Loading ...
Sorry, an error occurred while loading the content.

24619Re: puzzle for a counterexample

Expand Messages
  • bhelmes_1
    Nov 1, 2012
      Dear David,

      thank you for your efforts,

      i have the slight suspision that there may be counterexamples
      for n=3 mod 4 but not for n=1 mod 4

      even if i could not prove it.

      I know that you are very fast in replying.
      If you find one counterexample of the form n=1 mod 4
      i would be glad to know it.

      Sorry that i bother you with such a lot of questions,
      but i hope that there will be some success.
      Take your time.

      p = 1 mod 4

      > > the improved test
      > > 1. gcd (a, p)=1
      > > 2. Let jacobi (a, p) = -1
      > > 3. let jacobi (a^2-a, p)=-1
      > > 4. a^[(p-1)/2]=-1 mod p
      > > 5. if (a+sqrt (a))^p = a-sqrt(a) mod p
      > > 6. s+t*sqrt(a):=(a+sqrt (a))^[(p+1)/2] implies that
      > > gcd (s, p)=1 or 0 or p, and gcd (t, p)=1 or 0 or p.

      Nice Greetings from the primes
      a lot of beautifull flowers in the mathematical universe
    • Show all 66 messages in this topic