/x^0 + nC1 * x^(m-1) * 1/x + ... mC(m-1) * x * 1/x^(m-1) + mCm * x^0 * 1/x^m$$
Simplifying and rearranging, we get:
$$x^m + x^(-m) + mx^(m-2) + mx^(2-m)... $$
Clearly, this is a linear combination of terms of the for $x^a + 1/x^a$
Now assume the original hypothesis is true for all k < n. Then the terms $$mx^(m-2) + mx^(2-m)...$$ are integers, leaving $$x^m + 1/x^m$$. However, since $(1+1/x)^n$ is an integer by hypothesis, so is $$x^m + 1/x^m$$.
