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Re: sufficent proof for primes p:=x^2+x+1

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  • Bernhard Helmes
    A nice Sunday evening, ... generate. Perhaps you need some data more. I checked the primes p:=x^2+x+1 where x is a power of 2. x:=1; while TRUE do
    Message 1 of 6 , Mar 12 12:14 PM
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      A nice Sunday evening,

      > Thank you Bernhard for that data, way beyond what I could
      generate.

      Perhaps you need some data more. I checked the primes p:=x^2+x+1
      where x is a power of 2.

      x:=1;
      while TRUE do
      p:=(x+1)*x+1;
      if p mod 3 = 1 then
      erg:=powermod (2, (p-1)/3, p);
      res:=powermod (erg, 3, p);
      if res=1 and (erg=x or erg=x^2) and isprime (p)=FALSE then
      print (ifactor (x), p, ifactor (x+1));
      end_if;
      end_if;
      x:=x*2;
      end_while;


      27 4
      2 , 18014398643699713, 3 19 87211

      81
      2 , 5846006549323611672814741748716771307882079584257,

      5
      3 19 163 87211 135433 272010961

      171
      2 ,

      895897896871121684222976912227377711248658198893860113275530965754434
      39658\
      67181184471134228436276477953


      3 2
      , 3 19 571 174763 160465489 19177458387940268116349766612211

      243
      2 ,

      199791907220223502808422222706762643567910281130558153654986045416023
      79129\
      859977620592666523272986616017227171838989504031362210844729986994352
      9473


      6
      , 3 19 163 1459 87211 135433 139483 272010961
      10429407431911334611


      all of the examples are where x+1 is a power of 3 with > 3

      > And there is at least one interesting subplot: why is it that when
      > p=3 the composites are generated when x+1 is a power of two (with
      > just one known exception),

      I found 3 other exception 2^81, 2^171, 2^243 besides.
      The numbers get a little bigger. :-)

      Nice Greetings from the primes
      Bernhard
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