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ParthKohli

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

  • one year ago
  • one year ago

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  1. ParthKohli
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    Give it a start, Kingy.

    • one year ago
  2. KingGeorge
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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.

    • one year ago
  3. ParthKohli
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    \[2n,n^2 - 1, n^2 + 1 \]?

    • one year ago
  4. ParthKohli
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    I know one. \(3,4,5\).

    • one year ago
  5. KingGeorge
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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.

    • one year ago
  6. ParthKohli
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    Another one! \(5,12,13\)!

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

    • one year ago
  8. ParthKohli
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    Can you tell me the formula(s) for it, please?

    • one year ago
  9. KingGeorge
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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 \(m>n\).

    • one year ago
  10. ParthKohli
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    I see :) Are there any more?

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

    • one year ago
  12. ParthKohli
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    Can this formula generate all such triplets, or you have to use some other formulae to generate more?

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

    • one year ago
  14. ParthKohli
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    Oh wait. It does generate all.

    • one year ago
  15. KingGeorge
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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.

    • one year ago
  16. KingGeorge
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    Ccorrection, \(m>n\ge1\).

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

    • one year ago
  18. ParthKohli
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    Well. I'd just use the first one :)

    • one year ago
  19. KingGeorge
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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.

    • one year ago
  20. satellite73
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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

    • one year ago
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  21. ParthKohli
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    If \(\gcd(m,n) = 1\), then it automatically means that \(m - n\) is odd.

    • one year ago
  22. satellite73
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    no, try 5 and 3

    • one year ago
  23. ParthKohli
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    Oh grr.

    • one year ago
  24. satellite73
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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

    • one year ago
  25. ParthKohli
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    Yes, I am looking at it.

    • one year ago
  26. satellite73
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    it requires no more than algebra

    • one year ago
  27. ParthKohli
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    All set. I got that formula well.

    • one year ago
  28. KingGeorge
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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.

    • one year ago
  29. ParthKohli
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    Thank you, Your Majesty and Your Bikesty!

    • one year ago
  30. satellite73
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    have fun. there are several steps to the proof, but it is all there

    • one year ago
  31. Ishaan94
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    what do you mean by primitive triplets?

    • one year ago
  32. KingGeorge
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    \(\gcd(a,b,c)=1\).

    • one year ago
  33. satellite73
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    "primitive" no common factors

    • one year ago
  34. Ishaan94
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    oh

    • one year ago
  35. satellite73
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    so 3, 4, 5 but not 6, 8, 10

    • one year ago
  36. ParthKohli
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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.

    • one year ago
  37. mukushla
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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

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

    • one year ago
  39. mukushla
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    np...:)

    • one year ago
  40. ParthKohli
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    Haha\[ \Huge \ddot \smile \]

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

    • one year ago
  42. ParthKohli
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    Well. Isn't that the same thing?

    • one year ago
  43. KingGeorge
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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.

    • one year ago
  44. Ishaan94
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    i understand that. but is it necessary for m-n to be odd?

    • one year ago
  45. KingGeorge
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    If you want a primitive triplet, it is.

    • one year ago
  46. KingGeorge
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    If they are both even or both odd, \(\gcd(a,b,c)\ge2\)

    • one year ago
  47. Ishaan94
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    how do i prove it?

    • one year ago
  48. KingGeorge
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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.

    • one year ago
  49. Ishaan94
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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?

    • one year ago
  50. KingGeorge
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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.

    • one year ago
  51. satellite73
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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

    • one year ago
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