## generalised 6-selfridge double fermat+lucas

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• Hi, I will use capital letters to represent 2 by 2 matrices and lower case for integers. Consider: R^2-r*R+1==0 I call this trivial if r=0 (mod d) or r=+-1
Message 1 of 5 , Sep 14, 2010
Hi,

I will use capital letters to represent 2 by 2 matrices and lower case for integers.

Consider:

R^2-r*R+1==0

I call this "trivial" if r=0 (mod d) or r=+-1 (mod d) for some proper divisor "d" of a given "n", because the equation is cyclic.

Now on to the double equations:

M^2-x*M+1==0
N^2-y*N+1==0

I do not want x=+-y (mod d) because they will be identical for that divisor of "n".

The composite test for "n" is:

First find x and y:
gcd(x^3-x,n)==1
gcd(y^3-y,n)==1
gcd(x^2-y^2,n)==1
jacobi(x^2-4,n)==-1
jacobi(Y^2-4,n)==-1

Secondly, check
x^(n-1) == 1 (mod n)
y^(n-1) == 1 (mod n)
M^(n+1) == I (mod n)
N^(n+1) == I (mod n)

I have checked n<2*10^4 with gcd(30,n)==1,

Paul
• ... Back to the drawing board: n=41159; x=3547; y=3225; ... Paul
Message 2 of 5 , Sep 14, 2010
--- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
>
> Hi,
>
> I will use capital letters to represent 2 by 2 matrices and lower case for integers.
>
> Consider:
>
> R^2-r*R+1==0
>
> I call this "trivial" if r=0 (mod d) or r=+-1 (mod d) for some proper divisor "d" of a given "n", because the equation is cyclic.
>
> Now on to the double equations:
>
> M^2-x*M+1==0
> N^2-y*N+1==0
>
> I do not want x=+-y (mod d) because they will be identical for that divisor of "n".
>
> The composite test for "n" is:
>
> First find x and y:
> gcd(x^3-x,n)==1
> gcd(y^3-y,n)==1
> gcd(x^2-y^2,n)==1
> jacobi(x^2-4,n)==-1
> jacobi(Y^2-4,n)==-1
>
> Secondly, check
> x^(n-1) == 1 (mod n)
> y^(n-1) == 1 (mod n)
> M^(n+1) == I (mod n)
> N^(n+1) == I (mod n)
>
> I have checked n<2*10^4 with gcd(30,n)==1,
>

Back to the drawing board:
n=41159;
x=3547;
y=3225;

:-(

Paul
• ... 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 3 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 4 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 5 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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