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• ... {tst(n,x,a)=2
Message 1 of 8 , Dec 3 2:27 PM
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"paulunderwooduk" <paulunderwood@...> wrote:

> I wriggle with this test:
> kronecker(x,n)==-1

{tst(n,x,a)=2<x&&x<(n+1)/2&&1<a&&a<(n+1)/2&&
kronecker(x^2-4,n)==-1&&gcd(a^3-a,n)==1&&gcd(a,x)==1&&
Mod(Mod(1,n)*(L+a),L^2-x*L+1)^(n+1)==(a^2+1+a*x)&&
Mod(Mod(1,n)*(L-a),L^2-x*L+1)^(n+1)==(a^2+1-a*x)&&!isprime(n);}

{if(tst(429749641620836200211,69208918734832269040,
200208504884456069347),print("fooled"));}

fooled

David
• ... Well done. Thanks, Paul
Message 2 of 8 , Dec 3 3:11 PM
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>
>
>
> "paulunderwooduk" <paulunderwood@> wrote:
>
> > I wriggle with this test:
> > kronecker(x,n)==-1
>
> {tst(n,x,a)=2<x&&x<(n+1)/2&&1<a&&a<(n+1)/2&&
> kronecker(x^2-4,n)==-1&&gcd(a^3-a,n)==1&&gcd(a,x)==1&&
> Mod(Mod(1,n)*(L+a),L^2-x*L+1)^(n+1)==(a^2+1+a*x)&&
> Mod(Mod(1,n)*(L-a),L^2-x*L+1)^(n+1)==(a^2+1-a*x)&&!isprime(n);}
>
> {if(tst(429749641620836200211,69208918734832269040,
> 200208504884456069347),print("fooled"));}
>
> fooled
>

Well done. Thanks,

Paul
• ... After I worked out to how to forge one counterexample, the gremlins were able to find more than 21,000: {tst(n,x,a)=2
Message 3 of 8 , Dec 4 7:49 AM
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"paulunderwooduk" <paulunderwood@...> wrote:

> > {if(tst(429749641620836200211,69208918734832269040,
> > 200208504884456069347),print("fooled"));}
> > fooled
> Well done. Thanks

After I worked out to how to forge one counterexample,
the gremlins were able to find more than 21,000:

{tst(n,x,a)=2<x&&x<(n+1)/2&&1<a&&a<(n+1)/2&&
kronecker(x^2-4,n)==-1&&gcd(a^3-a,n)==1&&gcd(a,x)==1&&
kronecker(x,n)==-1&& \\ added wriggle from Paul
Mod(2,n)^(n-1)==1&&n%12==11&& \\ by construction
Mod(Mod(1,n)*(L+a),L^2-x*L+1)^(n+1)==(a^2+1+a*x)&&
Mod(Mod(1,n)*(L-a),L^2-x*L+1)^(n+1)==(a^2+1-a*x)&&!isprime(n);}

print(sum(k=1,#F,tst(F[k][1],F[k][2],F[k][3]))" counterexamples");}

21728 counterexamples

It will be observed that, in each case, n = 11 mod 12 is a
base-2 Fermat pseudoprime. That was for convenience and
is not an intrinsic limitation.

David
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