--- 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