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Math Help - X+(0*0)=x

  1. #1
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    X+(0*0)=x

    working on the famous diophantienne equation in mutualy prime numbers x^k+y^k=z^k I've been finding an interresting result for k = 2:

    show that for such mutualy primes number x,y,z / x^2+y^2=z^2:
    for the multiples existing positives integers k,c,d for wich:
    k*x+1= abs(2x-z)*c and k*(z-x)+1= abs(2x-z)*d
    we have abs (c*(z-x)+d*(z-(z+x)))= abs(c*(z-x)+d(-x))=1

    (so z-x and x are prime! (wich is not the interesting part of all this stuff))
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  2. #2
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    Quote Originally Posted by SkyWatcher View Post
    working on the famous diophantienne equation in mutualy prime numbers x^k+y^k=z^k I've been finding an interresting result for k = 2:

    show that for such mutualy primes number x,y,z / x^2+y^2=z^2:
    for the multiples existing positives integers k,c,d for wich:
    k*x+1= abs(2x-z)*c and k*(z-x)+1= abs(2x-z)*d
    we have abs (c*(z-x)+d*(z-(z+x)))= abs(c*(z-x)+d(-x))=1

    (so z-x and x are prime! (wich is not the interesting part of all this stuff))
    I can post the full solution to the Diophantine equation x^2+y^2=z^2. It is not that complicated if you want.
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  3. #3
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    i dont think the full solution in mutual prime numbers:
    x=(m^2-n^2) y=2mn z=(m^2+n^2) whit m>n and m and n prime
    would lead 'naturaly' to the solution of the assertion i propose
    i let it unproved (for the moment) to excerce the sagacity of the readers who could be interested (and i must say i find my solution of this problem very interresting...)
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