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Math Help - Pythagorean triples

  1. #1
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    Pythagorean triples

    Show that the area A of a triangle whose sides form a pythagorean triple is always divisible by 6.

    Hint: For an integer x, x^2 is of the form 4k or 1+8k, and when 3 doesn't divide x, x^2 is of the form 1+3k.

    I don't know what to do. I figured the strategy would be to show that it is divisible by 2 and also by 3, so that it is divisible by 6=2*3, but I don't know how to go about this or if it is a correct approach.

    thanks
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  2. #2
    Senior Member Dinkydoe's Avatar
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    I hope you can do a little modular arithmatic. You're idea is indeed the right one:

    Let (a,b,c) be such that a^2+b^2= c^2
    The area of the triangle is given by: A = \frac{1}{2}ab

    Observe that: c^2\equiv 0,1 mod 4. And c^2\equiv 0,1 mod 3. (you can easily check this by looking at quadratic residu-classes modulo 4, and modulo 3).

    Case 1.
    Assume c^2=a^2+b^2\equiv 0 mod 4 \Longrightarrow a^2\equiv b^2\equiv 0 mod 4 \Longrightarrow ab\equiv 0 mod 4.

    Assume c^2= a^2+b^2\equiv 0 mod 3 \Longrightarrow a^2\equiv b^2\equiv 0 mod 3 \Rightarrow ab\equiv 0 mod 3.

    Case 2.
    Assume c^2 = a^2+b^2\equiv 1mod 4. This implies a^2+b^2\equiv 1 mod 8, or a^2+b^2\equiv 5 mod 8

    Now we'll look at quadratic residues modulo 8 wich are \left\{0,1,4\right\}

    Since there is no quadratic residu that is 5 mod 8. It follows that c^2=a^2+b^2\equiv 1 mod 8  \Rightarrow a^2\equiv 0 mod 8, b^2\equiv 1 mod 8. Hence ab\equiv 0 mod 8.


    Assume c^2= a^2+b^2\equiv 1 mod 3 \Longrightarrow a^2\equiv 0 mod 3, b^2\equiv 1 mod 3 \Rightarrow ab\equiv 0 mod 3

    So in both cases we have 4|ab and 3|ab\Rightarrow  6|A
    Last edited by Dinkydoe; January 27th 2010 at 02:57 PM.
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  3. #3
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    Thanks so much!
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  4. #4
    Senior Member Dinkydoe's Avatar
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    Please check my corrections:

    I argued that if c^2 = a^2+b^2\equiv 1 mod 4 that the only possiblility is that c^2=a^2+b^2\equiv 1 mod 8

    This is an important observation!
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  5. #5
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    I'm sorry, I have one more question. I understand your proof, but wouldn't it be easier to do it with mod 3 and mod 2 instead of mod 3 and mod 4, or is there some reason why you can't consider it with mod 2?

    Thanks.
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  6. #6
    Senior Member Dinkydoe's Avatar
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    We want to proof that 2|(\frac{ab}{2}) hence we must prove 4|ab. That's why ab\equiv 0 mod 2 isn't enough.
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  7. #7
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    Sorry for all of the questions, but then wouldn't we need to show that 6 divides ab instead of 3 since we need to show that 3 divides (ab)/2.
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  8. #8
    Senior Member Dinkydoe's Avatar
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    Sorry for all of the questions, but then wouldn't we need to show that 6 divides ab instead of 3 since we need to show that 3 divides (ab)/2
    No. if 3|ab then it follows 3|\frac{ab}{2}, since 2,3 are coprime. Do I need to proof this for you? ;p

    If we show that 3|ab and 4|ab then we've shown that 12|ab. Hence 6|{\frac{ab}{2}}.
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  9. #9
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    Ok, I've got it now. Thanks for your patience.
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