<s>[INST] If for a real $x, x +\frac1x$ is an integer, prove $x^{2017}+\frac1{x^{2017}}$ is also. 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.
 [/INST] 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.
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