 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-4/6e/ii/solution

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

For fixed

m, l

D_{n} = F_{n + l} F_{n + m} - F_{n} F_{n + m + l} .

n = 0

D_{0} = F_{l} F_{m}

. Applying the Fibonacci recurrence to every term and cancelling gives

D_{n + 1} = - D_{n}

. Induction therefore proves

F_{n + l} F_{n + m} - F_{n} F_{n + m + l} = (- 1)^{n} F_{m} F_{l} .

Applying this identity to the relevant pairs of indices and using

F_{r + 1} = F_{r} + F_{r - 1}

gives

F_{j + k}^{2} - F_{j - k}^{2} = F_{2 k} F_{2 j}

and

F_{j + k + 1}^{2} + F_{j - k}^{2} = F_{2 k + 1} F_{2 j + 1} .

Thus, for

j \geq k \geq 0

(F_{j + k} - F_{j - k}) (F_{j + k} + F_{j - k}) = F_{2 k} F_{2 j}

and

F_{j + k + 1}^{2} + F_{j - k}^{2} = F_{2 k + 1} F_{2 j + 1} .

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