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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Is there any generalized formula to generate pythagoream triples – Aman Sharma · 2 years ago

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– Steven Zheng · 2 years ago

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

– Aman Sharma · 2 years ago

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

– Steven Zheng · 2 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?Log in to reply

– Aman Sharma · 2 years ago

Find out the pythagorean triple \((x,y,z)\) such that \(xyz=16320\)Log in to reply

– Steven Zheng · 2 years ago

Number of solutions \(a,b\) such that \(2ab({a}^{2}+{b}^{2})({a}^{2}-{b}^{2}) = 16320\).Log in to reply

– Aman Sharma · 2 years ago

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

– Steven Zheng · 2 years ago

Euclid's formula deal with generating Pythagorean triples that are integers.Log in to reply

Good – Ayanlaja Adebola · 2 years ago

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