- 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 - --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
>

Back to the drawing board:

> 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,

>

n=41159;

x=3547;

y=3225;

:-(

Paul > Back to the drawing board:

n = q*r with (r-1)/(q^2-1)=1/12

> n=41159;

> x=3547;

> y=3225;

and each kronecker getting its minus sign from

the smaller prime factor q.

I do recommend such semiprimes for your study :-)

David- --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
>

I found one that was not a semiprime, but the ratio rule still applies:

> > 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 :-)

>

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 - --- In primenumbers@yahoogroups.com,

"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