20211023, 21:49  #1 
Mar 2016
2^{2}·3·5·7 Posts 
2*2 Matrix with determinant 1
A peaceful night for you,
do I see it right that 2*2 matrix with determinant 1 build a subgroup in linear algebra and that calculation in the group is better for factoring than calculation in all 2*2 matrix ? Do I see right that the matrix (2 0) (0 1) has the determinant 2 and that an exponentation with 2*p of Mp mod Mp will result in a matrix with determinant 1 Is it rigth that there are 3 paulimatrixs https://en.wikipedia.org/wiki/Pauli_matrices with determinant 1 and that a parallel multiplication of a start matrix with determinant 1 and a pauli matrix will speed up a possible factorisation ? Would be nice to get a reply. P.S. I do not know why or how, but I am sure we will find the next Mp before next christmas. Last fiddled with by bhelmes on 20211023 at 21:51 
20211024, 07:45  #2 
Dec 2012
The Netherlands
714_{16} Posts 
Over any commutative ring R, the 2x2 matrices with determinant 1 form a subgroup \(SL_2(R)\)
of the unit group of the ring of all 2x2 matrices. The Pauli matrices have determinant 1, however. It is important to decide carefully which commutative ring R you are working over: the integers, the Gaussian integers, the complex numbers, etc. Some of these can be constructed using matrices: for example, matrices of the form \[ \left(\begin{array}{rr}x & y \\ y & x\end{array}\right)\] over the real numbers form a copy of the complex numbers. 
20211024, 23:21  #3 
Feb 2017
Nowhere
6237_{10} Posts 
Let's see. det(M^{k}) = (det(M))^{k}.
So if det(M) = 2, det(M^{2p}) = (2)^{2p} = 2^{2p} == 1 (mod 2^{p}  1). Check. I don't see any relation to factorization. 
20220214, 05:10  #4  
Mar 2016
110100100_{2} Posts 
Quote:
(n²+1; n) (n; 1) =(a; b) (c; d) The determinant is one. For n=0 it includes the neutral element concerning the multiplication, is closed under exponentation, means for M^n the elements of b and c are equal and Is closed under multiplication for the element a=1 mod 4. It has obviously an inverse matrix and the multiplication of M*N seems to be kommutativ. It is not clear for me how you calculate the group order of this subsubgroup if you limit all elements modulo f and limit the multiplication to the exponentation of one M Perhaps someone could correct me and add a little bit more knowledge. Thanks in advance. P.S. I managed to speed up my computer ; 7 days instead 8 days for an exponent at the wawefront. (137 Watt instead 120 Watt) (you could easily speed up the calculation in the bios (Asus mainboard) ) 

20220214, 15:04  #5  
Feb 2017
Nowhere
3^{4}·7·11 Posts 
Quote:
Since M is a real symmetric matrix, it is diagonalizable, with diagonal elements the roots of F(x) = 0. If f = p, a prime number, and gcd(p,D) = 1, then the multiplicative order of M (mod p) divides p  1 if (D/p) = +1 (quadratic character). Because F(x) is a quadratic with constant term 1, we can say that the multiplicative order divides p + 1 if (D/p) = 1. So we can say that if p does not divide n or n^2 + 4, the multiplicative order (mod p) divides p  (D/p). EDIT: Assuming p does not divide D, we can drop the square factor n^2 from the quadratic residue symbol, and use instead of (D/p). Last fiddled with by Dr Sardonicus on 20220215 at 15:14 Reason: formatting; and as indicated 

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