Some Proofs About Fibonacci Numbers: Relation 5

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We now prove the Cassini’s Identity, which states that ∀n ≥ 2:

Fn+1Fn− 1 − Fn2 = (− 1)n

(7)

We apply Mathematical Induction over n. For n = 2:

2 2 2 2 Fn+1Fn −1 − Fn = F3F1 − F2 = 2⋅1 − 1 = 1 = (− 1) .

So, the identity is true for n = 2. Say, it is true for some n = k,k ≥ 2. Then,

Fk+1Fk −1 − Fk2 = (− 1)k

Now we need to prove it for n = k + 1:

2 2 Fn+1Fn− 1 − Fn = Fk+2Fk − Fk+1 {expanding Fk+2, and one Fk+1 in the square term } = (Fk + Fk+1)Fk − Fk+1(Fk− 1 + Fk) = Fk2 + Fk+1Fk − Fk+1Fk −1 − Fk+1Fk 2 = Fk − Fk+1Fk −1 = (− 1)(Fk+1Fk −1 − Fk2) {induction hypothesis for n = k} k = (− 1)(− 1) = (− 1)k+1

Thus, if the identity is true for n = k, it is also true for n = k + 1. Hence, the identity must hold for all n ≥ 2.

Example: for n = 6, F₇F₅ − F₆² = 13 ⋅ 5 − 8² = 1 = (−1)⁶