# 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
6 years ago

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Nice post

Is there any generalized formula to generate pythagoream triples

- 6 years ago

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

- 6 years ago

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

- 6 years ago

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?

- 6 years ago

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

- 6 years ago

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

- 6 years ago

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

- 6 years ago

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

- 6 years ago

Good

- 6 years ago

Nice post Is there any generalized formula to generate pythagoream triples.(?) There are several. There is another as yet unpublished formula which will bring an additional as yet unconsidered aspect of Pythagoras.

- 3 years, 1 month ago