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Re: How to prove this congruence problem???

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  • Bill
    it should be for all a . thanks ... (x^3
    Message 1 of 4 , Jun 4, 2003
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      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!
    • 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 2 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 3 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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