291. Find all integers $n$ for which the fraction $\frac{n-2}{n+1}$ takes integer values.
$\triangle$ To isolate the integer part of the fraction in the numerator, we add and subtract 1 (the "plus-minus" technique):

$$
\frac{n-2}{n+1}=\frac{(n+1)-1-2}{n+1}=\frac{(n+1)-3}{n+1}=1-\frac{3}{n+1}
$$

The fraction $\frac{3}{n+1}$ must be equal to an integer. Let's consider all cases.

1) Suppose $n+1=1$. Then $n=0$.
2) Suppose $n+1=-1$. Then $n=-2$.
3) When $n+1=3$, we get $n=2$.
4) When $n+1=-3$, we find that $n=-4$.

Answer: $2,0,-2,-4$