6E (Numbers and Sets)

For

n = 1

F_{2} F_{0} - F_{1}^{2} = - 1 = (- 1)^{1} .

C_{n} = F_{n + 1} F_{n - 1} - F_{n}^{2}

, then the recurrence gives

C_{n + 1} = F_{n + 2} F_{n} - F_{n + 1}^{2} = (F_{n + 1} + F_{n}) F_{n} - F_{n + 1}^{2} = F_{n}^{2} - F_{n + 1} F_{n - 1} = - C_{n} .

Induction yields the Fibonacci determinant identity known as Cassini's identity:

F_{n + 1} F_{n - 1} - F_{n}^{2} = (- 1)^{n} .

ii

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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