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My son notes that for Fibonacci numbers $F_n$, $$ (F_{n+1})^2-F_n \cdot F_{n+2} =(-1)^n $$

  1. I assume that this is true. Update: I see that the proof is already here: Prove the given property of the Fibonacci numbers , so never mind about that part.

  2. I'm really looking for a good geometrical intuition, since those two terms seem like likely areas in those spiral diagrams. Is there a direct way to think about this identity geometrically?

Community
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djsadinoff
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2 Answers2

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$$ (F_{n+1})^2-F_n \cdot F_{n+2} =(-1)^n $$

So, I am assuming by the spiral you are referring to the golden ratio. deMoivre found a closed-form expression of the Fibonacci numbers.

Binet's Formula: $F_n=\dfrac{\phi^n-\psi^n}{\sqrt{5}} $ where $\phi$ is the golden ratio given by $\dfrac{1+\sqrt{5}}{2}$ and $\psi=1-\phi=-\dfrac{1}{\phi}$ .

$$\left(\dfrac{\phi^{n+1}-(-\phi)^{-n-1}}{\sqrt{5}}\right)^2-\dfrac{\phi^n-(-\phi)^{-n}}{\sqrt{5}}\cdot\dfrac{\phi^{n+2}-(-\phi)^{-n-2}}{\sqrt{5}}$$

Simplifying this expression indeed does give the desired result:

$$(-1)^{n}$$

Note that the golden ratio also satisfies the equation below:

$$\phi^2=\phi+1$$

Matt
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Consider the following diagram, the Fibonacci spiral arrangement of squares up to $5^2$:

$\hspace{5cm}$enter image description here

I claim that $5^2 = 3\cdot8+1$. This is equivalent to $A+B = A+C+1$, which simplifies to $C = B-1$. In other words, it remains to show that $3^2 = 2\cdot5-1$, and the rest is induction.

Théophile
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