## Task B-3.3.

For which natural numbers $n$ is the value of the expression $\frac{n^{2}-4 n+4}{n+1}$ an integer?
Solution.

$\frac{n^{2}-4 n+4}{n+1}=\frac{(n+1)(n-5)+9}{n+1}=n-5+\frac{9}{n+1}$

The expression $n-5+\frac{9}{n+1}$ is an integer if $\frac{9}{n+1} \in \mathbb{Z}$.

This is possible only if $n+1$ is a divisor of the number 9, i.e., $n+1 \in\{ \pm 1, \pm 3, \pm 9\}$.

Then $n \in\{-2,0,-4,2,-10,8\}$.

Since $n \in \mathbb{N}$, the sought numbers are $n \in\{2,8\}$.