Following the example of my teacher:

> 
> Find all natural numbers $n$ such that $n-2$ divides $n+5$.
> 
> 
> $$n+5 = n-2+7$$
> 
> 
> As $n-2|n-2$, $n-2$ will divide $n+5$ if and only if $n-2|7$. Yet, $7$ has the divisors $-7$, $-1$, $1$, $7$ hence the equations:
> 
> 
> * $n-2=-7 \Leftrightarrow n = -5 $
> * $n-2=-1 \Leftrightarrow n = 1 $
> * $n-2=7 \Leftrightarrow n = 3 $
> * $n-2=-1 \Leftrightarrow n = 9 $
> 
> 
> So $S = \{ 1, 3, 9 \}$
> 
> 
> 

I decompose $n+1$ the exact same way:

$$n+1 = 3n+11 - 2(n+5)$$

But I'd get stuck as $2(n+5)$ since only $-2$, $-1$, $1$ and $2$ are divisors, which don't satisify the equation as I'd hoped:

* $3n+11 = -2 \Leftrightarrow n =\frac{-13}{3}$
* $3n+11 = -1 \Leftrightarrow n = -4$
* $3n+11 = 1 \Leftrightarrow n =\frac{-10}{3}$
* $3n+11 = 1 \Leftrightarrow n = -3$

Any clues?