 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-4/11f/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

The convergents begin

2, 3, \frac{5}{2}, \frac{8}{3}, \frac{37}{14}, \dots .

Since

8^{2} - 7 \cdot 3^{2} = 1,

a strictly positive solution is

(x, y) = (8, 3)

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