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Math Help - Can you help me about GCD.?

  1. #1
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    Can you help me about GCD.?

    I cannot proof 2 problem.....*-*!!!

    1.Prove that if gcd(a,b)=1,then gcd(a+b,ab)=1.


    2.If gcd(a,b)=1,and c|(a+b),then gcd(a,c)=gcd(b,c)=1.
    (I can show proof gcd(a,c)=1 but I can show gcd(b,c)=1.)



    ....Thank !
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  2. #2
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    Quote Originally Posted by konna View Post
    I cannot proof 2 problem.....*-*!!!

    1.Prove that if gcd(a,b)=1,then gcd(a+b,ab)=1.


    Say gcd(a+b,ab)=d\Longrightarrow:

    (1) a+b=kd

    (2) ab=md

    Now multiply (1) by b: ab+b^2=(kb)d and substract (2): b^2=(kb-m)d\Longrightarrow d\mid b^2\Longrightarrow d\mid b

    Now do as above but multiplying (1) by a and substracting (2) and you'll get that also d\mid a, so...

    Tonio


    2.If gcd(a,b)=1,and c|(a+b),then gcd(a,c)=gcd(b,c)=1.
    (I can show proof gcd(a,c)=1 but I can show gcd(b,c)=1.)



    ....Thank !
    .
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  3. #3
    Senior Member Shanks's Avatar
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    Notice that gcd(n,m)=1 iff the m-class is in the Z_n^* !
    since gcd(a,b)=1, we have gcd(a+b,a)=1, and gcd(a+b,b)=1 by the euler algrithem.
    thus gcd(a+b,ab)=1, that is because the statement "gcd(a+b,a)=1, and gcd(a+b,b)=1 " mean that a and b are in the Z_{a+b}^* group, so the product of a-class and b-class lies in the group, that is equivalent to say gcd(a+b,ab)=1.
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  4. #4
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    Thank you so much , I'm cheerful !

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