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Math Help - Greatest common divisor

  1. #1
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    Greatest common divisor

    Hi!
    I have a math problem:
    we know gcd(x;y;z)=1, x \neq y \neq z and x,y,z>1.
    gcd(\frac{(x+y)(y+z)(z+x)-(x-y)(y-z)(z-x)}{2}-xyz;
    \frac{(x+y)(y+z)(z+x)+(x-y)(y-z)(z-x)}{2}-xyz;x+y+z)
    may have value?
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  2. #2
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    Re: Greatest common divisor

    gcd(2,4,6)=2

    EDIT: whoops. sorry. gcd(x,y,z)=1 is given
    Oh well, looks like if you multply every thing out and use gcd[a(x,y,z)] you might get there.
    Last edited by Hartlw; September 16th 2013 at 08:38 AM.
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  3. #3
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    Re: Greatest common divisor

    Did algebra and got as far as:

    gcd(xz2+yx2+zy2, xy2+yz2+zx2, x+y+z)

    stuck
    Last edited by Hartlw; September 16th 2013 at 09:36 AM. Reason: typo
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  4. #4
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    Re: Greatest common divisor

    Thanks for the reply but I question the current, still do not know how to prove it. for numbers in the form 3n+1; 3k+1; 3m+1 we have gcd=3.
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  5. #5
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    Re: Greatest common divisor

    This is a problem in divisibility of polynomials. In the form of post #3, neither of the first two terms is divisible by x+y+z, gcd =1. Perlis has a nice chapter on this.

    From Google: gcd(a,b,c) = gcd(gcd(a,b),c)
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