3. For a given natural number $n$, solve the equation in the domain of positive real numbers

$$
\left\lfloor x \sqrt{n^{2}-1}\right\rfloor=n x-1 .
$$

(The symbol $\lfloor r\rfloor$ denotes the greatest integer that does not exceed the number $r$.)
3. A positive number $x$ is a solution to the equation $\mathrm{s}$ given by $n$, precisely when the number $n x$ is a natural number and the inequalities are satisfied

$$
n x-1 \leqq x \sqrt{n^{2}-1} < n x, \text{ since clearly } \sqrt{n^{2}-1}<\sqrt{n^{2}}=n. \text{ It remains to solve the left inequality (with respect to the unknown } x \text{). After a simple rearrangement, we get }

$$
\begin{gathered}
x\left(n-\sqrt{n^{2}-1}\right) \leqq 1 \\
x \leqq \frac{1}{n-\sqrt{n^{2}-1}}=n+\sqrt{n^{2}-1}
\end{gathered}
$$

We used the fact that the expression $n-\sqrt{n^{2}-1}$ is positive and in the product with the conjugate expression $n+\sqrt{n^{2}-1}$ gives the number 1. Multiplying both sides of the derived inequality by the number $n$, we get for the natural number $k=n x$ the equivalent condition

$$
k \leqq n^{2}+n \sqrt{n^{2}-1}
$$

which is satisfied precisely for $k \in\left\{1,2, \ldots, 2 n^{2}-1\right\}$, since for the second addend on the right side of the last inequality, the integer estimates clearly hold

$$
n^{2}-1 \leqq n \sqrt{n^{2}-1}<n^{2}
$$

(again using only the inequality $\sqrt{n^{2}-1}<n$). All solutions to the given equation are of the form $x=k / n$ and thus form the set of fractions

$$
\left\{\frac{1}{n}, \frac{2}{n}, \ldots, \frac{2 n^{2}-1}{n}\right\}
$$

For a complete solution, award 6 points, of which 3 points for deriving the inequality $x \leqq n+\sqrt{n^{2}-1}$. Award 1 point for noting that each solution $x$ must be of the form of the fraction $k / n$.