 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-3/9h/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

p_{- 2} = 0, p_{- 1} = 1, q_{- 2} = 1, q_{- 1} = 0, p_{n} = a_{n} p_{n - 1} + p_{n - 2}, q_{n} = a_{n} q_{n - 1} + q_{n - 2} .

Multiplying the matrices

(a_{j} 1 10)

gives

(p_{n - 1} q_{n - 1} p_{n - 2} q_{n - 2}), [a_{0}, \dots, a_{n - 1}, β] = \frac{β p _{n - 1} + p _{n - 2}}{β q _{n - 1} + q _{n - 2}} .

Their determinants also give

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

. If the infinite tail is

β = [a_{n + 1}, a_{n + 2}, \dots] > a_{n + 1}

, the same formula implies

θ - \frac{p _{n}}{q _{n}} = \frac{1}{q _{n} ( β q _{n} + q _{n - 1} )} < \frac{1}{q _{n} q _{n + 1}} .

q_{n + 2} \geq q_{n + 1} + q_{n}

. Hence they converge to

θ

M^{'} = D a - M

D^{'} = (53 - M^{'2}) / D

and

a^{'} = ⌊(7 + M^{'}) / D^{'} ⌋

, gives

53 = [7; \overline{3, 1, 1, 3, 14}] .

7/1, 22/3, 29/4, 51/7

. Direct substitution gives positive solutions to the requested Pell-type equations:

5 1^{2} - 53 \cdot 7^{2} = 4, 2 9^{2} - 53 \cdot 4^{2} = - 7.

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