For which integers $n$ are the fractions reducible:
a) $\frac{n^{2}+2 n+4}{n^{2}+n+3}$
b) $\frac{n^{3}-n^{2}-3 n}{n^{2}-n+3}$?
a) Since $\left(n^{2}+2 n+4, n^{2}+n+3\right)=(n+1,3)$, the fraction will be reducible when $(n+1,3)=3$.

b) Since $\left(n^{3}-n^{2}-3 n, n^{2}-n+3\right)=\left(n^{2}-n+3,6 n\right)$, the fraction can be reduced by either 2, or 3, or some divisor of the number $n$. The first case is impossible since the number $n^{2}-n+3$ is always odd. In the second case, $n$ must equal $3 k$ or $3 k+1$. In the third case,

$\left(n^{2}-n+3, n\right)=(n, 3)$, so $n$ must again be a number of the form $3 k$.

## Answer

a) For $n=3 k-1 ;$ b) for $n=3 k$ and $n=3 k+1$.