Problem 1540

1. Given the equation $ab - 1 = n^2$, we can rewrite it as:
$$ab = n^2 + 1$$
We need to prove that $a - b \geq \sqrt{4n - 3}$.

2. Start by squaring both sides of the inequality $a - b \geq \sqrt{4n - 3}$:
$$(a - b)^2 \geq 4n - 3$$

3. Expand the left-hand side:
$$a^2 - 2ab + b^2 \geq 4n - 3$$

4. Substitute $ab = n^2 + 1$ into the inequality:
$$a^2 - 2(n^2 + 1) + b^2 \geq 4n - 3$$
Simplify:
$$a^2 + b^2 - 2n^2 - 2 \geq 4n - 3$$

5. Rearrange the terms:
$$a^2 + b^2 \geq 2n^2 + 4n - 1$$

6. Consider the sum $a + b$. By the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality):
$$a + b \geq 2\sqrt{ab} = 2\sqrt{n^2 + 1}$$
Since $\sqrt{n^2 + 1} > n$, we have:
$$2\sqrt{n^2 + 1} > 2n$$
Therefore:
$$a + b > 2n$$

7. To satisfy the equality $a - b = \sqrt{4n - 3}$, we need:
$$a + b = 2n + 1$$

8. For equality to hold, $4n - 3$ must be a perfect square. Let $4n - 3 = k^2$ for some integer $k$:
$$4n = k^2 + 3$$
$$n = \frac{k^2 + 3}{4}$$
For $n$ to be an integer, $k^2 + 3$ must be divisible by 4. This implies $k$ must be odd.

9. Let $k = 2m + 1$ for some integer $m$:
$$n = \frac{(2m + 1)^2 + 3}{4} = \frac{4m^2 + 4m + 1 + 3}{4} = m^2 + m + 1$$

10. Substitute $n = m^2 + m + 1$ back into the expressions for $a$ and $b$:
$$a + b = 2(m^2 + m + 1) + 1 = 2m^2 + 2m + 3$$
$$ab = (m^2 + m + 1)^2 + 1$$

11. Let $a = (m + 1)^2 + 1$ and $b = m^2 + 1$:
$$a = m^2 + 2m + 2$$
$$b = m^2 + 1$$
$$a - b = (m^2 + 2m + 2) - (m^2 + 1) = 2m + 1$$
$$\sqrt{4n - 3} = \sqrt{4(m^2 + m + 1) - 3} = \sqrt{4m^2 + 4m + 4 - 3} = \sqrt{4m^2 + 4m + 1} = 2m + 1$$

Thus, the equality holds when $n = m^2 + m + 1$, $a = (m + 1)^2 + 1$, and $b = m^2 + 1$.

The final answer is $\boxed{ a - b \geq \sqrt{4n - 3} }$ and equality holds when $n = m^2 + m + 1$, $a = (m + 1)^2 + 1$, and $b = m^2 + 1$.