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.
 A: You made a mistake simplifying $n+1 \vert (n^2+3)-(n^2+2n+1)$.  The next step should be $n+1 \vert 2-2n$ and you end up with $n+1 \vert 4$.  Then just check all the cases.
If this were a different problem like "find all $n$ such that $n+1 \vert n^2+3n+2$" in which $n+1 \vert 0$ could be legitimately derived in the same way then this derivation carried out in reverse would establish that it is true for all $n$.  So one thing to try if you encounter something like $n \vert 0$ is see if it is possible to work backwards from there.
A: Hint $\ $ If $\,f(x)\,$ is a polynomial with integer coef's then $\,n\!+\!1\mid f(n)\iff n\!+\!1\mid f(-1)$
since, by division, $\, f(n) = q(n)(n\!+\!1) + r,\,$ so $\,f(-1) = r\,$ be evaluating at $\,n=-1.$
Therefore $\,n\!+\!1\mid f(n)\iff n\!+\!1\mid f(n)-q(n)(n\!+\!1) = r = f(-1)$
Remark $\ $ If you know modular arithmetic then the proof is more intuitive:
${\rm mod}\ n\!+\!1\!:\,\ \color{#c00}{n\equiv -1}\,\Rightarrow\, f(\color{#c00}n)\equiv f(\color{#c00}{-1})\ $ so $\ f(n)\equiv 0\iff f(-1)\equiv 0$  
where above we used the Polynomial Congruence Rule.
A: 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. 
A: 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).
