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When Fibonacci meets Pythagoras

There is a cool identity relating Fibonacci numbers and Pythagorean triples, which states that given four consecutive Fibonacci numbers $${F}_{n},{F}_{n+1},{F}_{n+2},{F}_{n+3}$$, the triple $$({F}_{n}{F}_{n+3}, 2{F}_{n+1}{F}_{n+2}, {{F}_{n+1}}^{2}+{{F}_{n+2}}^{2})$$ is a Pythagorean triple.

To prove this, let $${F}_{n+1} = a$$ and $${F}_{n+2} = b$$. Then $${F}_{n} = b-a$$ and $${F}_{n+3} = b+a$$. Substituting the abstract values to the equation ${({F}_{n} {F}_{n+3})}^{2} +{(2{F}_{n+1} {F}_{n+2})}^{2} = {({{F}_{n+1}}^{2}+{ {F}_{n+2}}^{2})}^{2}$ will give ${((b-a)(b+a))}^{2} +{(2ab)}^{2}={({a}^{2}+{b}^{2})}^{2}.$

With a little simplification, we arrive at Euclid's formula for Pythagorean triples: ${({b}^{2}-{a}^{2})}^{2} +{(2ab)}^{2}={({a}^{2}+{b}^{2})}^{2}.$

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
2 years, 10 months ago

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Nice post

Is there any generalized formula to generate pythagoream triples

- 2 years, 10 months ago

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Euclid's Formula. The link above should bring you to a proof.

- 2 years, 10 months ago

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How we can solve problems in which product of three pythagorean triples is given and we have to find the triple

- 2 years, 10 months ago

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I'm not sure if that is solvable without more given conditions. You might also need the sum of sides or maybe the radius of an incircle also. Can you come up with an example of your question?

- 2 years, 10 months ago

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Find out the pythagorean triple $$(x,y,z)$$ such that $$xyz=16320$$

- 2 years, 10 months ago

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Number of solutions $$a,b$$ such that $$2ab({a}^{2}+{b}^{2})({a}^{2}-{b}^{2}) = 16320$$.

- 2 years, 10 months ago

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Does there exist more the one triples.....also i read on a website that euclid's formula can't generate all pythagorean triples

- 2 years, 10 months ago

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Euclid's formula deal with generating Pythagorean triples that are integers.

- 2 years, 9 months ago

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Good

- 2 years, 10 months ago

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