## Re: generalised 6-selfridge double fermat+lucas

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• ... n = q*r with (r-1)/(q^2-1)=1/12 and each kronecker getting its minus sign from the smaller prime factor q. I do recommend such semiprimes for your study
Message 1 of 5 , Sep 14, 2010
> Back to the drawing board:
> n=41159;
> x=3547;
> y=3225;

n = q*r with (r-1)/(q^2-1)=1/12
and each kronecker getting its minus sign from
the smaller prime factor q.

I do recommend such semiprimes for your study :-)

David
• ... I found one that was not a semiprime, but the ratio rule still applies: n=41159=79*521;ratio=12 n=45629=103*443;ratio=24 n=64079=139*461;ratio=42
Message 2 of 5 , Sep 15, 2010
>
> > Back to the drawing board:
> > n=41159;
> > x=3547;
> > y=3225;
>
> n = q*r with (r-1)/(q^2-1)=1/12
> and each kronecker getting its minus sign from
> the smaller prime factor q.
>
> I do recommend such semiprimes for your study :-)
>

I found one that was not a semiprime, but the ratio rule still applies:
n=41159=79*521;ratio=12
n=45629=103*443;ratio=24
n=64079=139*461;ratio=42
n=96049=139*691;ratio=28
n=197209=199*991;ratio=40
n=228241=181*(13*97);ratio=26
n=287051=151*1901;ratio=12
n=330929=149*2221;ratio=10
n=638189=619*1031;ratio=372
n=853469=239*3571;ratio=16
n=875071=241*3631;ratio=16

Note: the above "n" are all +-1 (mod 5)

Paul
• ... Here the kronecker is (-1)^3 = -1 and we are into Arnault territory: {pu3(n,x)=gcd(x^3-x,n)==1&&kronecker(x^2-4,n)==-1&&
Message 3 of 5 , Sep 15, 2010
"paulunderwooduk" <paulunderwood@...> wrote:

> n=228241=181*(13*97)

Here the kronecker is (-1)^3 = -1 and we are into Arnault territory:

{pu3(n,x)=gcd(x^3-x,n)==1&&kronecker(x^2-4,n)==-1&&
Mod(x,n)^(n-1)==1&&Mod((x+s)/Mod(2,n),s^2+4-x^2)^(n+1)==1;}

{pu6(n,x,y)=gcd(x^2-y^2,n)==1&&pu3(n,x)&&pu3(n,y);}

{n=13*97*181; x=218; y=824; if(pu6(n,x,y),print("refuted"));}

refuted

Note that there are no "matrices" above: a double mod is enough.
Congratulations again, Paul, on testing your ideas to destruction.

Best regards

David
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