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