Question 63
a) Verify that for any positive integer $a$ with $a\gt 1$, 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 of 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.
$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{a}
$$
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 of 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.