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sufficent proof for primes of the kind p:=x^2+x+1

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  • Bernhard Helmes
    A beautifull evening, I try to give you a sufficent proof for primes p:=x^2+x+1 p:=x^2+x+1=x*(x+1)+1 with p = 3 mod 4 = 2 appears only one time as divisor of
    Message 1 of 3 , Nov 10, 2006
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      A beautifull evening,

      I try to give you a sufficent proof for primes p:=x^2+x+1

      p:=x^2+x+1=x*(x+1)+1 with p = 3 mod 4 => 2 appears only one time as
      divisor of p-1
      p mod 3 = 1 or in other words 3 | p-1
      p mod 9 > 1 => 3 appears only one time as divisor of p-1

      2 ^ ((p-1)/2) = p-1 mod p, must be verified

      then p is prime

      proof:

      x and x^2 are 3.roots of 1 because of the construction of the prime
      p

      t1 | x and t2 | x+1 t1, t2 are odd t1=/=t2 t1>3 t2>3

      2 ^ ((p-1)/t1)=1 and 2^ ((p-1)/t2) = 1
      then 2 ^ {[(p-1)/t1 - (p-1)/t2)] / t1*t2} = 1
      <=> 2 ^ [(t2-t1) *(p-1) / (t1*t2)] = 1 with t2-t1 = 2*r
      <=> 2 ^ [(2*r) *(p-1) / (t1*t2)] = 1
      A)<=> 4 ^ [ r *(p-1) / (t1*t2)] = 1

      2^ [(p-1)/2] = p-1 with squaring
      B) 4 ^[(p-1)/2] = 1

      A) and B) are contradicted because (p-1)/2 is odd and r*(p-1) is
      even

      => with the theorem of Pocklington follows that p is a prime.

      A factorisation of p-1 is not needed.

      Please let me know if the proof is o.k.

      Friendly greetings from the primes

      Bernhard Helmes

      www.devalco.de
    • leavemsg1
      ... Hi, Bernhard. I think that x=15 is a counter-example... 2^((240/2)) == 1 mod 241. Is it? Bill
      Message 2 of 3 , Nov 14, 2006
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        --- In primenumbers@yahoogroups.com, "Bernhard Helmes" <bhelmes@...>
        wrote:
        >
        > A beautifull evening,
        >
        > I try to give you a sufficent proof for primes p:=x^2+x+1
        >
        > p:=x^2+x+1=x*(x+1)+1 with p = 3 mod 4 => 2 appears only one time as
        > divisor of p-1
        > p mod 3 = 1 or in other words 3 | p-1
        > p mod 9 > 1 => 3 appears only one time as divisor of p-1
        >
        > 2 ^ ((p-1)/2) = p-1 mod p, must be verified
        >
        > then p is prime
        >

        Hi, Bernhard.
        I think that x=15 is a counter-example... 2^((240/2)) == 1 mod 241.
        Is it? Bill

        > proof:
        >
        > x and x^2 are 3.roots of 1 because of the construction of the prime
        > p
        >
        > t1 | x and t2 | x+1 t1, t2 are odd t1=/=t2 t1>3 t2>3
        >
        > 2 ^ ((p-1)/t1)=1 and 2^ ((p-1)/t2) = 1
        > then 2 ^ {[(p-1)/t1 - (p-1)/t2)] / t1*t2} = 1
        > <=> 2 ^ [(t2-t1) *(p-1) / (t1*t2)] = 1 with t2-t1 = 2*r
        > <=> 2 ^ [(2*r) *(p-1) / (t1*t2)] = 1
        > A)<=> 4 ^ [ r *(p-1) / (t1*t2)] = 1
        >
        > 2^ [(p-1)/2] = p-1 with squaring
        > B) 4 ^[(p-1)/2] = 1
        >
        > A) and B) are contradicted because (p-1)/2 is odd and r*(p-1) is
        > even
        >
        > => with the theorem of Pocklington follows that p is a prime.
        >
        > A factorisation of p-1 is not needed.
        >
        > Please let me know if the proof is o.k.
        >
        > Friendly greetings from the primes
        >
        > Bernhard Helmes
        >
        > www.devalco.de
        >
      • Dario Alpern
        ... as ... But p=241 is not congruent to 3 (mod 4). So it is not a counter- example. Best regards, Dario Alpern Buenos Aires - Argentina
        Message 3 of 3 , Dec 13, 2006
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          --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...>
          wrote:
          >
          > --- In primenumbers@yahoogroups.com, "Bernhard Helmes" <bhelmes@>
          > wrote:
          > >
          > > A beautifull evening,
          > >
          > > I try to give you a sufficent proof for primes p:=x^2+x+1
          > >
          > > p:=x^2+x+1=x*(x+1)+1 with p = 3 mod 4 => 2 appears only one time
          as
          > > divisor of p-1
          > > p mod 3 = 1 or in other words 3 | p-1
          > > p mod 9 > 1 => 3 appears only one time as divisor of p-1
          > >
          > > 2 ^ ((p-1)/2) = p-1 mod p, must be verified
          > >
          > > then p is prime
          > >
          >
          > Hi, Bernhard.
          > I think that x=15 is a counter-example... 2^((240/2)) == 1 mod 241.
          > Is it? Bill
          >

          But p=241 is not congruent to 3 (mod 4). So it is not a counter-
          example.

          Best regards,

          Dario Alpern
          Buenos Aires - Argentina
          http://www.alpertron.com.ar/ENGLISH.HTM
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