Determine the integers $n$ such that there is an odd number of solutions $(x, y)$ such that:

$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{n}
$$
The equation can be rewritten as:

$$
(x-n)(y-n)=n^{2}
$$

There are as many solutions as there are divisors of $n$. For each divisor $a$ of $n$, we can associate another divisor $n / a$, which is always different if and only if $n$ is not a square. Therefore, $n$ has an even number of divisors if and only if $n$ is not a square. Thus, the previous equation has an odd number of solutions if and only if $n$ is a square.