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Re: sufficent test for prime mod 4 = 3 ?

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  • djbroadhurst
    ... It s very easy to fool your test, in that case: {tst(n,b)=n%4==3&&kronecker(b,n)==-1&& Mod(b,n)^((n-1)/2)==-1&& Mod(Mod(1,n)*(b+x),x^2+1)^n==b-x;}
    Message 1 of 4 , Oct 24, 2010
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      --- In primenumbers@yahoogroups.com,
      "bhelmes_1" <bhelmes@...> wrote:

      > there might be counterexamples if you do not take the smallest b

      It's very easy to fool your test, in that case:

      {tst(n,b)=n%4==3&&kronecker(b,n)==-1&&
      Mod(b,n)^((n-1)/2)==-1&&
      Mod(Mod(1,n)*(b+x),x^2+1)^n==b-x;}

      if(tst(337*3079,298117),print(fooled));

      fooled

      David
    • djbroadhurst
      ... Wrong again. You are still using 1+3=4 selfridges. Please see Theorem 3.5.8 of Crandall and Pomerance. David
      Message 2 of 4 , Oct 24, 2010
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        --- In primenumbers@yahoogroups.com,
        "bhelmes_1" <bhelmes@...> wrote:

        > The good news is that the test "only" need 3 Selfridges

        Wrong again. You are still using 1+3=4 selfridges.
        Please see Theorem 3.5.8 of Crandall and Pomerance.

        David
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