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Math Help - Elliptic Curve y^2 = x^3 +17; show N_p = p

  1. #16
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by vernongetzler View Post
    Hat’s off. Well done, as we know that “hard work always pays off”, after a long struggle with sincere effort it’s done.
    ========
    vernon
    Exercises
    Huh? The link confuses me. Is your post about the link, or about what's been done on this thread?
    Last edited by chiph588@; March 31st 2010 at 01:54 PM.
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  2. #17
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    double posted by mistake...
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  3. #18
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    Quote Originally Posted by chiph588@ View Post
    Let's assume  b_1,b_2\not|p and  g is a primitive root (  b_1\equiv0\mod{p} is trivial).
    Why are you assuming that  b_1,b_2\not|p ? Do you actually mean  p\not|b_1,b_2 ?

    Then  b_1\equiv g^n,\;b_2\equiv g^m \mod{p} , so  g^{3n}\equiv g^{3m}\mod{p}\implies 3n\equiv 3m\mod{\phi(p)} .
    Why  b_1\equiv g^n,\;b_2\equiv g^m \mod{p} ? What are n and m?

    Since  (3,p-1)=1 \implies n\equiv m\mod{\phi(p)} \implies m=n+a(p-1)

    So  b_2\equiv g^m=g^{n+a(p-1)} \equiv 1\cdot g^n\equiv b_1\mod{p} .
    On the last line, I think you're using Fermat's little theorem, but how do you know that p\not|g ?

    Thanks for clarifying this!
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  4. #19
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    This is my version of the definition of primitive root:
    a is a primitive root mod m iff the order of a mod m is equal to phi(m) (where the order of a mod m is the smallest positive integer h such that a^h is congruent to 1 mod m.)

    But from here I really don't see how we can get b1=g^n (mod p)...
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  5. #20
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by kingwinner View Post
    This is my version of the definition of primitive root:
    a is a primitive root mod m iff the order of a mod m is equal to phi(m) (where the order of a mod m is the smallest positive integer h such that a^h is congruent to 1 mod m.)

    But from here I really don't see how we can get b1=g^n (mod p)...
    A primitive root will 'generate' every element in the set  \{1,2,\cdots,p-1\} .

    i.e.  \{g,g^2,\cdots, g^{p-1}\} = \{1,2,\cdots,p-1\}
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