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## Re: 7-selfridge puzzle

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• ... 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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