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Natural Numbers & the Pythagorean Theorem

  1. Dec 8, 2009 #1

    Char. Limit

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    I was pondering square numbers today, and I noticed something interesting: every natural number contains information to construct a Pythagorean triple. Let me show what I mean for an odd natural number, also using the binomial theorem for square quadratic equations (equations of the form [tex](x+y)^2[/tex])

    Firstly, for a square number to be odd, it must satisfy the equation [tex]m^2=2n+1=n+(n+1)[/tex]. In other words, an odd square number is always equal to two integers one number apart. Seems obvious, huh? Well, it's important too. Bear with me. Now, here's another equation:

    [tex]n^2+2n+1=(n+1)^2[/tex]

    Now, let's rearrange that last equation:

    [tex]2n+1+n^2=(n+1)^2[/tex]

    Now, is it not true that any positive number is equal to the square of its square root ([tex]x=(\sqrt{x})^2[/tex])? So, this is true:

    [tex]2n+1+n^2=(n+1)^2=(\sqrt{2n+1})^2+n^2[/tex]

    Also, if [tex]x^2=2n+1[/tex], then [tex]x=\sqrt{2n+1}[/tex] and [tex]n=\frac{x^2-1}{2}[/tex]. Substituting some values gives us:

    [tex]x^2+(\frac{x^2}{2}-\frac{1}{2})^2=(\frac{x^2}{2}-\frac{1}{2}+1)^2=(\frac{x^2}{2}+\frac{1}{2})^2[/tex]

    In summary, a pythagorean triple can be formed with any odd number, half of its square minus one-half, and half of its square plus one-half. I have a marvelous equation for the even natural numbers, but it is too narrow to fit in the margin, and besides, I'm sure everyone here has done this before. Lastly, here are 7 examples:
    [tex]3^2+4^2=5^2[/tex]
    [tex]5^2+12^2=13^2[/tex]
    [tex]7^2+24^2=25^2[/tex]
    [tex]9^2+40^2=41^2[/tex]
    [tex]11^2+60^2=61^2[/tex]
    [tex]13^2+84^2=85^2[/tex]
    [tex]15^2+112^2=113^2[/tex]
     
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  3. Dec 9, 2009 #2
    a pythagorian triple can be constructed from every multiple of 4.

    4^2+3^2=5^2
    8^2+15^2=17^2
    12^2+35^2=37^2

    let x^2=z^2-y^2=(z+y)(z-y) let x^2=4*k. Then we can make z-y=4 "by hand" and solve for z+y=k.

    If x is an odd number, then we can make z-y=1 and solve for z+y=x^2.

    If x=2 mod 4 then we get a pythagorean triple but it is not primitive.

    In general, if x=mn, m and n relatively prime, then we can make z-y=n^2 and z+y=m^2.
     
  4. Dec 9, 2009 #3

    Char. Limit

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    Er... is that what I did? I'm still not sure if anyone has used the method I used...
     
  5. Dec 9, 2009 #4
    You notice the difference between the even leg and the hypotenuese is always one. And the sum of the even leg and the hypotenuese is the square of the odd leg. I guess there are a lot of roads that take us here.

    I did a science project on pythogorean triples in high school about thirty years ago. I haven't seen anything published on the topic. Maybe because it is just high school math.
     
  6. Dec 9, 2009 #5
    Let (x,y,z) be a pythagorean triple (PT). Then I believe all pythagorean triples can be expressed as (2mn,m^2-n^2,m^2+n^2) m and n positive integers with m>n. I personally classify primitive PTs into three groups. The regular odds, which is the list you ennumerate, is when m-n=1. The regular evens, my list, is when m is even and n=1. And all other primitive PTs are the irregulars. For example m=5 n=2 gives us (20,21,29)
     
    Last edited: Dec 9, 2009
  7. Dec 9, 2009 #6

    Char. Limit

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    Yes.

    Also, a similar idea can be used for even numbers...

    I can't remember the proof (I'm not where it is right now), but the equation generally goes like this:

    [tex]x^2+(\frac{x^2}{4}-1)^2=(\frac{x^2}{4}+1)^2[/tex]

    The first seven triples are:

    [tex]4^2+3^2=5^2[/tex]
    [tex]6^2+8^2=10^2[/tex]
    [tex]8^2+15^2=17^2[/tex]
    [tex]10^2+24^2=26^2[/tex]
    [tex]12^2+35^2=37^2[/tex]
    [tex]14^2+48^2=50^2[/tex]
    [tex]16^2+63^2=65^2[/tex]

    Notice that for every multiple of two that isn't a multiple of four, the prime produced isn't a primitive prime.
     
  8. Dec 20, 2009 #7
    Having already shown that every odd number is the leg of a primitive pythagorean triple. The corrollary, every number that's twice an odd number is the leg of a non-primitive pythagorean triple, is rather trivial. In your examples, every multiple of two that isn't a multiple of four is the leg of twice the regular odd primitive pythagorean triple. (6,8,10)=2*(3,4,5), (10,24,26)=2*(5,12,13), (14,48,50)=2*(7,24,25).
     
  9. Dec 21, 2009 #8

    Char. Limit

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    The odd thing that I saw was that even numbers not divisible by four didn't get a primitive triple, but multiples of four do, using this formula. Thus, an odd number times four can be given two primitive triples using this formula, but an odd number times two cannot.

    Just look at 12. You have (12,16,20) and you have (12,35,37).


    I don't understand why nearly everything I've said or done here so far has been discredited in some fashion. I know I can't be THAT stupid, or I wouldn't even be here.

    (Cynicism rules!)
     
  10. Dec 23, 2009 #9
    I suspect some Greek has 3000 years of priority on both of us. I'm just trying to suggest further fields of research. Everything you write is true. I don't mean to discredit you. At the same time, you're not going to get a Field prize studying primitive pythagorean triples.

    Let's look at the odd leg. Odd numbers that are semiprimes have an additional PPT. i. e. (21,20,29) (119,120,169) since 21=7x3 and 119=7x17.
     
  11. Dec 24, 2009 #10

    Char. Limit

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    Sorry about being offended, I was agitated by something completely unrelated that night.

    No Prize for the high-school student? Oh well. (lol)

    Hmm, what's the pattern there, I wonder? Both of the other legs are one away, but in different directions... but I'm sure if I dredged up enough examples, an equation could be written.

    I've never heard of semiprimes before today... is it just a coincidence that both examples are 7-divisible?
     
  12. Dec 24, 2009 #11
    A semiprime is the product of two primes, neither necessarily 7. There's an irregular ppt (65,72,97). 65=13*5. We get the other sides by simultaneously solving z+y=169 and z-y=25. Incidentally, we can always write two linear equations in two unknowns as a quadratic equation so I just find my notation easier to calculate with.

    The two examples of irregular ppts off the top of my head were almost isosceles ppts. To be an ai ppt m~(1+sqrt(2))n . m=7 n=3 and m=17 n=7. The next ai ppt (to the best of my knowledge) is (696, 697, 985). 697=17*41. So we can construct a sequence 1,3,7,17,41... that gives us the ai ppts. Can you prove this?

    The even leg alternates between being larger than the odd leg and smaller.
     
  13. Dec 24, 2009 #12

    Char. Limit

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    I could, maybe, after explanation on both the tilde (what does that mean?) and what "ai ppts" are.

    I get what a ppt is though.
     
  14. Dec 24, 2009 #13
    tilde- almost equal. aippt- almost isoscoles primitive pythagorean triple. The odd leg is one more or less than the even leg.

    Let me define the sequence 1,2,5,12,29,70.... x1=1 x2=2 xn=2*xn-1+xn-2.

    Then (2*xn*xn-1,x2n-x2n-1,x2n+x2n-1) is an aippt.

    Proof is by induction. If n=2 we get (3,4,5) which is an aippt. The formula gives a ppt for any integers, so it suffices to show that it is ai. I'll leave that as an exercise for the reader.
     
  15. Dec 24, 2009 #14

    Char. Limit

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    You know, this isn't a book, you don't leave exercises for readers...

    But, I'll take a bit of a look...

    So, if the one leg is multiplied by two, it must be even, therefore, the other must be odd. Not that that helps much...

    I also know that [tex]2 x_n x_{n-1} \pm 1 = x^2_n-x^2_{n-1}[/tex]...

    Just working on the right side, I set [tex]x_n[/tex] equal to [tex]2 x_{n-1} + x_{n-2}[/tex]...

    And I get [tex] 3 x^2_{n-1} + 4 x_{n-1} x_{n-2} + x^2_{n-2} [/tex]

    Let's try getting an [tex]x_{n-2}[/tex] in the left side.

    [tex]4 x^2_{n-1} + 2 x_{n-1} x_{n-2} \pm 1[/tex]

    Setting them equal to each other and doing some canceling gives me...

    [tex]x^2_{n-1} \pm 1 = x_{n-2} (2 x_{n-1} + x_{n-2}) = x_{n-2} x_n[/tex]

    Huh? How did [tex]x_n[/tex] reappear?
     
  16. Jan 13, 2010 #15
    I have a proof but the margin is too small to contain it. I just don't see the value added in typing up a page of high school algebra. And if there's any high school math teachers in the peanut gallery, it makes a neat problem for proof by induction.
     
  17. Feb 16, 2010 #16
    3=2*1+1 , 4= 2*1*2 , 5= 2*1*2+1
    5=2*2+1 , 12= 2*2*3 , 13= 2*2*3+1
    7=2*3+1 , 24= 2*3*4 , 25= 2*3*4+1
    9=2*4+1 , 40= 2*4*5 , 41= 2*4*5+1
    11=2*5+1 , 60= 2*5*6 , 61= 2*5*6+1
    13=2*6+1 , 84= 2*6*7 , 85= 2*6*7+1
    15=2*7+1 , 112= 2*7*8 , 113= 2*7*8+1
    ...
    (2*i+1)^2 + (2*i*(i+1))^2 = (2*i*(i+1)+1)^2

    Ref: see (10) in:
    http://www.math.rutgers.edu/~erowland/pythagoreantriples.html [Broken]

    and also:
    http://www.friesian.com/pythag.htm
     
    Last edited by a moderator: May 4, 2017
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