Problem 1127

5. (1) For each positive integer $n$, there exists a unique positive integer $r$ such that
$$r^{2} \leqslant n < (r+1)^{2} \text{.}$$

Let $n = r^{2} + s$. Then $0 \leqslant s \leqslant 2r$.
Depending on the parity of $s$, we consider two cases.
(i) $s$ is even.

Consider the fraction $\frac{r^{2} + \frac{s}{2}}{r} = r + \frac{s}{2r}$.
Let $a = r^{2} + \frac{s}{2}, b = r$.
Then $0 < b = r \leqslant \sqrt{n}$.
By $n = r^{2} + s \leqslant \left(\frac{a}{b}\right)^{2} = r^{2} + s + \left(\frac{s}{2r}\right)^{2}$
$\leqslant r^{2} + s + 1 = n + 1$
$\Rightarrow \sqrt{n} \leqslant \frac{a}{b} \leqslant \sqrt{n+1}$.
(ii) $s$ is odd.

Consider the fraction
$$\frac{(r+1)^{2} - r + \frac{s-1}{2}}{r+1} = r + 1 - \frac{2r + 1 - s}{2(r+1)} \text{.}$$

Let $a = (r+1)^{2} - r + \frac{s-1}{2}, b = r+1$.
Then $0 < b = r+1 \leqslant \sqrt{n} + 1$.
By $n = r^{2} + s = (r+1)^{2} - (2r + 1 - s)$
$$\begin{array}{l} \leqslant (r+1)^{2} - (2r + 1 - s) + \left(\frac{2r + 1 - s}{2(r+1)}\right)^{2} \\ = \left(\frac{a}{b}\right)^{2} \leqslant (r+1)^{2} - (2r + 1 - s) + 1 \\ = n + 1 \\ \Rightarrow \sqrt{n} \leqslant \frac{a}{b} \leqslant \sqrt{n+1} \text{.} \end{array}$$
(2) It suffices to prove: For each positive integer $r$, there does not exist a fraction $\frac{a}{b}$ (where $a, b$ are integers) satisfying $b \leqslant \sqrt{r^{2} + 1}$, such that
$$\sqrt{r^{2} + 1} \leqslant \frac{a}{b} \leqslant \sqrt{r^{2} + 2} \text{.}$$

Assume there exists a fraction $\frac{a}{b}$ satisfying the above conditions.
$$\begin{array}{l} \text{By } b \leqslant \sqrt{r^{2} + 1} < r + 1 \Rightarrow b \leqslant r \\ \Rightarrow (br)^{2} < b^{2}(r^{2} + 1) \leqslant a^{2} \leqslant b^{2}(r^{2} + 2) \\ \quad \leqslant b^{2}r^{2} + 2br < (br + 1)^{2} \text{.} \end{array}$$

The above inequality shows that $a^{2}$ lies between two consecutive perfect squares, which is a contradiction.

Therefore, there exist infinitely many positive integers $n = r^{2} + 1$ such that no fraction satisfying the conditions exists.