$$\Huge \mathsf{\text{Primitive Pythagorean Triplets.}} $$

Question

parthkohli · Answer

Give it a start, Kingy.

kinggeorge · Answer

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 · Answer

$$2n,n^2 - 1, n^2 + 1 $$?

parthkohli · Answer

I know one. $3,4,5$.

kinggeorge · Answer

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 · Answer

Another one! $5,12,13$!

kinggeorge · Answer

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

parthkohli · Answer

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

kinggeorge · Answer

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 · Answer

I see :)
Are there any more?

kinggeorge · Answer

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

parthkohli · Answer

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

anonymous · Answer

another one 

$$(x,y,z)=(2k+1,2k^2+2k,2k^2+2k+1)$$

parthkohli · Answer

Oh wait. It does generate all.

kinggeorge · Answer

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 · Answer

Ccorrection, $m>n\ge1$.

anonymous · Answer

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

parthkohli · Answer

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

kinggeorge · Answer

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 · Answer

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

parthkohli · Answer

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

anonymous · Answer

no, try 5 and 3

parthkohli · Answer

Oh grr.

anonymous · Answer

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

parthkohli · Answer

Yes, I am looking at it.

anonymous · Answer

it requires no more than algebra

parthkohli · Answer

All set. I got that formula well.

kinggeorge · Answer

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 · Answer

Thank you, Your Majesty and Your Bikesty!

anonymous · Answer

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

anonymous · Answer

what do you mean by primitive triplets?

kinggeorge · Answer

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

anonymous · Answer

"primitive" no common factors

anonymous · Answer

oh

anonymous · Answer

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

parthkohli · Answer

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

anonymous · Answer

@ParthKohli method of obtaining the formula i gave u...we will talk about later if u want... http://openstudy.com/study#/updates/500a7eb3e4b0549a892eeeee

parthkohli · Answer

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

anonymous · Answer

np...:)

parthkohli · Answer

Haha$$ \Huge \ddot \smile $$

anonymous · Answer

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

parthkohli · Answer

Well. Isn't that the same thing?

kinggeorge · Answer

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

anonymous · Answer

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

kinggeorge · Answer

If you want a primitive triplet, it is.

kinggeorge · Answer

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

anonymous · Answer

how do i prove it?

kinggeorge · Answer

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 · Answer

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 · Answer

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 · Answer

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