 Solution

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

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

Let

D_{n} = p_{n - 1} q_{n - 2} - p_{n - 2} q_{n - 1} = (- 1)^{n - 2}

. Solving the given Möbius transformation for the complete quotient and rationalizing gives

θ_{n} = \frac{p _{n - 2} - d q _{n - 2}}{d q _{n - 1} - p _{n - 1}} = \frac{D _{n} d + d q _{n - 1} q _{n - 2} - p _{n - 1} p _{n - 2}}{p _{n - 1}^{2} - d q _{n - 1}^{2}} .

Therefore

s_{n} = \frac{p _{n - 1}^{2} - d q _{n - 1}^{2}}{D _{n}}, r_{n} = \frac{d q _{n - 1} q _{n - 2} - p _{n - 1} p _{n - 2}}{D _{n}} .

D_{n}

\pm 1

, so both are integers. The algebraic norm denominator is nonzero because

d

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