Find all pairs of positive integers $x, y$ such that $x$ and $y$ are coprime, $x<y$ and $2000\left(\frac{x}{y}+\frac{y}{x}\right)$ is an odd integer.

## List 7
Solution 1: We have:

$$
2000\left(\frac{x}{y}+\frac{y}{x}\right)=16 \times 125\left(\frac{x^{2}+y^{2}}{x y}\right)
$$

Since $x$ and $y$ are coprime, we conclude that $xy$ and $x^{2}+y^{2}$ have no common factors (prove this). Therefore, $x$ and $y$ are divisors of $2000=16 \times 125=2^{4} \times 5^{3}$. Moreover, 16 must divide $xy$ because the expression must be odd. Thus, $16=2^{4}$ divides $x$ or $y$.

1st case: 16 divides $x$.

If $x>16$, then $x$ is at least $16 \times 5=80$, since $x$ divides 2000. In this case, $x>y$ because $xy$ divides 2000. Therefore, $x=16$. Since $x$ and $y$ are coprime, if $x=16$ then the possibilities are: $y=16 \times 5$ and $x=1$; $y=16 \times 25$ and $x=1$; $y=16 \times 125$ and $x=1$.

If $y=16$ then the possibilities for $x$ are: $x=1$; $x=5$.

Therefore, the pairs $(x, y)$ satisfying the conditions of the problem are:

$$
(16,25) ;(16,125) ;(5,16) ;(1,16) ;(1,80) ;(1,400) ;(1,2000)
$$

However, since we can swap $x$ and $y$ due to the symmetry of the expression, we also have the solutions:

$$
(25,16) ;(125,16) ;(16,5) ;(16,1) ;(80,1) ;(400,1) ;(2000,1)
$$

## Solution 2:

We have $N=2000\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{16 \times 125 \times\left(x^{2}+y^{2}\right)}{x y}=\frac{2^{4} \times 5^{3}}{x y} \times\left(x^{2}+y^{2}\right)$.

Since $N$ is odd, it follows that $\frac{2^{4} \times 5^{3}}{x y}$ and $x^{2}+y^{2}$ are odd. The options for this are:
$x y=2^{4}, 2^{4} \times 5,2^{4} \times 5^{2}, 2^{4} \times 5^{3}$ and, $x$ and $y$ have different parities. Let's determine $x$ and $y$ for each of these options:

| $x y$ | $x$ | $y$ |
| :---: | :---: | :---: |
| $2^{4}$ | 1 | $2^{4}$ |
| $2^{4} \cdot 5$ | 1 | $2^{4} \cdot 5$ |
|  | $2^{4}$ | 5 |
| $2^{4} \cdot 5^{2}$ | 1 | $2^{4} \cdot 5^{2}$ |
|  | $2^{4}$ | $5^{2}$ |
| $2^{4} \cdot 5^{3}$ | 1 | $2^{4} \cdot 5^{3}$ |
|  | $2^{4}$ | $5^{3}$ |

Therefore, the pairs $(x, y)$ satisfying the conditions of the problem are:

$$
(1,16) ;(1,80) ;(16,5) ;(1,400) ;(16,25) ;(1,2000) ;(16,125)
$$

However, since we can swap $x$ and $y$ due to the symmetry of the expression, we also have the solutions:

$$
(16,1) ;(80,1) ;(5,16) ;(400,1) ;(25,16) ;(2000,1) ;(125,16)
$$

## List 7