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} , F_{0 }= 1 , F_{1} = 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}

Re: Relation between Fibonacci Numbers and Pythagorean Triples

Putting everything in terms of just two of the numbers and expanding both sides (of $\displaystyle x^{2}+y^{2}=z^{2})$ works.

$\displaystyle F_{i+3}=F_{i+2}+F_{i+1}= F_{i+2}+(F_{i+2}-F_{i})=2F_{i+2}-F_{i}$

and

$\displaystyle F_{i+1}=F_{i+2}-F_{i}.$

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

Re: Relation between Fibonacci Numbers and Pythagorean Triples

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}) = b^{2} - a^{2} where

b = F_{i+2} and a = F_{i+1}.

Now (b^{2} - a^{2})^{2} + (2ab)^{2} = (b^{2} + a^{2})^{2}

Hence proved...

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