Problem 1368

1. We start with the given condition that the fractions $\frac{a}{b}$ and $\frac{a-1}{b-1}$ differ by 1. This can be expressed as:
$$\left| \frac{a}{b} - \frac{a-1}{b-1} \right| = 1$$

2. We consider the two possible cases for the absolute value equation:
$$\frac{a}{b} - \frac{a-1}{b-1} = 1 \quad \text{or} \quad \frac{a-1}{b-1} - \frac{a}{b} = 1$$

3. Case 1:
$$\frac{a}{b} - \frac{a-1}{b-1} = 1$$
Simplify the left-hand side:
$$\frac{a}{b} - \frac{a-1}{b-1} = \frac{a(b-1) - b(a-1)}{b(b-1)} = \frac{ab - a - ba + b}{b(b-1)} = \frac{b}{b(b-1)} = \frac{1}{b-1}$$
Thus, we have:
$$\frac{1}{b-1} = 1 \implies b-1 = 1 \implies b = 2$$
Substitute $b = 2$ back into the equation:
$$\frac{a}{2} - \frac{a-1}{1} = 1 \implies \frac{a}{2} - (a-1) = 1 \implies \frac{a - 2(a-1)}{2} = 1 \implies \frac{a - 2a + 2}{2} = 1 \implies \frac{2 - a}{2} = 1 \implies 2 - a = 2 \implies a = 0$$
This is a contradiction since $a$ and $b$ are positive integers greater than 1. Therefore, this case is not possible.

4. Case 2:
$$\frac{a-1}{b-1} - \frac{a}{b} = 1$$
Simplify the left-hand side:
$$\frac{a-1}{b-1} - \frac{a}{b} = \frac{(a-1)b - a(b-1)}{b(b-1)} = \frac{ab - b - ab + a}{b(b-1)} = \frac{a - b}{b(b-1)}$$
Thus, we have:
$$\frac{a - b}{b(b-1)} = 1 \implies a - b = b(b-1) \implies a = b^2$$
Substitute $a = b^2$ back into the fractions:
$$\frac{a}{b} = \frac{b^2}{b} = b \quad \text{and} \quad \frac{a-1}{b-1} = \frac{b^2 - 1}{b-1} = \frac{(b-1)(b+1)}{b-1} = b+1$$
Both $\frac{a}{b} = b$ and $\frac{a-1}{b-1} = b+1$ are integers.

Therefore, the condition that $\frac{a}{b}$ and $\frac{a-1}{b-1}$ differ by 1 implies that both fractions are integers.

$\blacksquare$