If for a real $x, x +\frac1x$ is an integer, prove $x^{2017}+\frac1{x^{2017}}$ is also.

Question

For some real $x$, let $y=x +\frac1x$, with $y\in \mathbb{Z}$.

$y=x +\frac1x\implies x^2 -xy +1 =0\implies x= \frac{y\pm \sqrt{y^2 -4}}2$, so $x+\frac1x = \frac{y\pm \sqrt{y^2 -4}}2+ \frac2{y\pm \sqrt{y^2 -4}}$

This implies: $ y = \frac{(y\pm \sqrt{y^2 -4})^2+4}{2\cdot (y\pm \sqrt{y^2 -4})}$

Not sure of proper logic (request guidance on this part), but hope that can take positive & negative signs for $ \sqrt{y^2 -4}$ alternately, in both numerator & denominator simultaneously.

Case (a) : positive $ \sqrt{y^2 -4}$ :
$ y = \frac{(y+ \sqrt{y^2 -4})^2+4}{2\cdot (y+ \sqrt{y^2 -4})}=\frac{y^2 + y\sqrt{y^2-4}}{y+\sqrt{y^2-4}}\implies y$

Case (b) : negative $ \sqrt{y^2 -4}$ :
$ y = \frac{(y- \sqrt{y^2 -4})^2+4}{2\cdot (y-\sqrt{y^2 -4})}=\frac{y^2 - y\sqrt{y^2-4}}{y-\sqrt{y^2-4}}\implies y$

There can be formed no conclusion with above, except vindicating that the roots are correct.

Now to show further that $y'= x^{2017}+\frac1{x^{2017}}$ is also an integer, there seems no way 'algebraically (i.e., direct multiplication) ' except possibly by modulus arithmetic. The direct route is not clear to me, as modulus is to be different (higher with each step) powers of $x^i+\frac1{x^i}, i\in \mathbb{Z+}$.

A variant of the above idea can be with strong induction, which considers all powers of $x, \frac1x$ till $2017$, or any positive integral power $i$ as follows:

Step 1: Base case of $n=1$ holds true, i.e. for some real $x, y = x+\frac1x$ is an integer.
Step 2: Suppose hypothesis holds for all $i\le n$ for the induction hypothesis step, i.e. :$x^n+\frac1{x^n}$ is an integer too.
Step 3: Need prove that for $i = n+1, x^i +\frac 1{x^i}$ is also an integer.

$$(x^n+\frac1{x^n})(x+\frac1x)=x^{n+1}+\frac1{x^{n-1}}+x^{n-1}+\frac1{x^{n+1}}=(x^{n+1}+\frac1{x^{n+1}})+ (\frac1{x^{n-1}}+x^{n-1})$$ so $$x^{n+1}+\frac1{x^{n+1}}=(x^n+\frac1{x^n})(x+\frac1x)-(x^{n-1}+1/x^{n-1})$$ with the r.h.s. being an integer, hence proved.

Have two doubts:
1. Is it possible to prove by finding the complex roots of $x^i +\frac1{x^i}$ and proving using finding roots, by changing / extending domain of $x$ to be in $\mathbb{C}$.
2. No idea for negative integer values of powers of $x, \frac1x$ by the strong induction approach.

Answer 1

Alt. hint: $\;a=x\,$ and $\,b=\dfrac{1}{x}\,$ are roots of the quadratic $\,z^2-y z + 1\,$ where $\,y = a+b\,$.

If $\,y=x+\dfrac{1}{x}\,$ is an integer, then the quadratic is a monic polynomial with integer coefficients, and therefore $\,a^n+b^n=x^n+\dfrac{1}{x^n}\,$ is an integer for all $\,\forall n\,$ by Newton's identities.

Answer 2

Let $A_n = (x + 1/x)^n$.

Then $A_n$ is a palindromic polynomial with integer coefficients.

Let $Y_k = x^{k} + 1/x^{k}$.

Then $Y_{k+1} = A_{k+1} - \sum_{j = 1}^k c_k Y_k$ where $c$ is the list of integer coefficients necessary to make the construction.

So if $x$ is such that $A_1$ (and thus $Y_1$) is an integer, then all $A_n$ must be integers and so all $Y_n$ must also be integers.