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I'm looking for this question in MSE. But,I could not find a similar one.

Is ${x+\dfrac 1x}$ equal an integer number, only $x=1$ ? If, $x>0$.

I can prove this only when $x\in \mathbb{N}$ but not for $x \in \mathbb{R}^+$.

Remark: the duplicate link just answer the case where $x \in \mathbb{Q}$.

Siong Thye Goh
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4 Answers4

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Let $x>0$,$$x + \frac{1}{x} = k \in \mathbb{Z}$$

$$x^2 + 1 = kx$$

$$x^2-kx+1=0$$

$$x = \frac{k\pm\sqrt{k^2-4}}{2}$$

Now, let $k=3$, then $x=\frac{3+\sqrt{5}}{2}$, hence $x$ need not be equal to $1$.

Siong Thye Goh
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    Are $11$ seconds sufficient to read and understand this post ? – Peter Oct 02 '17 at 12:00
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    @peter why is that your or the OP's attention span ? –  Oct 02 '17 at 12:01
  • I was just surprised that an upvote appeared $11$ seconds after the post, nothing more. – Peter Oct 02 '17 at 12:02
  • @Peter I have seen upvotes coming even after 5 seconds. Sometimes people upvote first and then read the post because even first sight of lost sometimes give an exact idea of mathematical consistency of posy to someone who already know the answer. – Vidyanshu Mishra Oct 02 '17 at 12:05
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    @Peter I was the first to upvote, and after 10 seconds, it appeared to me that this answer points out that $x+\frac1x=3$ has a non-integer solution, so I upvoted the answer. Then I checked it, of course (but that was after the 11 seconds...) – 5xum Oct 02 '17 at 12:09
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The solutions of

$$x+\frac1x=n$$ are

$$\frac{n\pm\sqrt{n^2-4}}2.$$

For $x\in\mathbb R$, there are infinitely many solutions.

If you restrict yourself to $x\in\mathbb Z$, obviously the inverse of an integer other than $\pm1$ is not an integer.

If you restrict $x$ to $\mathbb Q$, by the formula above, $n^2-4$ must be a perfect square (the square root of an integer is never a proper fraction). Then

$$n^2-4=m^2$$ has no other solutions than $n=\pm2,m=0$, because the difference between two different squares is at least $2n+1$.

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For any positive integer $n > 2$, the equation $$x + {\small{\frac{1}{x}}} = n$$ has two positive real roots (the roots of $x^2 - nx + 1 = 0$), so without further restrctions on $x$, it's not true that for $x > 0$, the expression $x + {\large{\frac{1}{x}}}$ is an integer only if $x=1$.

But if you require $x$ to be a positive integer, with $x > 1$, then $$x < x + {\small{\frac{1}{x}}} < x + 1$$ so $x + {\large{\frac{1}{x}}}$ is trapped between two consecutive integers, hence can't be an integer.

Next suppose $x$ is a positive rational number, with $x \ne 1$, such that $x + {\large{\frac{1}{x}}}$ is a positive integer, equal to $n$, say. \begin{align*} \text{Then}\;\;&x + {\small{\frac{1}{x}}} = n\\[4pt] \implies\;&x^2 + 1 = nx\\[4pt] \implies\;&x^2 -nx + 1 = 0\\[4pt] \end{align*} but then, by the rational root test, $x$ must be an integer, which we already know is impossible (since $x > 0,\;x \ne 1$).

To recap:

  • For $x > 0$, if you allow irrational values of $x$, the expression $x + {\large{\frac{1}{x}}}$ can be made equal to any positive integer $n > 2$, for some some positive irrational number $x$.
  • For positive rational values of $x$, the expression $x + {\large{\frac{1}{x}}}$ is an integer if and only if $x=1$.
quasi
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Assume $n>1$, then $n<n+\frac{1}{n}<n+1$

AlvinL
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