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Divisibility rules and proof by contradiction

  1. Jul 7, 2012 #1
    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
    [itex]5|(x+y-z)[/itex] and [itex]2*5^{4}k=(x+y)(z-y)(z-x)((x+y)^2+(z-y)^2+(z-x)^2)[/itex]
    or equivalently [itex]5^{4}k=(x+y)(z-y)(z-x)((x+y-z)^2-xy+xz+yz)[/itex]
    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, [itex]2((x+y)^2+(z-y)^2+(z-x)^2)=((x+y-z)^2-xy+xz+yz)[/itex] 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. jcsd
  3. Jul 7, 2012 #2
    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
     
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