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GCD relative prime question

  1. Feb 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Let a and b be relatively prime. Show that gcd(a+b,ab) = 1

    2. Relevant equations
    ax+by = 1 for some integers x and y

    3. The attempt at a solution
    I set gcd(a+b,ab) = d. I'm now trying to show that d = 1 using elementary algebra techniques.
    a+b = rd
    ab = sd
    ax + by = 1


    I'm kind of stuck here.. am I on the right track? Do I just need to aggresively rearrange stuff until I can express (a+b) and (ab) as a linear combination that equals one? or perhaps arrange them in such away that I show d divides both a and b individually and therefore it must be one since they are relatively prime? any hints?

    Update:

    a+b = rd ---> b = rd -a

    ab = a(rd-a) = (ard)-(a^2) = sd
    dividing both sides by d...

    ar - (a^2 / d) = s
    ar is an integer and s is an integer, so a^2 / d must be an integer, therefore d|a^2.

    I employed a similar argument to show that d|b^2. Since d|a^2 and d|b^2 and gcd(a,b) = 1, we can use the fact that gcd(a^2,b^2) = 1 to conclude that d = 1.

    Is this a solid argument?
     
    Last edited: Feb 19, 2015
  2. jcsd
  3. Feb 20, 2015 #2

    wabbit

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    Gold Member

    Your argument looks perfectly sound to me. You can make it a little more elegant by going : ## ard-a^2=rd \implies a^2=(ar-r)d \implies d|a^2 ## , but it's really the same thing.
    Note however that this is not the method suggested in the formulation of the problem : you didn't make use of ## ax+by=1 ##. If you want to try that as well, do you know where that equation comes from ?
     
    Last edited: Feb 20, 2015
  4. Feb 20, 2015 #3
    Oh wow, didn't even notice that I didn't use the relatively prime information, haha. Yeah, I know about the equation. Thanks for the post yo!
     
  5. Feb 20, 2015 #4

    wabbit

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    Gold Member

    No problem. I can see you don't like going by the book, you tackle the problem your own way, which is cool : )
     
  6. Feb 20, 2015 #5
    Thanks ^_^ I appreciate it. I'm pretty new to this math thing but I want to learn it's mysteries so I appreciate the complement :D
     
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