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Re: [PrimeNumbers] Modulo 1 question

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  • Kevin Acres
    Hi Jack, Your counter example threw me there for a minute. I need to tighten the constraints a little. I should specify that I am looking for the first value
    Message 1 of 9 , Oct 6 6:14 PM
      Hi Jack,

      Your counter example threw me there for a minute. I need to tighten
      the constraints a little.

      I should specify that I am looking for the first value of x other than
      1. So in your example the first x that fits the requirements is 4.

      Regards,

      Kevin.

      >
      > > What I am looking for is a proof that for
      >
      > > x^2 = 1 mod p
      >
      > > where p is an odd composite and x is not 1 that x is always even.
      >
      > Well, first, the choice of 'p' to represent an odd composite
      > is, well, a bit *odd*... :)
      >
      > Other than that, how about a counter-example:
      >
      > x = 11
      > p = 15
      >
      >
      >
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    • Jack Brennen
      ... In that case, the smallest counterexample is for modulus 45. The smallest x 1 such that x^2 == 1 (mod 45) is x=19.
      Message 2 of 9 , Oct 6 6:39 PM
        > I should specify that I am looking for the first value of x other than
        > 1. So in your example the first x that fits the requirements is 4.

        In that case, the smallest counterexample is for modulus 45.

        The smallest x>1 such that x^2 == 1 (mod 45) is x=19.
      • Kevin Acres
        Hi Jack, Thanks for that. I must admit that I wasn t looking at low numbers, but larger bi-primes. I ll have to revisit my stratgey to handle composites which
        Message 3 of 9 , Oct 6 6:53 PM
          Hi Jack,

          Thanks for that. I must admit that I wasn't looking at low numbers,
          but larger bi-primes.

          I'll have to revisit my stratgey to handle composites which are the
          product of more than two (not necessarily unique) primes.

          Regards,

          Kevin.

          >
          > > I should specify that I am looking for the first value of x other
          than
          > > 1. So in your example the first x that fits the requirements is
          4.
          >
          > In that case, the smallest counterexample is for modulus 45.
          >
          > The smallest x>1 such that x^2 == 1 (mod 45) is x=19.
          >
          >
          >
          >
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          >
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          >
          >
          > Yahoo! Groups Links
          >
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        • Kevin Acres
          Hi Jens, I think that my point is that every odd composite has an even x where x^2 = 1 mod p. In the case of p = 55 then x is 34. So what I am getting at,
          Message 4 of 9 , Oct 7 3:35 AM
            Hi Jens,

            I think that my point is that every odd composite has an even 'x' where
            x^2 = 1 mod p. In the case of p = 55 then x is 34.

            So what I am getting at, is that every odd composite will have a case where
            x is even (and less than p-1), but may not have a case where x is odd.

            The counter example that I need to disprove this is where an odd composite
            (p) does not conform to the statement x^2 = 1 mod p where x is less than p-1.

            Regards,

            Kevin.






            At 07:53 PM 7/10/2004, you wrote:
            >----- Original Message -----
            >From: "Kevin Acres" <research@...>
            >To: "Jack Brennen" <jack@...>; "Primenumbers"
            ><primenumbers@yahoogroups.com>
            >Sent: Thursday, October 07, 2004 3:53 AM
            >Subject: Re: [PrimeNumbers] Modulo 1 question
            >
            >
            > > Hi Jack,
            > >
            > > Thanks for that. I must admit that I wasn't looking at low numbers,
            > > but larger bi-primes.
            > >
            > > I'll have to revisit my stratgey to handle composites which are the
            > > product of more than two (not necessarily unique) primes.
            > >
            > > Regards,
            > >
            > > Kevin.
            >
            >Smallest bi-prime counter example:
            >21^2 == 1 (mod 55)
            >
            >If you can program in any language, try experimenting before posting.
            >
            >--
            >Jens Kruse Andersen (offline)
          • jbrennen
            ... Okay, let s look at this a couple of things at a time... First, I ll change the modulus from p to m because p is usually reserved for primes: x^2 ==
            Message 5 of 9 , Oct 7 6:51 AM
              --- Kevin Acres wrote:
              > So what I am getting at, is that every odd composite will have
              > a case where x is even (and less than p-1), but may not have a
              > case where x is odd.

              Okay, let's look at this a couple of things at a time...

              First, I'll change the modulus from 'p' to 'm' because 'p' is
              usually reserved for primes:

              x^2 == 1 (modulo m) {m is an odd composite}

              Note that if there exists any solution x, with 1 < x < m-1, then
              there exist two such solutions. This is easy to see, because
              if we have:

              a^2 == 1 (modulo m)

              then x=(m-a) is also a solution:

              (m-a)^2 == m^2-2*a*m+a^2 == m*(m-2*a)+a^2

              m*(m-2*a)+a^2 == a^2 (modulo m)

              m*(m-2*a)+a^2 == 1 (modulo m)

              Now, since m is odd, the pair (a,m-a) contains both an odd
              number and an even number.

              So, we can restate: If there exists any solution x, with
              1 < x < m-1, then there exist both odd and even solutions
              which satisfy that constraint.

              Assume that m is a product of two odd numbers j and k which
              are greater than one and which are coprime.

              To solve x^2 == 1 (modulo m), we can solve:

              u^2 == 1 (modulo j)
              v^2 == 1 (modulo k)

              and then by the Chinese Remainder Theorem, each unique pair
              of solutions (u,v) leads to a unique solution x. There are
              at least 4 solutions to this system of modular equivalences:

              (u,v) == (1,1)
              (u,v) == (1,k-1)
              (u,v) == (j-1,1)
              (u,v) == (j-1,k-1)

              The first of these four leads to the solution x == 1, and the
              last leads to the solution x == m-1. The other two also lead
              to solutions to the original equation, and the Chinese Remainder
              Theorem states that those solutions will fall in the range:
              0 <= x <= m-1, and that they will be distinct from the other
              solutions. A little thought shows that these two solutions are
              of the form x==a and x==(m-a) as discussed earlier, and that
              the existence of both odd and even solutions is directly shown.

              The only remaining problem is in the case where m cannot be
              expressed as a product of coprime odd numbers > 1. If m is
              an odd composite, the only way this can happen is if m is a
              perfect power of a prime.

              I won't go into the proof by induction that x^2 == 1 (mod p^n)
              has only the two trivial solutions x == 1 and x == p^n-1, but
              it's not too hard to find.

              In any case, the final conclusion is that:

              x^2 == 1 (mod m) {m an odd number}

              has both odd and even non-trivial solutions (1 < x < m-1)
              if and only if the modulus m is divisible by at least two
              distinct primes.

              Hope that helps.
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