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

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## Comments

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TopNewestNice post

Is there any generalized formula to generate pythagoream triples

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

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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

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$(x,y,z)$ such that $xyz=16320$

Find out the pythagorean tripleLog in to reply

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

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Good

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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 There is another as yet unpublished formula which will bring an additional as yet unconsidered aspect of Pythagoras.

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