Find all natural numbers $n$ such that $n+1$ divides $3n+11$ 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?

Set $n+1=m$
$3n+11=3(m-1)+11=3m+8\equiv8\pmod m$
