ParthKohli
  • ParthKohli
\[\Huge \mathsf{\text{Primitive Pythagorean Triplets.}} \]
Mathematics
katieb
  • katieb
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ParthKohli
  • ParthKohli
Give it a start, Kingy.
KingGeorge
  • KingGeorge
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
  • ParthKohli
\[2n,n^2 - 1, n^2 + 1 \]?

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ParthKohli
  • ParthKohli
I know one. \(3,4,5\).
KingGeorge
  • KingGeorge
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.
ParthKohli
  • ParthKohli
Another one! \(5,12,13\)!
KingGeorge
  • KingGeorge
Correct, (3,4,5) is a triplet as well as (5,12,13) and (7,24,25) among infinite others.
ParthKohli
  • ParthKohli
Can you tell me the formula(s) for it, please?
KingGeorge
  • KingGeorge
The most well known one is \[2mn, \quad m^2-n^2, \quad m^2+n^2\]Given any \(m,n\) such that \(m>n\).
ParthKohli
  • ParthKohli
I see :) Are there any more?
KingGeorge
  • KingGeorge
What you gave above, only works for \(m\) even. So it generates an infinite number, but not all of them.
ParthKohli
  • ParthKohli
Can this formula generate all such triplets, or you have to use some other formulae to generate more?
anonymous
  • anonymous
another one \[(x,y,z)=(2k+1,2k^2+2k,2k^2+2k+1)\]
ParthKohli
  • ParthKohli
Oh wait. It does generate all.
KingGeorge
  • KingGeorge
There is also \[(a,b,c)=\left(mn,\;\;\frac{m^2-n^2}{2},\;\;\frac{m^2+n^2}{2}\right)\]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
  • KingGeorge
Ccorrection, \(m>n\ge1\).
anonymous
  • anonymous
if you want them "primitive" makes sure \(m,n\) are opposite parity, and no common factors
ParthKohli
  • ParthKohli
Well. I'd just use the first one :)
KingGeorge
  • KingGeorge
Satellite is correct. The first formula I wrote only gives primitive triplets for \(m,n\) such that \(m-n\) is odd. And \(\gcd(m,n)=1\). However, any two positive integers will give you a triplet.
anonymous
  • anonymous
algebra shows you that \((m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2\) if you want a proof that every such triple is generated by such an \(m\) and \(n\) try working through the attachment
1 Attachment
ParthKohli
  • ParthKohli
If \(\gcd(m,n) = 1\), then it automatically means that \(m - n\) is odd.
anonymous
  • anonymous
no, try 5 and 3
ParthKohli
  • ParthKohli
Oh grr.
anonymous
  • anonymous
really, if you have time, work through the worksheet i sent you you will see why all triples come from this formula
ParthKohli
  • ParthKohli
Yes, I am looking at it.
anonymous
  • anonymous
it requires no more than algebra
ParthKohli
  • ParthKohli
All set. I got that formula well.
KingGeorge
  • KingGeorge
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.
ParthKohli
  • ParthKohli
Thank you, Your Majesty and Your Bikesty!
anonymous
  • anonymous
have fun. there are several steps to the proof, but it is all there
anonymous
  • anonymous
what do you mean by primitive triplets?
KingGeorge
  • KingGeorge
\(\gcd(a,b,c)=1\).
anonymous
  • anonymous
"primitive" no common factors
anonymous
  • anonymous
oh
anonymous
  • anonymous
so 3, 4, 5 but not 6, 8, 10
ParthKohli
  • ParthKohli
Whoa! Primitive triplets are \(a,b,c\) such that \(\gcd(a,b,c) = 1\). Other words, they all are co-prime.
anonymous
  • anonymous
@ParthKohli method of obtaining the formula i gave u...we will talk about later if u want... http://openstudy.com/study#/updates/500a7eb3e4b0549a892eeeee
ParthKohli
  • ParthKohli
@mukushla I'd see that tutorial tomorrow; tired somewhat at the moment. Looks impressive by looking at it though.
anonymous
  • anonymous
np...:)
ParthKohli
  • ParthKohli
Haha\[ \Huge \ddot \smile \]
anonymous
  • anonymous
why do we need to have opposite parity? and what's with m-n being odd?
ParthKohli
  • ParthKohli
Well. Isn't that the same thing?
KingGeorge
  • KingGeorge
\(m-n\) is odd if and only if they have opposite parity. So either \(m\) or \(n\) is odd, but not both.
anonymous
  • anonymous
i understand that. but is it necessary for m-n to be odd?
KingGeorge
  • KingGeorge
If you want a primitive triplet, it is.
KingGeorge
  • KingGeorge
If they are both even or both odd, \(\gcd(a,b,c)\ge2\)
anonymous
  • anonymous
how do i prove it?
KingGeorge
  • KingGeorge
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, \(m^2-n^2\) and \(m^2+n^2\) will both be even as well since even-even=even and odd-odd=odd. Hence, \(a,b,c\) are all even, and their gcd must be divisible by 2.
anonymous
  • anonymous
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?
KingGeorge
  • KingGeorge
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.
anonymous
  • anonymous
you can work by cases assume \(m\) and \(n\) are not divisible by 5 then prove that either \(m^2-n^2\) or \(m^2+n^2\) must be for example, if \(m\equiv n\) mod 5, then \(m^2-n^2\equiv 0\) mod 5

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