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Greatest common divisor

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppsoe a, b[tex]\in[/tex]natural numbers, and d=GCD(a,b). Then d^2=GCD(a^2,b^2). I need to find where the proof goes wrong.

    2. Relevant equations

    3. The attempt at a solution
    By hypothesis, we have that d divides a and d divides b, so there are integres s and t with a=ds and b=dt. Then a^2=d^2s^2 and so d^2 divides a^2. Similarily d^2 divides b^2. Thus d^2 is a common divisor of a^2 and b^2, as desired.
  2. jcsd
  3. Nov 17, 2008 #2


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    Science Advisor

    Re: d^2=GCD(a^2,b^2)

    No, not "as desired". What you "desire" is the greatest common divisor. Saying that d^2 is a common divisor does not mean it is the GREATEST common divisor.
  4. Nov 17, 2008 #3
    Re: d^2=GCD(a^2,b^2)

    Ok, that made a lot more sense now that I see that!
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