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Gcd(a,b,c)lcm(a,b,c)=abc => a,b,c relatively prime in pairs

  1. Apr 10, 2010 #1
    Claim: If gcd(a,b,c)lcm(a,b,c) = abc, then gcd(a,b)=gcd(b,c)=gcd(a,c)=1.

    I'm trying to understand why this is true...
    How can we prove it?

    Any help is appreciated!
     
  2. jcsd
  3. Apr 10, 2010 #2
    Start by the prime factorizations of a,b and c, then use the expressions for the gcd and lcm in terms of these.
     
  4. Apr 12, 2010 #3
    What do you mean?
    I've written out the prime factorizations of a,b, and c. But I don't know what to do next...
     
  5. Apr 12, 2010 #4
    If the prime factorizations of a, b and c are:

    [tex]a=2^{e_1}3^{e_2}\cdots p^{e_i}[/tex]

    [tex]b=2^{f_1}3^{f_2}\cdots p^{f_i}[/tex]

    [tex]c=2^{g_1}3^{g_2}\cdots p^{g_i}[/tex]

    (If a particular prime factor doesn't appear in the factorization, its exponent is zero)

    Then you should know that:

    [tex]gcd\left(a,b,c\right)=2^{min \left\{e_1,f_1,g_1\right\}}3^{min \left\{e_2,f_2,g_2\right\}}\cdots p^{min\left\{e_i,f_i,g_i\right\}}[/tex]

    And:

    [tex]lcm\left(a,b,c\right)=2^{max \left\{e_1,f_1,g_1\right\}}3^{max \left\{e_2,f_2,g_2\right\}}\cdots p^{max\left\{e_i,f_i,g_i\right\}}[/tex]

    Now plug these in your equality and see what must happen for the exponents to agree.
     
  6. Apr 12, 2010 #5
    I think we'll then have min{ei,fi,gi}+max{ei,fi,gi}=ei+fi+gi, but why does this imply gcd(a,b)=gcd(b,c)=gcd(a,c)=1?
     
  7. Apr 13, 2010 #6
    For a given i, what must happen to the ei's, fi's and gi's for that equality to be true? For example, can they all be > 0?
     
  8. Apr 25, 2010 #7

    disregardthat

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    Science Advisor

    Use that [tex]\text{gcd}(a,b,c)=\text{gcd}(\text{gcd}(a,b),c)[/tex], and likewise for [tex]\text{lcm}[/tex]. Also, the fact that [tex]\text{gcd}(a,b)\text{lcm}(a,b)=ab[/tex] might come in handy. You can extract a lot of information from the equation using this, and you do not have to go the way through their respective prime factorizations.
     
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