For what integer $n$ is $1+n$ divisible by $1+n^2$?

Question

For what integer $n$ is $1+n$ divisible by $1+n^2$?

I tried it in modulo $3$ and $8$ but to no avail.

Trial and error gave me answers $0$,$1$,$-2$ and $-3$.

Answer 1

Supposing you mean $n^2+1$ is divisible by $n+1$, we have:

$$n^2+1=(n^2-1)+2= (n-1)(n+1)+2$$ We know that $(n-1)(n+1)$ is divisible by $n+1$, and the problem states that $n^2+1$ is also divisible by $n+1$, thus $n+1$ divides their difference $(n^2+1)-(n^2-1)=2$. So acceptable values for $n+1$ are $1$, $2$, $-1$, $-2$, thus $n=0, 1, -2, -3$ are all integer solutions for this problem.

Answer 2

HINT:

$$\dfrac{n^2+1}{n+1}=\dfrac{(n+1-1)^2+1}{n+1}=n+1-2+\dfrac2{n+1}$$

OR

$$\dfrac{n^2+1}{n+1}=\dfrac{(n+1)(n-1)+2}{n+1}=n-1+\dfrac2{n+1}$$

Answer 3

The only solutions are 0 and 1