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ParthKohli
Group Title
\[\Huge \mathsf{\text{Primitive Pythagorean Triplets.}} \]
 2 years ago
 2 years ago
ParthKohli Group Title
\[\Huge \mathsf{\text{Primitive Pythagorean Triplets.}} \]
 2 years ago
 2 years ago

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ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Give it a start, Kingy.
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.4
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.
 2 years ago

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

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.4
I 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.
 2 years ago

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

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

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

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

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

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

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

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

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Oh wait. It does generate all.
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.4
There 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.
 2 years ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
if you want them "primitive" makes sure \(m,n\) are opposite parity, and no common factors
 2 years ago

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.4
Satellite 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.
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
algebra 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
 2 years ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
no, try 5 and 3
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
really, if you have time, work through the worksheet i sent you you will see why all triples come from this formula
 2 years ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
it requires no more than algebra
 2 years ago

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

KingGeorge Group TitleBest ResponseYou've already chosen the best response.4
There 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.
 2 years ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
have fun. there are several steps to the proof, but it is all there
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
what do you mean by primitive triplets?
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.4
\(\gcd(a,b,c)=1\).
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
"primitive" no common factors
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
so 3, 4, 5 but not 6, 8, 10
 2 years ago

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

mukushla Group TitleBest 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
 2 years ago

ParthKohli Group TitleBest 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.
 2 years ago

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

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
why do we need to have opposite parity? and what's with mn being odd?
 2 years ago

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

KingGeorge Group TitleBest 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.
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
i understand that. but is it necessary for mn to be odd?
 2 years ago

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

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

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
how do i prove it?
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.4
Note 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.
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.0
that 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?
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.4
When 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.
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you 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
 2 years ago
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