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Math Help - proof with carmichael numbers

  1. #1
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    proof with carmichael numbers

    Let k be a positive integer such that 6k + 1 = p_1,  12k + 1 = p_2, \mbox { and } 18k + 1 = p_3 are all prime numbers, and put  m = p_1p_2p_3
    Show that (p_i - 1) \mid (m - 1) \mbox{ for }i = 1,2,3
    Deduce that if (a, p_i) = 1 then  a^{m - 1} \equiv 1  (mod \mbox{ }p_i), i = 1,2,3.
    Conclude that if (a,m) = 1 \mbox{ then } a^{m-1} \equiv 1 \mbox{ (mod m)} that is, that m is a Carmichael number.

    I am stumped...
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  2. #2
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    Maybe you should have gone to Norin's office hours?
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    That's pretty funny. Now stop stalking each other, guys.
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  4. #4
    MHF Contributor chiph588@'s Avatar
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     m=1296k^3+396k^2+36k+1

    It's easy to see that  6k\mid m-1 ,  12k\mid m-1 ,  18k\mid m-1 .

     m-1=k_i(p_i-1)
     a^{m-1} = (a^{p_i-1})^{k_i} \equiv 1^{k_i} =1 \mod{p_i} by Fermat's Little Theorem.


    From above we now know  a^{m-1}-1 = n_ip_i .
    Therefore  (a^{m-1}-1)^3 = n_1n_2n_3m \Longrightarrow (a^{m-1}-1)^3 \equiv 0 \mod{m}.

     m\mid(a^{m-1}-1)^3 \Longrightarrow (a^{m-1}-1)^3 = km = kp_1p_2p_3 = (k')^3p_1^3p_2^3p_3^3 since we have an integer cubed.

    Therefore  m^3 \mid (a^{m-1}-1)^3 \Longrightarrow m\mid (a^{m-1}-1).
    i.e.  a^{m-1}-1 \equiv 0 \mod{m} \Longrightarrow a^{m-1}\equiv 1 \mod{m} .
    Last edited by chiph588@; January 19th 2010 at 01:14 PM.
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