4. (7 points) Natural numbers $m, n$ are such that the fraction $\frac{m}{n}$ is irreducible. Is the fraction $\frac{n}{2 m+3 n}$ reducible?
Solution. Let $n$ be divisible by $d$ and $2 m+3 n$ be divisible by $d, d \neq 1$. Then $3 n$ is divisible by $d$ and $2 m$ is divisible by $d$.

From the fact that $2 m$ is divisible by $d$ it follows that 2 is divisible by $d$, i.e., $d=2$ (since $m$ is not divisible by $d$, as the fraction $\frac{m}{n}$ is irreducible by the condition).

Since $n$ is divisible by $d$ and $d=2$, then $n=2 k$, i.e., the fraction $\frac{n}{2 m+3 n}$ is reducible for even $n$.

Answer. Reducible for even $n$.