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ParthKohli
 4 years ago
\[\Huge \mathsf{\text{Primitive Pythagorean Triplets.}} \]
ParthKohli
 4 years ago
\[\Huge \mathsf{\text{Primitive Pythagorean Triplets.}} \]

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ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Give it a start, Kingy.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.4Well, they're triplets of the form \((a,b,c)\) such that \(a^2+b^2=c^2\) and \(\gcd(a,b,c)=1\(. Unknown to many high school students, there are in fact several formulas for generating an infinite number of these triplets.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0\[2n,n^2  1, n^2 + 1 \]?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0I know one. \(3,4,5\).

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.4I just can't type that correctly can I? Well, they're triplets of the form \((a,b,c)\) such that \(a^2+b^2=c^2\) and \(\gcd(a,b,c)=1\). Unknown to many high school students, there are in fact several formulas for generating an infinite number of these triplets.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Another one! \(5,12,13\)!

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.4Correct, (3,4,5) is a triplet as well as (5,12,13) and (7,24,25) among infinite others.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Can you tell me the formula(s) for it, please?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.4The most well known one is \[2mn, \quad m^2n^2, \quad m^2+n^2\]Given any \(m,n\) such that \(m>n\).

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0I see :) Are there any more?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.4What you gave above, only works for \(m\) even. So it generates an infinite number, but not all of them.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Can this formula generate all such triplets, or you have to use some other formulae to generate more?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0another one \[(x,y,z)=(2k+1,2k^2+2k,2k^2+2k+1)\]

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Oh wait. It does generate all.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.4There is also \[(a,b,c)=\left(mn,\;\;\frac{m^2n^2}{2},\;\;\frac{m^2+n^2}{2}\right)\]This is a different formula that what I gave above since it only works with \((m,n)\) such that \(n>m\ge1\) and \(\gcd(m,n)=1\). This does not generate all of them.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.4Ccorrection, \(m>n\ge1\).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you want them "primitive" makes sure \(m,n\) are opposite parity, and no common factors

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Well. I'd just use the first one :)

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.4Satellite is correct. The first formula I wrote only gives primitive triplets for \(m,n\) such that \(mn\) is odd. And \(\gcd(m,n)=1\). However, any two positive integers will give you a triplet.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0algebra shows you that \((m^2n^2)^2+(2mn)^2=(m^2+n^2)^2\) if you want a proof that every such triple is generated by such an \(m\) and \(n\) try working through the attachment

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0If \(\gcd(m,n) = 1\), then it automatically means that \(m  n\) is odd.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0really, if you have time, work through the worksheet i sent you you will see why all triples come from this formula

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, I am looking at it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it requires no more than algebra

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0All set. I got that formula well.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.4There is also a multitude of fascinating properties of primitive triplets. For example: \(c\) is always odd. 2,3,4 divide exactly one of \(a\) or \(b\), 5 divides exactly one of \(a,b\) or \(c\). \(a+b+c\) is always even among others. These are some of the simpler facts.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you, Your Majesty and Your Bikesty!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0have fun. there are several steps to the proof, but it is all there

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what do you mean by primitive triplets?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"primitive" no common factors

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 3, 4, 5 but not 6, 8, 10

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Whoa! Primitive triplets are \(a,b,c\) such that \(\gcd(a,b,c) = 1\). Other words, they all are coprime.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@ParthKohli method of obtaining the formula i gave u...we will talk about later if u want... http://openstudy.com/study#/updates/500a7eb3e4b0549a892eeeee

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0@mukushla I'd see that tutorial tomorrow; tired somewhat at the moment. Looks impressive by looking at it though.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Haha\[ \Huge \ddot \smile \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why do we need to have opposite parity? and what's with mn being odd?

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Well. Isn't that the same thing?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.4\(mn\) is odd if and only if they have opposite parity. So either \(m\) or \(n\) is odd, but not both.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i understand that. but is it necessary for mn to be odd?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.4If you want a primitive triplet, it is.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.4If they are both even or both odd, \(\gcd(a,b,c)\ge2\)

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.4Note that \(2mn\) is always even. Also notice that if \(m,n\) are both odd, then \(m^2\) and \(n^2\) are both odd. If they are both even, \(m^2\) and \(n^2\) are both even. Now, \(m^2n^2\) and \(m^2+n^2\) will both be even as well since eveneven=even and oddodd=odd. Hence, \(a,b,c\) are all even, and their gcd must be divisible by 2.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that was so easy and so stupid of me to not get it. i am sorry for troubling you on this. if i am to prove 5 exactly divides one of the triplet, is assuming all of them to be non divisible by 5 the right way?

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.4When I first did it, I assumed \(a\) was not divisible by 5, and deduced that either \(b\) or \(c\) was divisible by 5. Then assume \(5\) divides \(a\) and show that it doesn't divide \(b\) or \(c\). This probably isn't the fastest way to do it thought.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you can work by cases assume \(m\) and \(n\) are not divisible by 5 then prove that either \(m^2n^2\) or \(m^2+n^2\) must be for example, if \(m\equiv n\) mod 5, then \(m^2n^2\equiv 0\) mod 5
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