Relation between Fibonacci Numbers and Pythagorean Triples

The problem is as follows:
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We all know that this is the Fibonacci recusion.

F_n = F_n-1 + F_n-2 , F₀ = 1 , F₁​ = 1

How to prove that the following triple (x,y,z) follow Pythagorean Theorem ?

x = (F_i) x (F_i+3)

y = 2 x (F_i+1) x (F_i+2)

z = (F_i+1)² + (F_i+2)²

Putting everything in terms of just two of the numbers and expanding both sides (of works.



and



Substitute and expand both sides, they turn out to be identical.

Thanks but I've found another simple proof for this...

F_i+3 = F_i+2 + F_i+1

and

x = F_i x F_i+3 = (F_i+2 - F_i+1) x (F_i+2 + F_i+1) = b² - a² where

b = F_i+2 and a = F_i+1.

Now (b² - a²)² + (2ab)² = (b² + a²)²

Hence proved...

Anyways thanks for your proof too... (Clapping)