Let $n \in \mathbb{N}$. Find the set of common divisors of $2 n+3$ and $3 n+2$.
Solving this problem amounts to finding $d \geq 0$ the GCD of $2 n+3$ and $3 n+2$. Let's write the Euclidean division of $n$ by $5: n=5 k+r$ where $k \in \mathbb{Z}$ and $r \in$ $\{0 ; 1 ; 2 ; 3 ; 4\}$. By substituting $n$ with this expression, we have that $d \mid 10 k+(2 r+3)$ and $d \mid 15 k+(3 r+2)$, hence $d \mid 3 \times(10 k+(2 r+3))-2 \times(15 k+(3 r+2))=5$. Thus $d \in\{1 ; 5\}$ and therefore $d \mid(3 r+2)-(2 r+3)=r-1$. Since $r \in\{0 ; 1 ; 2 ; 3 ; 4\}$, we have $d=5 \Rightarrow r=1$. The converse is clear. Thus, $d=5 \Leftrightarrow n \equiv 1[5]$, and $d$ is 1 in the other cases.