
3. Find all pairs of $(x, y)$ of positive integers such that $2 x+7 y$ divides $7 x+2 y$.


Solution: Let $d=\operatorname{gcd}(x, y)$. Then $x=d x_{1}$ and $y=d y_{1}$. We observe that $2 x+7 y$ divides $7 x+2 y$ if and only if $2 x_{1}+7 y_{1}$ divides $7 x_{1}+2 y_{1}$. This means $2 x_{1}+7 y_{1}$ should divide $49 x_{1}+14 y_{1}$. But $2 x_{1}+7 y_{1}$ divides $4 x_{1}+14 y_{1}$. Hence $2 x_{1}+7 y_{1}$ divides $45 x_{1}$. Similarly, we can show that $2 x_{1}+7 y_{1}$ divides $45 y_{1}$. Hence $2 x_{1}+7 y_{1}$ divides $\operatorname{gcd}\left(45 x_{1}, 45 y_{1}\right)=45 \operatorname{gcd}\left(x_{1}, y_{1}\right)=45$. Hence

$$
2 x_{1}+7 y_{1}=9,15 \text { or } 45
$$

If $2 x_{1}+7 y_{1}=9$, then $x_{1}=1, y_{1}=1$. Similarly, $2 x_{1}+7 y_{1}=15$ gives $x_{1}=4$, $y_{1}=1$. If $2 x_{1}+7 y_{1}=45$, then we get

$$
\left(x_{1}, y_{1}\right)=(19,1),(12,3),(5,5)
$$

Thus all solutions are of the form

$$
(x, y)=(d, d),(4 d, d),(19 d, d),(12 d, 3 d),(5 d, 5 d)
$$
