Does it hold/can you prove that if $\frac{1}{x} + x$ is an integer, then $x = 1$? I am trying to show that if $\frac{1}{x} + x$ is an integer, then $x = 1$, where $x$ is a positive integer. Not sure where to begin
 A: Let's use $n$ instead of $x$, as this is how positive integers are typically denoted.
You want to prove the statement $\frac1n+n\in\mathbb{N} \implies n=1$.
Instead, prove the equivalent statement $n\neq1 \implies \frac1n+n\not\in\mathbb{N}$:
$n\neq1\implies$
$n>1\implies$
$\frac1n<1\implies$
$\frac1n\not\in\mathbb{N}\implies$
$\frac1n+n\not\in\mathbb{N}$
A: Suppose $x\gt 1$. Then, $\frac1x\lt1$.
Since $x$ is an integer, if $x+\frac1x$ was an integer, then, $$x+\frac1x-x=\frac1x$$ would be an integer.
But $0\lt\frac1x\lt1$. Thus, as no integer lies between $0$ and $1$, $\frac1x$ cannot be an integer.
The only possibility left is $x=1$, which is easily verified.
A: In simple terms, since $x$ is integer , for $x+\frac{1}{x}$ to be integer, $\frac{1}{x}$ should be integer , since $\text{integer}+\text{unknown} = \text{integer}$ $\implies$ $\text{unknown} = \text{integer}$, $\frac{1}{x}$ is integer only when $x$ divides $1$, which is when $x=1$.
A: Suppose that $\frac{1}{x}-x=n$ for some integer $n$. Then $1=x(n-x)$. Hence $x\mid 1$. Since $x$ is a postive integer that divides 1, we conclude that $x=1$.
