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Thread: proof with carmichael numbers

  1. #1
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    proof with carmichael numbers

    Let k be a positive integer such that $\displaystyle 6k + 1 = p_1, 12k + 1 = p_2, \mbox { and } 18k + 1 = p_3$ are all prime numbers, and put $\displaystyle m = p_1p_2p_3$
    Show that $\displaystyle (p_i - 1) \mid (m - 1) \mbox{ for }i = 1,2,3$
    Deduce that if $\displaystyle (a, p_i) = 1$ then $\displaystyle a^{m - 1} \equiv 1 (mod \mbox{ }p_i), i = 1,2,3$.
    Conclude that if $\displaystyle (a,m) = 1 \mbox{ then } a^{m-1} \equiv 1 \mbox{ (mod m)}$ 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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    $\displaystyle m=1296k^3+396k^2+36k+1 $

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

    $\displaystyle m-1=k_i(p_i-1) $
    $\displaystyle 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 $\displaystyle a^{m-1}-1 = n_ip_i $.
    Therefore $\displaystyle (a^{m-1}-1)^3 = n_1n_2n_3m \Longrightarrow (a^{m-1}-1)^3 \equiv 0 \mod{m}$.

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