Problem 3. Determine the number of elements of the set

$$
M=\left\{(x, y) \in \mathbb{N}^{*} \times \mathbb{N}^{*} \left\lvert\, \frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}=\frac{1}{\sqrt{2016}}\right.\right\}
$$
Solution Let $(x, y) \in M$. Since $\sqrt{2016}=12 \sqrt{14}$, we have

$$
\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}=\frac{1}{12 \sqrt{14}} \Leftrightarrow \frac{1}{\sqrt{14 x}}-\frac{1}{\sqrt{14 y}}=\frac{1}{2016} \Leftrightarrow \frac{1}{\sqrt{14 x}}=\frac{1}{\sqrt{14 y}}+\frac{1}{2016}
$$

By squaring, we obtain $\frac{1}{14 x}=\frac{1}{14 y}+\frac{1}{2016}+\frac{1}{1008 \sqrt{14 y}}$, from which $\sqrt{14 y} \in \mathbb{Q}$ and, consequently, $\sqrt{14 x} \in \mathbb{Q}$.

Then there exist non-zero natural numbers $a, b$ such that $x=14 a$ and $y=14 b$.

Substituting into the relation $\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}=\frac{1}{12 \sqrt{14}}$, we get $\frac{1}{a}-\frac{1}{b}=\frac{1}{12}$, from which $a=\frac{12 b}{b+12}$.

From $\frac{12 b}{b+12} \in \mathbb{N}$ it follows that $b+12 \mid 144$, so $b \in\{4,6,12,24,36,60,132\}$, from which $a \in\{3,4,6,8,9,10,11\}$.

In conclusion, the set $M$ admits 7 elements.