Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer. Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer.
Attempt:
We have
\begin{equation*}
\frac{3n^2+4n+5}{2n+1} = \frac{4n^2+4n+1 - (n^2-4)}{2n+1} = 2n+1 - \frac{n^2-4}{2n+1}.
\end{equation*}
So, we must have $(2n+1) \mid (n^2-4)$, so $n^2-4 = k(2n+1)$, for some $k \in \Bbb Z$. But, I did not be able to find $n$ from here.
Any ideas? Thanks in advanced.

Another approach:
By direct division we find:
$3n^2+4n+5=(\frac 32 n+\frac 54)(2n+1)+\frac{15}4$
multiplying both sides by $4$ and simplify we get:
$12n^2+16n+5=(6n+5)(2n+1)$
Or:
$\frac{4(3n^2+4n+5)}{2n+1}=6n+5+\frac {15}{2n+1}$
Therefore $2n+1$ must divide $15$ that is it must be equal to one of its divisors which are $\pm 1$, $\pm 3$, $\pm 5$ and $\pm 15$
1): $2n+1=1\rightarrow n=0$, $2n+1=-1\rightarrow n=-1$
2): $2n+1=3\rightarrow n=1$, $2n+1=-3 \rightarrow n=-2$
3): $2n+1=5\rightarrow n=2$, $2n+1=-5 \rightarrow n=-3$
4): $2n+1=15 \rightarrow n=7$, $2n+1=-15 \rightarrow n=-8$  so solutions are:
$0, \pm 1, \pm 2, -3, 7, -8$
