<s>[INST] 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.
 [/INST] Using effectively polynomial long division, we have
$$ \frac{n^2+3}{n+1} = \frac{n(n+1)-n+3}{n+1} \\
= n + \frac{-n+3}{n+1} = n - \frac{n+1-1-3}{n+1} = n-1+\frac{4}{n+1}, $$
so $n+1 \mid 4$ (and hence is $0,1,2$ or $4$, and the first case is excluded).
</s>