Exercise 1. Find all integers $x, y \geqslant 1$ such that

$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{2017}
$$
Solution to Exercise 1 The equation is equivalent to $2017(x+y)=xy$, so $2017|xy$. Since 2017 is prime, $2017|x$ or $2017|y$. Without loss of generality, since $x$ and $y$ play the same role, we can assume that $2017|x$. Let $x' \geq 1$ such that $x=2017x'$. We then have $2017x' + y = x'y$, which simplifies to

$$
2017x' = y(x' - 1)
$$

Thus, $2017|y$ or $2017|x' - 1$, as before.

If $2017|y$, we write $y=2017y'$, and thus $x' = y'(x' - 1)$, or $x' + y' = x'y'$. We see that there is no solution if $x' = 1$ or $y' = 1$. If $x' = 2$, we find $y' = 2$ and vice versa. If $x', y' \geq 3$, then $x'y' \geq \max(3x', 3y') > x' + y'$, so there is no solution. Thus, we find a unique solution: $x' = y' = 2$, which gives $x = y = 4034$. We verify that this satisfies the original equation.

If $2017 \mid x' - 1$, we write $x' = 2017k + 1$ with $k \in \mathbb{N}$. We then get $2017k + 1 = yk$, or $(y - 2017)k = 1$. This implies that $k = 1$ and thus $y - 2017 = 1$. Therefore, $y = 2018$ and $x = 2017 \times 2018$. We verify that this satisfies the initial equation.

Conclusion: There are three solutions, namely the integer pairs $(4034, 4034)$, $(2017 \times 2018, 2018)$, and $(2018, 2017 \times 2018)$. Indeed, since $x$ and $y$ play the same role, the second case we treated gives two solutions.