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?
 A: $n+1$ certainly divides $3n+3$. If it divides $3n+11$, it must also
divide $(3n+11)-(3n+3)$.
A: You can use polynomial long division, which is useful if you have a more complex expression:
$$
\require{enclose}
\begin{array}{rll}
    3 && \\[-3pt]
   n+1 \enclose{longdiv}{\ 3n+11}\kern-.2ex \\[-3pt]
      \ \ - \underline{\ (3n+3)} &&  \\[-3pt]
      8 &&  \\[-3pt]
  \end{array}
$$
Therefore $\frac{3n+11}{n+1} = 3 + \frac{8}{n+1}$. Can you continue?
A: Set $n+1=m$
$3n+11=3(m-1)+11=3m+8\equiv8\pmod m$
A: When the example has
$$  n + 5 = n-2 + 7  $$
it might have been more clearly written 
$$  n + 5 = 1\cdot(n-2) + 7  $$
to highlight the quotient, $1$, and remainder, $7$, of the division $\frac{n+5}{n-2}$.
For your problem, 
$$  3n + 11 = 3 \cdot (n+1) + 8  \text{.}  $$
That is, the quotient is $3$ and the remainder is $8$.  So you want to inspect the choices of $n$ such that $(n+1) \mid 8$.
A: $\!\! \bmod n\!+\!1\!:\,\ \color{#c00}{n\equiv -1}\,\Rightarrow\, 3\,\color{#c00}n+11\equiv 3(\color{#c00}{-1})+11\equiv 8\ $ by Congruence Sum & Product Rules $\ \ \ \ \ $
