Find every n $\in \mathbb{N}$ such that $n+1 \mid n^2+3$

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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).

Answer 4

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.