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Re: Primality with respect to multiplication modulo n.

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  • djbroadhurst
    ... No problem, Dimeter. ... Thanks. Where you write ... which avoids having to use your detailed checks, value by value. ... and then we have only to write
    Message 1 of 7 , Feb 4, 2011
      --- In primenumbers@yahoogroups.com,
      "Dimiter Skordev" <skordev@...> wrote:

      > I decided to submit the statement as a problem to the American
      > Mathematical Monthly. I actually submitted there a little more
      > precise form of that statement, but yesterday I withdraw my
      > submission, since I observed that already on January 11
      > David Broadhurst has posted a proof of the statement from my
      > message (I am sorry for observing David's message so late!).

      No problem, Dimeter.

      > I am now free to post my proof

      Thanks. Where you write

      > Suppose n is some of the numbers 2, 3, 4, 6. It easy to check
      > that, whenever x and y are natural numbers less than n and
      > different from 1, the number x*y is not congruent to n-1 modulo n.

      I had written:

      > For n = 3, 4, or 6, the group of units (Z/Zn)* has
      > precisely 2 elements

      which avoids having to use your detailed checks, value by value.

      That gave the key to my proof of the main theorem:

      > Theorem 3: Lemmas 1 and 2 exhaust the situations in which a
      > prime p < n is a "prime mod n", as defined by Skordev.

      > Proof: First, suppose that n is coprime to p. If n = 5 or
      > n > 6, then the group of units (Z/Zn)* has at least 4 elements

      and then we have only to write tidy words, as you did.

      Best regards

      David
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