Hi! I have a math problem: we know , and . may have value?
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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; Sep 16th 2013 at 07:38 AM.
Did algebra and got as far as: gcd(xz^{2}+yx^{2}+zy^{2}, xy^{2}+yz^{2}+zx^{2}, x+y+z) stuck
Last edited by Hartlw; Sep 16th 2013 at 08:36 AM. Reason: typo
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.
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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