Question: > 
> Find all $k \in \mathbb{R}$ such that both $\frac{12}{k+1}$ and $\frac{6}{k-1}$ are positive integers
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Clearly $k$ is a rational number, so I let $k=\frac{a}{b}$ where $a, b \in \mathbb{Z}$ and $b \neq 0$, this gives $\frac{12b}{a+b}$ and $\frac{6b}{a-b}$ are positive integers. I tried different things here but couldn't make progress. I want to verify that $2, 3$ and $\frac{7}{5}$ are the only possible values for $k$. How to proceed from here? Thanks in advance.

Answer: 
Without loss of generality $b>0$. Since $a>b$, $12/(k+1)<6$. Setting it to each of $1,\,\cdots,\,5$ obtains $k$, hence $6/(k-1)$.
Alternatively,

$$\begin{cases} \frac {12}{k+1}=m \\ \frac{6}{k-1}=n \end{cases}$$

$$ \frac{12}{m}-1=\frac 6n +1$$

$$m=6-\frac{18}{n+3}$$

$$n=3,6,15$$ which follows, $$m=3,4,5.$$

Then, you can find all values of $k$, from the formula

$$k=\frac 6n+1.$$