Determine the positive integers $x$ and $y$ such that

$$
\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\frac{3}{2}
$$
The idea is to bound $x$ and $y$ to reduce to a finite number of cases to test.

We start by noting that, by symmetry of the problem and without loss of generality, we can assume that $x \leqslant y$.

Consequently, since

$$
00
$$

and thus

$$
0\frac{6}{5}$

$$
0<x<\frac{1}{\frac{6}{5}-1}=5
$$

It is therefore sufficient to consider $x=1,2,3$ or 4.

- For $x=1$ we obtain the equation

$$
\left(1+\frac{1}{1}\right)\left(1+\frac{1}{y}\right)=\frac{3}{2}
$$

which simplifies to

$$
\frac{1}{y}=-\frac{1}{4}
$$

which has no positive integer solution.

- For $x=2$ we obtain the equation

$$
\left(1+\frac{1}{2}\right)\left(1+\frac{1}{y}\right)=\frac{3}{2}
$$

which simplifies to

$$
\frac{1}{y}=0
$$

which has no integer solution.

- For $x=3$ we obtain the equation

$$
\left(1+\frac{1}{3}\right)\left(1+\frac{1}{y}\right)=\frac{3}{2}
$$

which simplifies to

$$
\frac{1}{y}=\frac{1}{8}
$$

with the solution $y=8$.

- For $x=4$ we obtain the equation

$$
\left(1+\frac{1}{4}\right)\left(1+\frac{1}{y}\right)=\frac{3}{2}
$$

which simplifies to

$$
\frac{1}{y}=\frac{1}{5}
$$

with the solution $y=5$.

By symmetry, the set of solutions is therefore

$$
\{(3,8),(4,5),(8,3),(5,4)\} .
$$