Subject 4. Determine the natural numbers $n$ knowing that the fraction $\frac{3 n+1}{2 n-7}$ is reducible.
## Solution:

If the fraction is reducible, then there exists $d \neq 1$ such that $d \mid 3n + 1$ and $d \mid 2n - 7$. From here, we have $d \mid 2(3n + 1) - 3(2n - 7)$, which means $d \mid 23$, hence $d = 23$. Now, $23 \mid 3n + 1$ and $23 \mid 2n - 7$ imply that $23 \mid n + 8$, i.e., $n + 8 = 23k$, for $k \in \mathbb{N}^{*}$. It is verified that for $n = 23k - 8, k \in \mathbb{N}^{*}$, the fraction is reducible.

| Details of Solution | Associated Scoring |
| :---: | :---: |
| Specifies that there exists $d \neq 1$ such that $d \mid 3n + 1$ and $d \mid 2n - 7 \ldots$. | $1 \text{p}$ |
| Deduces that $d \mid 2(3n + 1) - 3(2n - 7) \Rightarrow d \mid 23 \Rightarrow d = 23$ | $2 \text{p}$ |
| From $23 \mid 3n + 1$ and $23 \mid 2n - 7 \Rightarrow 23 \mid n + 8$ i.e., $n + 8 = 23k$, for $k \in \mathbb{N}^{*}$. | $2 \text{p}$ |
| Specifies that the fraction is irreducible for $n = 23k - 8, k \in \mathbb{N}^{*}$ | $2 \text{p}$ |