\[\Huge \mathsf{\text{Primitive Pythagorean Triplets.}} \]

- ParthKohli

\[\Huge \mathsf{\text{Primitive Pythagorean Triplets.}} \]

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- ParthKohli

Give it a start, Kingy.

- 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

\[2n,n^2 - 1, n^2 + 1 \]?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- ParthKohli

I know one. \(3,4,5\).

- 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

Another one! \(5,12,13\)!

- KingGeorge

Correct, (3,4,5) is a triplet as well as (5,12,13) and (7,24,25) among infinite others.

- ParthKohli

Can you tell me the formula(s) for it, please?

- 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

I see :)
Are there any more?

- KingGeorge

What you gave above, only works for \(m\) even. So it generates an infinite number, but not all of them.

- ParthKohli

Can this formula generate all such triplets, or you have to use some other formulae to generate more?

- anonymous

another one
\[(x,y,z)=(2k+1,2k^2+2k,2k^2+2k+1)\]

- ParthKohli

Oh wait. It does generate all.

- 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

Ccorrection, \(m>n\ge1\).

- anonymous

if you want them "primitive" makes sure \(m,n\) are opposite parity, and no common factors

- ParthKohli

Well. I'd just use the first one :)

- 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

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

If \(\gcd(m,n) = 1\), then it automatically means that \(m - n\) is odd.

- anonymous

no, try 5 and 3

- ParthKohli

Oh grr.

- anonymous

really, if you have time, work through the worksheet i sent you
you will see why all triples come from this formula

- ParthKohli

Yes, I am looking at it.

- anonymous

it requires no more than algebra

- ParthKohli

All set. I got that formula well.

- 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

Thank you, Your Majesty and Your Bikesty!

- anonymous

have fun. there are several steps to the proof, but it is all there

- anonymous

what do you mean by primitive triplets?

- KingGeorge

\(\gcd(a,b,c)=1\).

- anonymous

"primitive" no common factors

- anonymous

oh

- anonymous

so 3, 4, 5 but not 6, 8, 10

- ParthKohli

Whoa!
Primitive triplets are \(a,b,c\) such that \(\gcd(a,b,c) = 1\).
Other words, they all are co-prime.

- 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

@mukushla I'd see that tutorial tomorrow; tired somewhat at the moment.
Looks impressive by looking at it though.

- anonymous

np...:)

- ParthKohli

Haha\[ \Huge \ddot \smile \]

- anonymous

why do we need to have opposite parity? and what's with m-n being odd?

- ParthKohli

Well. Isn't that the same thing?

- KingGeorge

\(m-n\) is odd if and only if they have opposite parity. So either \(m\) or \(n\) is odd, but not both.

- anonymous

i understand that. but is it necessary for m-n to be odd?

- KingGeorge

If you want a primitive triplet, it is.

- KingGeorge

If they are both even or both odd, \(\gcd(a,b,c)\ge2\)

- anonymous

how do i prove it?

- 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

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

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

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.