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Math Help - About Pythagorean triple...

  1. #1
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    Post About Pythagorean triple...

    Hi! Can anyone provide a proof for the following:
    The following formula
    a=k(m^2-n^2)
    b=k(2mn)
    c=k(m^2+n^2)
    generates all the positive integers that satisfy a^2+b^2=c^2.
    Thanks!
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by TerenceCS View Post
    Hi! Can anyone provide a proof for the following:
    The following formula
    a=k(m^2-n^2)
    b=k(2mn)
    c=k(m^2+n^2)
    generates all the positive integers that satisfy a^2+b^2=c^2.
    Thanks!
    I believe we can use an argument similar to the one in this thread post #22.
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by TerenceCS View Post
    Hi! Can anyone provide a proof for the following:
    The following formula
    a=k(m^2-n^2)
    b=k(2mn)
    c=k(m^2+n^2)
    generates all the positive integers that satisfy a^2+b^2=c^2.
    Thanks!
    What undefined suggests is good! Here is another way.

    You can suppose that a,b,c are relatively prime. Then it is impossible for both a,b to be odd (because then c^2\equiv 2 \mod 4, which is impossible), and impossible for both of them to be even, which would contradict the hypothesis that a,b,c are relatively prime. Hence, say a is even and b,c are odd. Write (c-b)(c+b)=a^2. Now both c-b,c+b are even, say c-b=2u,c+b=2v; moreover u,v are relatively prime. Since a^2=4uv is a square, both u,v must be squares, so we have c-b=2m^2. c+b=2n^2, i.e. c=m^2+n^2, b=n^2-n^2, a=2mn.
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