# Homework Help: Divisibility rules and proof by contradiction

1. Jul 7, 2012

### tt2348

I posted this in the number theory forum to no success... so I figured maybe the homework help people would have some input

Let x,y,z be integers with no common divisor satisfying a specific condition, which boils down to
$5|(x+y-z)$ and $2*5^{4}k=(x+y)(z-y)(z-x)((x+y)^2+(z-y)^2+(z-x)^2)$
or equivalently $5^{4}k=(x+y)(z-y)(z-x)((x+y-z)^2-xy+xz+yz)$
I want to show that GCD(x,y,z)≠1, starting with the assumption 5 dividing (x+y), (z-y), or (z-x) results in x,y or z being divisible by 5. then it's easy to show that 5 divides another term, implying 5 divides all three.
I run into trouble assuming 5 divides the latter part, $2((x+y)^2+(z-y)^2+(z-x)^2)=((x+y-z)^2-xy+xz+yz)$ and showing the contradiction from that point.
Any hints?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 7, 2012

### Yukoel

Hello tt2348,
Leave the square terms unchanged and consider all possible values of the square of any integer mod 5 .There are three squares here() .How will you get them to get an expression divisible by 5 ?There lies the answer that one of the square terms has to be divisible by 5.
Hoping this helps.
regards
Yukoel