Origins of Euclid's formula for Pythagorean triples

Hi, because of a question on this week's on Brilliant problem set, I've come across the Euclid's formula for generating Pythagorean triples for the first time.

The formula says that (in my own words) for a right angled triangle with sides a,b and hypotenuse c,

a2+b2=c2    (m2n2)2+(2mn)2=(m2+n2)2a^2+b^2 = c^2 \iff (m^2-n^2)^2 + (2mn)^2 = (m^2 + n^2)^2

for any 2 integers m>nm > n. After staring at this formula for a while, I realised it's actually a simple and straightforward case of expansion on LHS and RHS, followed by a check on whether the terms matched on both sides.

This made me wonder if Euclid came up with this theorem because of some eureka moment (you know, like maybe one day he's just sitting alone at his desk and this idea that (m2n2)+(2mn)2=(m2+n2)2 (m^2-n^2) + (2mn)^2 = (m^2 + n^2)^2 can be applied to the Pythagoras theorem just pops into his head) or was he driven by any other sort of motivation to derive this formula?

And this brings me to another question, suppose Euclid didn't leave any proof of the derivation of this formula because he thought that it was such a straight forward application of the idea of (m2n2)2+(2mn)2=(m2+n2)2 (m^2-n^2)^2 + (2mn)^2 = (m^2 + n^2)^2 to a2+b2=c2a^2 + b^2 = c^2. Would there had been any practical need to come up with a derivation from a2+b2=c2a^2 + b^2 = c^2?

I apologize if I sound like an idiot asking such questions, but well...I just couldn't get these questions off my mind :p

Note by Ghim Siang
7 years ago

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6 votes

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I've added a note/comment that the expansion doesn't prove that that every pythagorean triple must be of the form (m2n2,2mn,m2+n2) (m^2 - n^2, 2mn, m^2 + n^2) In fact, this statement is not true! What are the values of mm and nn that generate the triple (9,12,15) (9, 12, 15) ? We require m2+n2=15 m^2 + n^2 = 15 , which you can verify has no solutions.

All that the expansion shows, is that (m2n2,2mn,m2+n2) (m^2-n^2, 2mn, m^2 + n^2) is a Pythagorean triple. In fact, the classification is valid for primitive pythagorean triple, which are triples that have no common divisor. In the above example, the terms have a common divisor of 3.

Calvin Lin Staff - 7 years ago

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This is correct. A more general formula that applies to all pythagorean triples (as far as I know) is a=k(m2n2),b=k2mn,c=k(m2+n2) a = k \cdot (m^2 - n^2), b = k \cdot 2mn, c = k \cdot (m^2 + n^2) , where k is a positive integer.

Michael Tong - 7 years ago

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Ghim, are you asking for the derivation or the 'need' of the derivation? Sorry, but I couldn't understand your point.

Piyal De - 7 years ago

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well, i was asking about the 'need' for the derivation from a^2 + b^2 = c^2 to the euclid's formula, and i guess alyosha's explaination pretty much makes sense!

Ghim Siang - 7 years ago

Log in to reply Hope this could help

Rekarlo Jäger - 2 years ago

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