 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-3/10g/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-05

67

begin

8, 41/5, 90/11, 131/16, 221/27

. The last one yields

x = 221, y = 27, 22 1^{2} - 67 \cdot 2 7^{2} = 48841 - 48843 = - 2.

Changing either sign gives another valid integer solution.

 New to topics? Read the docs here!