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.

Re: Greatest common divisor

Did algebra and got as far as:

gcd(xz^{2}+yx^{2}+zy^{2}, xy^{2}+yz^{2}+zx^{2}, x+y+z)

stuck

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.

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)