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Show that $X+\dfrac{1}{X}$ is not an integer number for any rational $X$ and $X \neq 1, X \neq -1$

I think we can substitue $X=\dfrac{P}{Q}$ but I don't know if I can now assume that $\gcd(P,Q)=1$

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If $(P,Q)=D$ set $$\frac Pp=\frac Qq=D$$ – lab bhattacharjee Dec 30 '13 at 19:15
If $P$ and $Q$ have a common divisor $D$, just note that $$X = \frac{P/D}{Q/D}$$ – user61527 Dec 30 '13 at 19:16
Yes, if $x$ is rational, we let $x=\frac{a}{b}$, where $a$ and $b$ are relatively prime integers. The proof that this forces $\frac{a}{b}=\pm 1$ should then go smoothly. – André Nicolas Dec 30 '13 at 19:22
up vote 2 down vote accepted

$$x + \frac{1}{x} = n =>x^2-nx+1=0$$ The equation $x^2-nx+1=0$ has rational root $=> \delta =n^2-4=k^2$, where $ k$ is integer. $=>n^2-k^2=4=>(n+k)(n-k)=4=>n+k=+-2$ and $ n-k=+-2$ $=> n=+-2 => x=+-1$ false!

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The argument has a big gap. Why can you assume that $k$ is an integer? That crucial inference requires justification. – Bill Dubuque Dec 30 '13 at 20:18
It is known that if $p$ is a integer and $\sqrt{p} is rational when \sqrt{p} is integer. For additional arguments see… – medicu Dec 30 '13 at 20:34
Exactly! If that powerful property is used in your proof then it is essential to explicitly mention that. Otherwise a reader cannot determine how that crucial inference was made (which is the only difficult part of the proof). In particular, with nothing said, one cannot determine if the inference was made rigorously. Beginning students may not be aware of this subtlety, so it is essential to make it clear. – Bill Dubuque Dec 30 '13 at 20:37

Hint $\ n = x+1/x \Rightarrow x^2\!-nx+1 = 0\,$ so $\,x\,$ is an integer dividing $1$, by the Rational Root Test.

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