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

1. ParthKohli Group Title

Give it a start, Kingy.

2. KingGeorge Group Title

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. 3. ParthKohli Group Title $2n,n^2 - 1, n^2 + 1$? 4. ParthKohli Group Title I know one. \(3,4,5$$.

5. KingGeorge Group Title

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.

6. ParthKohli Group Title

Another one! $$5,12,13$$!

7. KingGeorge Group Title

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

8. ParthKohli Group Title

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

9. KingGeorge Group Title

The most well known one is $2mn, \quad m^2-n^2, \quad m^2+n^2$Given any $$m,n$$ such that $$m>n$$.

10. ParthKohli Group Title

I see :) Are there any more?

11. KingGeorge Group Title

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

12. ParthKohli Group Title

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

13. mukushla Group Title

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

14. ParthKohli Group Title

Oh wait. It does generate all.

15. KingGeorge Group Title

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.

16. KingGeorge Group Title

Ccorrection, $$m>n\ge1$$.

17. satellite73 Group Title

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

18. ParthKohli Group Title

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

19. KingGeorge Group Title

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.

20. satellite73 Group Title

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

21. ParthKohli Group Title

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

22. satellite73 Group Title

no, try 5 and 3

23. ParthKohli Group Title

Oh grr.

24. satellite73 Group Title

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

25. ParthKohli Group Title

Yes, I am looking at it.

26. satellite73 Group Title

it requires no more than algebra

27. ParthKohli Group Title

All set. I got that formula well.

28. KingGeorge Group Title

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.

29. ParthKohli Group Title

30. satellite73 Group Title

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

31. Ishaan94 Group Title

what do you mean by primitive triplets?

32. KingGeorge Group Title

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

33. satellite73 Group Title

"primitive" no common factors

34. Ishaan94 Group Title

oh

35. satellite73 Group Title

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

36. ParthKohli Group Title

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

37. mukushla Group Title

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

38. ParthKohli Group Title

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

39. mukushla Group Title

np...:)

40. ParthKohli Group Title

Haha$\Huge \ddot \smile$

41. Ishaan94 Group Title

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

42. ParthKohli Group Title

Well. Isn't that the same thing?

43. KingGeorge Group Title

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

44. Ishaan94 Group Title

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

45. KingGeorge Group Title

If you want a primitive triplet, it is.

46. KingGeorge Group Title

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

47. Ishaan94 Group Title

how do i prove it?

48. KingGeorge Group Title

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.

49. Ishaan94 Group Title

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?

50. KingGeorge Group Title

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.

51. satellite73 Group Title

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