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Math Help - Carmichael numbers

  1. #16
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    A new direction...

    Iíve got you to lcm(a_1,a_2,a_3)|gcd(a_2a_3+a_1a_3+a_1a_2 , a_1+a_2+a_3)

    Now, letís keep going. Define g=gcd(a_1,a_2,a_3) and let u_1=a_1/g , u_2=a_2/g , u_3=a_3/g .

    Then u_1u_2u_3g|gcd((u_2u_3+u_1u_3+u_1u_2 )g^2, (u_1+u_2+u_3)g)

    Or, alternatively:

    u_1|u_2u_3(gk)+(u_2+u_3)
    u_2|u_1u_3(gk)+(u_1+u_3)
    u_3|u_2u_1(gk)+(u_2+u_1)

    In order to satisfy this for all k, u_1u_2u_3|g and u_1u_2u_3|u_1+u_2+u_3 . This calls for three relatively prime integers whose product divides their sum. I believe 1,2,3 is the only such arrangement, leading us into a dead end.

    However, after looking at some raw data, it looks like gcd(p_1-1,p_2-1,p_3-1) and subsequent p_1/g, p_2/g, p_3/g are all over the map. This leads me to pose another conjecture:

    Weak: For any three relatively prime u_1<u_2<u_3, there exists a k such that (u_1k+1) (u_2k+1) (u_3k+1) is a three-factor Carmichael Number.

    Strong: For any three relatively prime u_1<u_2<u_3, there exists an a,b such that (u_1k+1) (u_2k+1) (u_3k+1) is a three-factor Carmichael Number iff k \equiv b (\bmod a) and u_ik+1 is prime for all i.

    Examples:
    For u=(1,2,3), a=6, b=0 generates Carmichaels 1729, 294409, 56052361, 118901521 Ö
    For u=(1,3,5), a=30, b=12 generates Carmichaels 29341, 1152271, 64377911, 775368901 Ö
    For u=(1,4,7), a=28, b=4 generates Carmichaels 2465, 6189121, 19384289, 34153717249 Ö
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  2. #17
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    Quote Originally Posted by Media_Man View Post
    Iíve got you to lcm(a_1,a_2,a_3)|gcd(a_2a_3+a_1a_3+a_1a_2 , a_1+a_2+a_3)

    Now, letís keep going. Define g=gcd(a_1,a_2,a_3) and let u_1=a_1/g , u_2=a_2/g , u_3=a_3/g .

    Then u_1u_2u_3g|gcd((u_2u_3+u_1u_3+u_1u_2 )g^2, (u_1+u_2+u_3)g)
    I'll have to take some time to think about your other conjectures, however you should know that we do not in general have the equality lcm(a_1,a_2,a_3)=u_1u_2u_3g, where g=gcd(a_1,a_2,a_3). Notice that having the property gcd(u_1,u_2,u_3)=1 does not necessarily imply that gcd(u_i,u_j)=1 for each distinct i,j.
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  3. #18
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    Quote Originally Posted by Media_Man View Post

    However, after looking at some raw data, it looks like gcd(p_1-1,p_2-1,p_3-1) and subsequent p_1/g, p_2/g, p_3/g are all over the map.
    Suppose n=p_1p_2p_3 is a Carmichael and let D =gcd(p_1-1,p_2-1,p_3-1). Try finding out what the number of (approximately) distinct values the ratio n/D can take for all Carmichaels n bounded above by some integer N (ex N=10,000, N=100,000 etc).

    What are the smallest and/or average values of n/D for various N?
    Last edited by jamix; April 21st 2009 at 10:56 PM.
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