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ParthKohli Group Title

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

  • 2 years ago
  • 2 years ago

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  1. ParthKohli Group Title
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    Give it a start, Kingy.

    • 2 years ago
  2. KingGeorge Group Title
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    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.

    • 2 years ago
  3. ParthKohli Group Title
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    \[2n,n^2 - 1, n^2 + 1 \]?

    • 2 years ago
  4. ParthKohli Group Title
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    I know one. \(3,4,5\).

    • 2 years ago
  5. KingGeorge Group Title
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    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.

    • 2 years ago
  6. ParthKohli Group Title
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    Another one! \(5,12,13\)!

    • 2 years ago
  7. KingGeorge Group Title
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    Correct, (3,4,5) is a triplet as well as (5,12,13) and (7,24,25) among infinite others.

    • 2 years ago
  8. ParthKohli Group Title
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    Can you tell me the formula(s) for it, please?

    • 2 years ago
  9. KingGeorge Group Title
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    The most well known one is \[2mn, \quad m^2-n^2, \quad m^2+n^2\]Given any \(m,n\) such that \(m>n\).

    • 2 years ago
  10. ParthKohli Group Title
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    I see :) Are there any more?

    • 2 years ago
  11. KingGeorge Group Title
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    What you gave above, only works for \(m\) even. So it generates an infinite number, but not all of them.

    • 2 years ago
  12. ParthKohli Group Title
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    Can this formula generate all such triplets, or you have to use some other formulae to generate more?

    • 2 years ago
  13. mukushla Group Title
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    another one \[(x,y,z)=(2k+1,2k^2+2k,2k^2+2k+1)\]

    • 2 years ago
  14. ParthKohli Group Title
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    Oh wait. It does generate all.

    • 2 years ago
  15. KingGeorge Group Title
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    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.

    • 2 years ago
  16. KingGeorge Group Title
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    Ccorrection, \(m>n\ge1\).

    • 2 years ago
  17. satellite73 Group Title
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    if you want them "primitive" makes sure \(m,n\) are opposite parity, and no common factors

    • 2 years ago
  18. ParthKohli Group Title
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    Well. I'd just use the first one :)

    • 2 years ago
  19. KingGeorge Group Title
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    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.

    • 2 years ago
  20. satellite73 Group Title
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    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

    • 2 years ago
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  21. ParthKohli Group Title
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    If \(\gcd(m,n) = 1\), then it automatically means that \(m - n\) is odd.

    • 2 years ago
  22. satellite73 Group Title
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    no, try 5 and 3

    • 2 years ago
  23. ParthKohli Group Title
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    Oh grr.

    • 2 years ago
  24. satellite73 Group Title
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    really, if you have time, work through the worksheet i sent you you will see why all triples come from this formula

    • 2 years ago
  25. ParthKohli Group Title
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    Yes, I am looking at it.

    • 2 years ago
  26. satellite73 Group Title
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    it requires no more than algebra

    • 2 years ago
  27. ParthKohli Group Title
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    All set. I got that formula well.

    • 2 years ago
  28. KingGeorge Group Title
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    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.

    • 2 years ago
  29. ParthKohli Group Title
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    Thank you, Your Majesty and Your Bikesty!

    • 2 years ago
  30. satellite73 Group Title
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    have fun. there are several steps to the proof, but it is all there

    • 2 years ago
  31. Ishaan94 Group Title
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    what do you mean by primitive triplets?

    • 2 years ago
  32. KingGeorge Group Title
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    \(\gcd(a,b,c)=1\).

    • 2 years ago
  33. satellite73 Group Title
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    "primitive" no common factors

    • 2 years ago
  34. Ishaan94 Group Title
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    oh

    • 2 years ago
  35. satellite73 Group Title
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    so 3, 4, 5 but not 6, 8, 10

    • 2 years ago
  36. ParthKohli Group Title
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    Whoa! Primitive triplets are \(a,b,c\) such that \(\gcd(a,b,c) = 1\). Other words, they all are co-prime.

    • 2 years ago
  37. mukushla Group Title
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    @ParthKohli method of obtaining the formula i gave u...we will talk about later if u want... http://openstudy.com/study#/updates/500a7eb3e4b0549a892eeeee

    • 2 years ago
  38. ParthKohli Group Title
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    @mukushla I'd see that tutorial tomorrow; tired somewhat at the moment. Looks impressive by looking at it though.

    • 2 years ago
  39. mukushla Group Title
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    np...:)

    • 2 years ago
  40. ParthKohli Group Title
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    Haha\[ \Huge \ddot \smile \]

    • 2 years ago
  41. Ishaan94 Group Title
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    why do we need to have opposite parity? and what's with m-n being odd?

    • 2 years ago
  42. ParthKohli Group Title
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    Well. Isn't that the same thing?

    • 2 years ago
  43. KingGeorge Group Title
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    \(m-n\) is odd if and only if they have opposite parity. So either \(m\) or \(n\) is odd, but not both.

    • 2 years ago
  44. Ishaan94 Group Title
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    i understand that. but is it necessary for m-n to be odd?

    • 2 years ago
  45. KingGeorge Group Title
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    If you want a primitive triplet, it is.

    • 2 years ago
  46. KingGeorge Group Title
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    If they are both even or both odd, \(\gcd(a,b,c)\ge2\)

    • 2 years ago
  47. Ishaan94 Group Title
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    how do i prove it?

    • 2 years ago
  48. KingGeorge Group Title
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    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.

    • 2 years ago
  49. Ishaan94 Group Title
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    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?

    • 2 years ago
  50. KingGeorge Group Title
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    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.

    • 2 years ago
  51. satellite73 Group Title
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    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

    • 2 years ago
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