<s>[INST] 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?
 [/INST] 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?
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