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Re: 7-selfridge puzzle

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  • paulunderwooduk
    ... Here is a counterexample: n=5777 a=131 b=2720 x=2927 Paul
    Message 1 of 3 , Dec 30, 2010
      --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
      >
      >
      >
      > --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@> wrote:
      > >
      > > Hi,
      > >
      > > consider the two equations:
      > >
      > > xX=(X-A)*(X-B)
      > > xY=(Y+A)*(Y+B)
      > >
      > > which can be rewitten as:
      > >
      > > X^2-(x+A+B)*X+A*B=0
      > > Y^2-(x-A-B)*Y+A*B=0
      > >
      > > with discriminants:
      > >
      > > D=(x+A+B)^2-4*A*B
      > > E=(X-A-B)^2-4*A*B
      > >
      > > If:
      > >
      > > kronecker(D,n)==-1
      > > kronecker(E,n)==-1
      > > gcd(x,n)==1
      > >
      > > and:
      > >
      > > (x+A+B)^(n-1)==1 (mod n)
      > > (x-A-B)^(n-1)==1 (mod n)
      > > (A*B)^(n-1)==1 (mod n)
      > > [x+A+B,-A*B;1,0]^(n+1)==[A*B,0;0,A*B] (mod n)
      > > [x-A-B,-A*B;1,0]^(n+1)==[A*B,0;0,A*B] (mod n)
      > >
      > > I conjecture n is prime.
      > >
      > > Note:
      > >
      > > gcd(x+A+B.n)==1
      > > gcd(x-A-B,n)==1
      > > gcd(A*B,n)==1
      > >
      > > are needed.
      > >
      > > I have tested the case A=B=1 for n<1.8*10^7 and carmichael numbers <2^32.
      > >
      > > Puzzle: Can you find a counterexample?
      > >
      >
      > I am adding:
      >
      > gcd(A+B,n)==1 (mod n)
      >

      Here is a counterexample:
      n=5777
      a=131
      b=2720
      x=2927

      Paul
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