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

Question

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?

Answer 1

Hint. Note that $$\frac{n^2+1}{n+1}=\frac{n^2-1+2}{n+1}=n-1+\frac{2}{n+1}.$$

Answer 2

Hint:

$n^2+1=(n+1)(n-1)+2$

$n+1$ divides $n^2+1$ if and only if it divides $2$

Answer 3

$\dfrac{n^2+1}{n+1}$ 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 $(n+1)$.

Answer 4

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.

Answer 5

$n+1|n^2+1$ $\Rightarrow n+1|n^2+1+2(n+1)$ $\Rightarrow n+1|(n^2+2n+1)+2$ $\Rightarrow n+1|(n+1)^2+2$ $\Rightarrow n+1|2 \hspace{1cm}(1)$

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.