 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-4/1i/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

29

begins with

a_{0} = 5

. Its complete quotients repeat after

29 = 5 + \frac{1}{2 + \frac{1}{1 + \frac{1}{1 + \frac{1}{2 + \frac{1}{10 + \dots}}}}} .

Thus

29 = [5; \overline{2, 1, 1, 2, 10}] .

The convergents through one period are

5, \frac{11}{2}, \frac{16}{3}, \frac{27}{5}, \frac{70}{13} .

The last one gives the negative Pell equation solution

7 0^{2} - 29 \cdot 1 3^{2} = 4900 - 4901 = - 1.

Therefore

x = 70, y = 13 .

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