Is ${x+\frac 1x}$ equal an integer number, only $x=1$ ? If, $x>0$. 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 $x \in \mathbb{Q}$.
 A: 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 $1$.
A: 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 $2n+1$.
A: 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 $x=1$.
 
A: Assume $n>1$, then $n<n+\frac{1}{n}<n+1$
