
NT1 Find all the pairs positive integers $(x, y)$ such that

$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{[x, y]}+\frac{1}{(x, y)}=\frac{1}{2}
$$

where $(x, y)$ is the greatest common divisor of $x, y$ and $[x, y]$ is the least common multiple of $x, y$.


## Solution

We put $x=d u$ and $y=d v$ where $d=(x, y)$. So we have $(u, v)=1$. From the conclusion we obtain $2(u+1)(v+1)=d u v$. Because $(v, v+1)=1, v$ divides $2(u+1)$.

Case 1. $u=v$. Hence $x=y=[x, y]=(x, y)$, which leads to the solution $x=8$ and $y=8$.

Case 2. $uv$. Because of the symmetry of $u, v$ and $x, y$ respectively we get exactly the symmetrical solutions of case 2 .

Finally the pairs of $(x, y)$ which are solutions of the problem are:

$(8,8),(9,24),(24,9),(5,20),(20,5),(12,15),(15,12),(8,12),(12,8),(6,12),(12,6)$.
