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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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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Good

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