20220922, 03:01  #12  
Mar 2016
2^{2}×3×5×7 Posts 
Is it possible to calculate one belonging rot. matrix from a vector ?
Quote:
let: p=31, u1=2, v1=3; so that the norm (u1,v1)=u1²+v1²=13=12⁻¹ mod 31 and 13²=20⁻¹ mod 31 Is it possible from linear algebra to calculate one belonging rotation matrix from this vector ? The calculated target is: 13* (27,2)* (2)=(5) (2,27) (3)=(9) 13* (14,17)* (2)=(4) (17,14) (3)=(11) 13* (24,8)* (2)=(6) (8,24) (3)=(15) http://localhost/devalco/unit_circle/system_tangens.php (the red coloured boxes right, all calculations checked and it seems to be all right.) My first try: Let p=31 Let M1=13*M1*= 13* (a*, b*) (b*, a*) 1. with det (M1)=1=det (13² * det (M1*)) so that det (M1*)=20 mod 31, therefore a*²+b*²=20 2. with M1*(u1,v1)=(u2,v2) with norm (u1,v1)=(u2,v2)=u1²+v1²=u2²+v2²=13 mod p so that 13* (a*, b*) = (u2) (b*, a*) (v2) This is more a fragment and should point in one direction, and as it is too late for me, I hope that some one could finish the calculation. 

20220925, 00:50  #13 
Feb 2017
Nowhere
1100001010011_{2} Posts 
It appears you are trying to solve
R[a;b] = [c;d] where R = [x,y;y,x]. Writing as a system of linear equations, x*a + y*b = c b*x  a*y = d which may be rewritten [a,b;b,a][x;y]=[c;d] which is solvable if a^2 + b^2 ≠ 0. EDIT: Feeding the formula to PariGP produces the same values you got: Code:
? matsolve([Mod(2,31),Mod(3,31);Mod(3,31),Mod(2,31)],Mod(1/13,31)*[Mod(5,31);Mod(9,31)]) %1 = [Mod(27, 31)] [Mod(2, 31)] ? matsolve([Mod(2,31),Mod(3,31);Mod(3,31),Mod(2,31)],Mod(1/13,31)*[Mod(4,31);Mod(11,31)]) %2 = [Mod(14, 31)] [Mod(17, 31)] ? matsolve([Mod(2,31),Mod(3,31);Mod(3,31),Mod(2,31)],Mod(1/13,31)*[Mod(6,31);Mod(15,31)]) %3 = [Mod(24, 31)] [Mod(8, 31)] Last fiddled with by Dr Sardonicus on 20220925 at 16:35 
20221002, 21:22  #14 
Mar 2016
2^{2}·3·5·7 Posts 
A prediction of a rotation matrix for Mp
A peaceful and pleasant night for you,
If Mp is a Mersenne number with exponent p I can predict the following rotation matrix M=(Mp2^[(p1)/2];2^[(p1)/2) (2^[(p1)/2]; Mp2^[(p1)/2]) with det (M)=1 Example: Mp=31 p=5 M=(27, 4) (4, 27) with det (M)=16+16=1 mod 31. This rotation matrix is not a result of two vectors (x1,y1), (x2, y2) with the same norm x1²+y²=x2²+y2²=n mod Mp, which needed to be found in advance, but it is completely predictive. I did not understand, why this matrix occurs one time for non quadratic residues and one time for quadratic residues: (27,4)* (2)=(4) (4,27) (3)=(11) with 22+32 = 13 and jac (13, 31)=1 (27,4)* (5)=(8) (4,27) (7)=(14) 52+72 = 12 and jac (12, 31)=1 (27,27)* (8)=(1) (27,27) (15)=(3) 82+152 = 10 and jac (10, 31)=1 Is there a mathematical explication for this phenomenon ? More examples and as usual a web interface under http://devalco.de/unit_circle/system_tangens.php Enjoy the predictable rotation matrix, 
20221004, 14:12  #15 
Feb 2017
Nowhere
13×479 Posts 
If 2*x^2 == 1 (mod N) and M = x[1,1;1,1] we have det(M) = 2*x^2 == 1 (mod N) and M^2 = x^2[0,2;2,0] == [0,1;1,0] (mod N), so that M^8 == [1,0;0,1] (mod N).
One can take N = 2^n  1 with n > 1 odd, and x = 2^((n1)/2). However, suitable x will exist for any N composed entirely of primes congruent to 1 or 7 (mod 8). 
20221124, 22:52  #16 
Mar 2016
110100100_{2} Posts 
A prediction of the rotation matrices
There should be more peace in the world !
The pythagorean triples can be used to make a surjective mapping to the rotation matrices concerning p if x²+y²=z²  : z² (x/z)²+(y/z)²=1 which is the determinant of the rotation matrix a=x/z mod p; b=y/z mod p M= (a, b) (b, a) Question: if M*(n, m)=(u, v), then n²+m²=u²+v² mod p and if jacobi (n²+m², p)=1 resp. n²+m² a non quadratic residue concerning p and p = 3 mod 4 is there a chance that [(n+mi)/(u,vi)]^r=1 mod p [r=(p+1)/2] and r maximal and not a right divisor of (p+1)/2 ? Or in other words: Is it possible to say anything about the group order of the resulting vector which is the divisor of the two vectors connected by the rotation matrix ? A peaceful 1. advent Of course a wonderful php scripting image with more infos you will need: http://devalco.de/unit_circle/system_tangens.php 
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