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

$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{n}
$$

## - Subsets -

In combinatorics, it is very important to know how to count the number of subsets of a set (in different senses). Here we summarize some results from this idea:

- Simply, the number of subsets of a set with $n$ elements is $2^{n}$.
- The number of subsets containing $k$ elements in a set with $n$ elements is

$$
\binom{n}{k}=\frac{n!}{k!(n-k)!}
$$

- The number of $k$-tuples (we consider the order of elements) of a set with $n$ elements. The Pascal's triangle formula allows calculating binomial coefficients recursively:

$$
\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}
$$

| $n \backslash k$ | 0 | 1 | 2 | 3 | 4 | 5 | $\cdots$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | 1 |  |  |  |  |  |  |
| 1 | 1 | 1 |  |  |  |  |  |
| 2 | 1 | 2 | 1 |  |  |  |  |
| 3 | 1 | 3 | 3 | 1 |  |  |  |
| 4 | 1 | 4 | 6 | 4 | 1 |  |  |
| 5 | 1 | 5 | 10 | 10 | 5 | 1 |  |
| $\vdots$ |  |  | $\vdots$ |  |  |  | $\ddots$ |

And binomial coefficients are also used in the calculation of the binomial theorem:

$$
(x+y)^{n}=\sum_{k=0}^{n}\binom{n}{k} x^{n-k} y^{k}
$$
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.