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ParthKohli

  • 2 years ago

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

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  1. ParthKohli
    • 2 years ago
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    Give it a start, Kingy.

  2. KingGeorge
    • 2 years ago
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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.

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

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

  5. KingGeorge
    • 2 years ago
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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.

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

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

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

  9. KingGeorge
    • 2 years ago
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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 \(m>n\).

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

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

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

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

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

  15. KingGeorge
    • 2 years ago
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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.

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

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

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

  19. KingGeorge
    • 2 years ago
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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.

  20. satellite73
    • 2 years ago
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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

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

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

  23. ParthKohli
    • 2 years ago
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    Oh grr.

  24. satellite73
    • 2 years ago
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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

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

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

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

  28. KingGeorge
    • 2 years ago
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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.

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

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

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

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

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

  34. Ishaan94
    • 2 years ago
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    oh

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

  36. ParthKohli
    • 2 years ago
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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.

  37. mukushla
    • 2 years ago
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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

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

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

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

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

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

  43. KingGeorge
    • 2 years ago
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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.

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

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

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

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

  48. KingGeorge
    • 2 years ago
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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.

  49. Ishaan94
    • 2 years ago
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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?

  50. KingGeorge
    • 2 years ago
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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.

  51. satellite73
    • 2 years ago
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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

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