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Re: [PrimeNumbers] Re: How to prove this congruence problem???

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  • Satoshi Tomabechi
    Consider p=7. Fermat s little theorem claims that a^6 = 1 mod 7. This implies that a^3 = 1 or -1 mod p for all a. Satoshi Tomabechi On Thu, 05 Jun 2003
    Message 1 of 4 , Jun 4, 2003
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      Consider p=7.
      Fermat's little theorem claims that a^6 = 1 mod 7.
      This implies that a^3 = 1 or -1 mod p for all a.

      Satoshi Tomabechi


      On Thu, 05 Jun 2003 03:20:55 -0000
      "Bill" <iampoliceman@...> wrote:

      > it should be "for all a". thanks
      >
      > --- In primenumbers@yahoogroups.com, "Bill" <iampoliceman@y...> wrote:
      > > Prove it's false.
      > >
      > > for all p (p is prime implies for a there exists an x such that
      > (x^3
      > > is congurence to a (mod p))
      > >
      > > thanks a lot!
      >
    • Mark Underwood
      In other words, Prove the following false: for any and all p and a, there always exists an x such that x^3 -a is divisible by p As Satoshi implied (although
      Message 2 of 4 , Jun 4, 2003
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        In other words, Prove the following false: for any and all p and a,
        there always exists an x such that x^3 -a is divisible by p


        As Satoshi implied (although mixing up the 'a' and the 'x' I think),
        by Fermats Little Theorem, if p = 7, 'a' must be of the form 7u, 7u+1
        or 7u-1. Clearly this contradicts the original condition that 'a'
        can be any number.

        Now on to disproving the Riemann hypothesis, hehe!

        Mark



        --- In primenumbers@yahoogroups.com, Satoshi Tomabechi
        <tomabeti@c...> wrote:
        > Consider p=7.
        > Fermat's little theorem claims that a^6 = 1 mod 7.
        > This implies that a^3 = 1 or -1 mod p for all a.
        >
        > Satoshi Tomabechi
        >
        >
        > On Thu, 05 Jun 2003 03:20:55 -0000
        > "Bill" <iampoliceman@y...> wrote:
        >
        > > it should be "for all a". thanks
        > >
        > > --- In primenumbers@yahoogroups.com, "Bill" <iampoliceman@y...>
        wrote:
        > > > Prove it's false.
        > > >
        > > > for all p (p is prime implies for a there exists an x such that
        > > (x^3
        > > > is congurence to a (mod p))
        > > >
        > > > thanks a lot!
        > >
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