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Math Help - Fermat's Theorem

  1. #1
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    Fermat's Theorem

    I was wondering if someone might be able to look over my proof's involving Fermat's Theorem.

    When n = 2p, where p is an odd prime, then a^(n-1)= a (mod n) for any integer a.
    So let n=2p
    We have a^(2p-1)
    a^(2p)*1/a
    a^2a^p*1/a
    a^2*1/aa^p
    a^(2-1)a^p
    aa^p
    Then from Fermat's Theorem, a^p=a mod p and thus a^(2p-1)=a(mod2p)



    For n = 195 = 3 * 5 * 13, we have a^(n-2) = a (mod n) for any integer a.
    Then we have a^193
    Since 193 is a prime, we have a^193=a mod 193 by Fermat's Theorem and thus a^193=a (mod195)
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  2. #2
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    Quote Originally Posted by mathematic View Post
    I was wondering if someone might be able to look over my proof's involving Fermat's Theorem.

    When n = 2p, where p is an odd prime, then a^(n-1)= a (mod n) for any integer a.
    So let n=2p
    We have a^(2p-1)
    a^(2p)*1/a


    This last step is legal only if a\neq 0\!\!\pmod n


    a^2a^p*1/a
    a^2*1/aa^p


    How?? Perhaps you meant a^2\cdot\frac{1}{a}\cdot a^p? Otherwise it's wrong.


    a^(2-1)a^p
    aa^p
    Then from Fermat's Theorem, a^p=a mod p and thus a^(2p-1)=a(mod2p)

    Don't get it: you had aa^p ,and applying flt, you get aa^p=a^2\!\!\pmod p . How

    does this translate into modulo 2p?

    Tonio




    For n = 195 = 3 * 5 * 13, we have a^(n-2) = a (mod n) for any integer a.
    Then we have a^193
    Since 193 is a prime, we have a^193=a mod 193 by Fermat's Theorem and thus a^193=a (mod195)
    .
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  3. #3
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    so if a^(2p)*1/a isn't valid, how might I work around that?


    Yeah I meant a^2*(1/a)*a^p
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