Find all positive integers $n$ such that $n^2 + 1$ is divisible by $n+ 1$ 
Find all positive integers $n$ such that $n^2 + 1$ is divisible by $n+ 1$

I know a lot about number theory, for example conjectures, theories, equations and so on. But I just can't solve this type of problems. Which algorithm should i use in order to solve them?
 A: Hint:
$n^2+1=(n+1)(n-1)+2$
$n+1$ divides $n^2+1$ if and only if it divides $2$
A: Another proof:
$$gcd(n+1,n^2+1)=gcd(n+1, n^2+1-n(n+1))=gcd(n+1,1-n)=gcd(n+1,1-n+(1+n))=gcd(n+1,2).$$
If $n$ is even, then $n+1$ is odd. Then $gcd(n+1,2)=1$ (is this clear?), this is:
$gcd(n+1,n^2+1)=1$ so that $n+1,n^2+1$ are relative primes.
If $n$ is odd, then $n+1$ is even, so $gcd(n+1,2)=2$, thus $gcd(n+1,n^2+1)=2$. 
Remember that: If $a>0$, then $gcd(a,b)=a \iff a \vert b. $ 
 Then $n+1 \vert n^2+1 \iff gcd(n+1,n^2+1)=n+1,$ then $n+1$ must be equal to $2$, then $n=1$ is the only positive integer that satisfies the property. 
A: $\dfrac{n^2+1}{n+1}$ is a natural number
$\Leftrightarrow \dfrac{n^2+2n+1-2n}{n+1}$ is a natural number
$\Leftrightarrow \dfrac{(n+1)^2-2n}{n+1}$ is a natural number
$\Leftrightarrow n+1-\dfrac{2n+2-2}{n+1}$ is a natural number
$\Leftrightarrow n+1-2-\dfrac{2}{n+1}$ is a natural number
$\Leftrightarrow (n+1)|2$
$\Leftrightarrow n=1$ because $n$ must be a positive odd number and $2$ is divisible by $(n+1)$.
A: $n+1|n^2+1$
$\Rightarrow n+1|n^2+1+2(n+1)$
$\Rightarrow n+1|(n^2+2n+1)+2$
$\Rightarrow n+1|(n+1)^2+2$
$\Rightarrow n+1|2  \hspace{1cm}(1)$
We have, $n+1≥2$ since $n≥1$, but from $(1)$ we have $n+1≤2$.
Therefore $n+1=2 \Rightarrow n=1$ and it satisfies $(1)$.
Hence there is only one such positive integer which is 1.
A: Hint. Note that
$$\frac{n^2+1}{n+1}=\frac{n^2-1+2}{n+1}=n-1+\frac{2}{n+1}.$$
