a) Verify that for any positive integer $a, \operatorname{with} a>1$, the equation

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

has at least three solutions of the form $(x, y)$, with $x$ and $y$ positive integers. For example, for $a=3$, the pairs $(6,6),(4,12)$ and $(12,4)$ are solutions.

b) Find the number of pairs of positive integers $(x, y)$ that are solutions to this equation when $a=2019$.

Hint: If the prime factorization of the positive integer $n$ is $p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}} \cdot \ldots \cdot p_{k}^{\alpha_{k}}$, then it has $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \ldots\left(\alpha_{k}+1\right)$ positive divisors.
Solution

a) We can find an equivalent equation:

$$
\begin{aligned}
\frac{1}{x}+\frac{1}{y} & =\frac{1}{a} \Leftrightarrow \\
(x-a)(y-a) & =a^{2}
\end{aligned}
$$

Since $1 / x$ and $1 / y$ are less than $1 / a$, it follows that $x-a$ and $y-a$ are positive. To find solutions to the last equation, consider the following systems:

$$
\left\{\begin{array} { l } 
{ x - a = 1 } \\
{ y - a = a ^ { 2 } }
\end{array} \quad \left\{\begin{array} { l } 
{ x - a = a } \\
{ y - a = a }
\end{array} \quad \left\{\begin{array}{l}
x-a=a^{2} \\
y-a=1
\end{array}\right.\right.\right.
$$

The solutions $(x, y)$ of them are, respectively, $\left(a+1, a+a^{2}\right),(2 a, 2 a)$ and $\left(a+a^{2}, a+1\right)$. If $a>1$, these solutions are distinct and satisfy the given equation.
b) In general, if $d$ is any divisor of $a^{2}$, there always exists a solution in positive integers for the system

$$
\left\{\begin{array}{l}
x-a=d \\
y-a=\frac{a^{2}}{d}
\end{array}\right.
$$

which is given by $(x, y)=\left(a+d, a+\frac{a^{2}}{d}\right)$. There is a correspondence between the pairs $(x, y)$ that are solutions to the original equation and the positive divisors of $a^{2}$, since for each solution the integer $x-a$ corresponds to some positive divisor $d$ of $a^{2}$. Since $2019^{2}=3^{2} \cdot 673^{2}$, the number of positive divisors is $(2+1) \cdot(2+1)=9$. Therefore, the number of solutions is 9.