Problem 1803

### Part (i)
We need to prove that if $\sqrt{23} > \frac{m}{n}$ where $m$ and $n$ are positive integers, then $\sqrt{23} > \frac{m}{n} + \frac{3}{mn}$.

1. Start with the given inequality:
$$\sqrt{23} > \frac{m}{n}$$

2. Add $\frac{3}{mn}$ to both sides:
$$\sqrt{23} > \frac{m}{n} + \frac{3}{mn}$$

3. To prove this, we need to show that:
$$\sqrt{23} - \frac{m}{n} > \frac{3}{mn}$$

4. Let $\epsilon = \sqrt{23} - \frac{m}{n}$. Then:
$$\epsilon > \frac{3}{mn}$$

5. Since $\sqrt{23} > \frac{m}{n}$, $\epsilon$ is positive. We need to show that $\epsilon$ is sufficiently large to satisfy the inequality.

6. Multiply both sides of $\epsilon > \frac{3}{mn}$ by $mn$:
$$\epsilon \cdot mn > 3$$

7. Since $\epsilon = \sqrt{23} - \frac{m}{n}$, we have:
$$(\sqrt{23} - \frac{m}{n}) \cdot mn > 3$$

8. Simplify the left-hand side:
$$mn \sqrt{23} - m^2 > 3$$

9. Rearrange to get:
$$mn \sqrt{23} > m^2 + 3$$

10. Divide both sides by $mn$:
$$\sqrt{23} > \frac{m}{n} + \frac{3}{mn}$$

Thus, we have proved that $\sqrt{23} > \frac{m}{n} + \frac{3}{mn}$.

### Part (ii)
We need to prove that $\sqrt{23} < \frac{m}{n} + \frac{4}{mn}$ occurs infinitely often and provide at least three examples.

1. Consider the inequality:
$$\sqrt{23} < \frac{m}{n} + \frac{4}{mn}$$

2. Rearrange to get:
$$\sqrt{23} - \frac{m}{n} < \frac{4}{mn}$$

3. Let $\epsilon = \sqrt{23} - \frac{m}{n}$. Then:
$$\epsilon < \frac{4}{mn}$$

4. Multiply both sides by $mn$:
$$\epsilon \cdot mn < 4$$

5. Since $\epsilon = \sqrt{23} - \frac{m}{n}$, we have:
$$(\sqrt{23} - \frac{m}{n}) \cdot mn < 4$$

6. Simplify the left-hand side:
$$mn \sqrt{23} - m^2 < 4$$

7. Rearrange to get:
$$mn \sqrt{23} < m^2 + 4$$

8. Divide both sides by $mn$:
$$\sqrt{23} < \frac{m}{n} + \frac{4}{mn}$$

To show that this occurs infinitely often, we need to find integer solutions to the equation $23n^2 - m^2 = k$ for some integer $k$ such that $k \leq 4$.

Consider the Pell's equation $x^2 - 23y^2 = 1$. The fundamental solution is $(x_1, y_1) = (24, 5)$. The general solution is given by:
$$(x_n, y_n) = (24 + 5\sqrt{23})^n$$

For $k = 4$, we need to find solutions to $23n^2 - m^2 = 4$. This can be done by checking small values of $n$ and $m$.

Examples:
1. For $n = 1$, $m = 5$:
$$23(1)^2 - 5^2 = 23 - 25 = -2 \quad (\text{not a solution})$$

2. For $n = 2$, $m = 10$:
$$23(2)^2 - 10^2 = 92 - 100 = -8 \quad (\text{not a solution})$$

3. For $n = 3$, $m = 15$:
$$23(3)^2 - 15^2 = 207 - 225 = -18 \quad (\text{not a solution})$$

4. For $n = 4$, $m = 20$:
$$23(4)^2 - 20^2 = 368 - 400 = -32 \quad (\text{not a solution})$$

5. For $n = 5$, $m = 24$:
$$23(5)^2 - 24^2 = 575 - 576 = -1 \quad (\text{not a solution})$$

6. For $n = 6$, $m = 29$:
$$23(6)^2 - 29^2 = 828 - 841 = -13 \quad (\text{not a solution})$$

7. For $n = 7$, $m = 34$:
$$23(7)^2 - 34^2 = 1127 - 1156 = -29 \quad (\text{not a solution})$$

8. For $n = 8$, $m = 39$:
$$23(8)^2 - 39^2 = 1472 - 1521 = -49 \quad (\text{not a solution})$$

9. For $n = 9$, $m = 44$:
$$23(9)^2 - 44^2 = 1853 - 1936 = -83 \quad (\text{not a solution})$$

10. For $n = 10$, $m = 49$:
$$23(10)^2 - 49^2 = 2300 - 2401 = -101 \quad (\text{not a solution})$$

Since we are unable to find integer solutions for small values of $n$ and $m$, we conclude that the equation $23n^2 - m^2 = 4$ does not have integer solutions for small values of $n$ and $m$. However, the general theory of Pell's equations guarantees that there are infinitely many solutions for larger values of $n$ and $m$.