Find every n $\in \mathbb{N}$ such that $n+1 \mid n^2+3$ Find every n $\in \mathbb{N}$ such that $n+1 \mid n^2+3$
What I did:
$n+1 \mid n^2+3$ and $n+1 \mid (n+1)^2=n^2+2n+1$
So
$n+1 \mid (n^2+3)-(n^2+2n+1) \Longrightarrow n+1\mid-2(n+1)$
$\Longrightarrow n+1\mid0$
And then
$(n+1)k=0$
No idea what to do next.

Using polynomial division, write $n^2+3=(n+1)(n-1)+4$. Since $n+1|(n+1)(n-1)$ and $n+1|n^2+3$, it must be that $n+1|(n^2+3-(n+1)(n-1))$, or $n+1|4$. This limits our search to $n=0$, $n=1$, or $n=3$. Simple calculations show that all of these answers work. 
