For which values of $n$ is the fraction

$$
\frac{3 n^{2}+2 n+4}{n+1}
$$

irreducible?
We perform the Euclidean division of $3 n^{2}+2 n+4$ by $\mathfrak{n}+1$:

$$
3 n^{2}+2 n+4=(3 n-1)(n+1)+5
$$

Let $d$ be the GCD of $3 n^{2}+2 n+4$ and $n+1$. According to the previous equality, $d$ divides 5. Therefore, $d=1$ or $d=5$. Now, 5 divides $n+1$ if and only if $n \equiv 4 \bmod 5$, and when $n \equiv 4 \bmod 5$, we have

$$
3 n^{2}+2 n+4 \equiv 3 \cdot 4^{2}+2 \cdot 4+4 \equiv 0 \quad \bmod 5
$$

Thus, the fraction is irreducible if and only if $\mathfrak{n} \not \equiv 4 \bmod 5$, in other words, if and only if $\mathrm{n}-4$ is not divisible by 5.