Question

number of integers n such that 1+n is the divisor of 1+n^2​

Answer 1

In other words, we need to find $n\in\mathbb{Z}$, for which $\dfrac{n^2+1}{n+1}$ is an integer.

$n\not=-1\\\\\dfrac{n^2+1}{n+1}=\dfrac{n^2+2n+1-2n}{n+1}=\dfrac{(n+1)^2-2n}{n+1}=n+1-\dfrac{2n}{n+1}\\\\\dfrac{2n}{n+1}=\dfrac{2n+2-2}{n+1}=\dfrac{2(n+1)-2}{n+1}=2-\dfrac{2}{n+1}$

So, $n+1$ must be an integer divisor of 2.

The integer divisors of 2 are: -2,-1,1,2

$n+1=-2 \vee n+1=-1 \vee n+1=1 \vee n+1=2\\n=-3 \vee n=-2 \vee n=0 \vee n=1$

Therefore, there are 4 integers such that $1+n$ is the divisor of $1+n^2$