# Greatest common divisor

• September 16th 2013, 05:30 AM
AndrewLowitz
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?
• September 16th 2013, 07:29 AM
Hartlw
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.
• September 16th 2013, 08:24 AM
Hartlw
Re: Greatest common divisor
Did algebra and got as far as:

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

stuck
• September 16th 2013, 11:40 AM
AndrewLowitz
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.
• September 18th 2013, 05:16 AM
Hartlw
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)