Find all $k \in \mathbb{R}$ such that both $\frac{12}{k+1}$ and $\frac{6}{k-1}$ are positive integers 
Find all $k \in \mathbb{R}$ such that both $\frac{12}{k+1}$ and $\frac{6}{k-1}$ are positive integers

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.
 A: Without loss of generality $b>0$. Since $a>b$, $12/(k+1)<6$. Setting it to each of $1,\,\cdots,\,5$ obtains $k$, hence $6/(k-1)$.
A: 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
$$k=\frac 6n+1.$$
A: Starting from $\frac{6}{k-1}=n \in \mathbb N$, we can generate all possible rational $k=\frac{n+6}{n}$ without regard to the second condition.
Considering the second condition, $\frac{12}{k+1}=m \in \mathbb N$, we substitute to get $m=\frac{12}{\frac{n+6}{n}+1}=\frac{12n}{2n+6}=\frac{6n}{n+3}$. We need to find positive integers $n$ that yield positive integers $m$.
Note that $\frac{6n}{n+3}$ asymptotically  approaches $6$, so we only have to look at values of $n$ such that $\frac{6n}{n+3} \le 5$, which by algebraic rearrangement yields $n \le 15$. We find $m\in \mathbb N \Rightarrow n= 3,6,15$, which in turn yields $k=3,2,\frac{7}{5}$, which is exactly what you found by trial and error.
