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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.

jiten
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2 Answers2

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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.

dxiv
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  • Seems a good alternate way, please elaborate the last line's part stating 'Newton's identities'. At least, I could not grasp that part. Thanks in anticipation. – jiten Jun 17 '18 at 02:47
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    @jiten Newton's identities allow expressing the power sums of the roots as an algebraic (polynomial) expression in the elementary symmetric functions, which in this case are just $e_0=1,$, $,e_1=y,$ and $,e_2=1,$. The explicit expressions are provided on the linked page under Expressing power sums in terms of elementary symmetric polynomials. The main point is that the RHS is an algebraic expression involving only integers, therefore so is the power sum on the LHS. – dxiv Jun 17 '18 at 02:58
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    Thanks a lot for the answer, & the link. It has added to my toolkit, – jiten Jun 17 '18 at 03:09
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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.

DanielV
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