A Universally derided "primality test".
Hi,
Information here. [url]http://romanvmprime.com/[/url] Regards, Roman V. Makarchuk 
[CODE][n,u]=[231, 65];Mod([1,1;1,u],n)^n==[u,1;1,1]
1 [/CODE] Theorem 1 also works for composites. Testing (n1)/2 times with different u is infeasible for large n. Theorem 2 is illdefined. What are a(u) and c(u). Can you give a numerical example how this works? 
Theorem 1 also works for composites.
Yes.There is a lot of exeption. For some values u. Carmichael numbers do it even for many of u, but there is not a problem. Not exist such exeption that have BOTH type of resudials for some different u, if so, this numbers is prime. Theorem 2 is illdefined. What are a(u) and c(u). Can you give a numerical example how this works?[/QUOTE] a(u), c(u), b(u)  polynomial of u, result of analytical powering of matrix. p=5 u^4+u^3+4u^2+5u+5 (1) 5u+2=7u, substitute u^429u^3+319u^21580u+2980 (2) (1)(2)=5(2u7)(3u^221u+85) (2u7)(3u^221u+85) = g(u) 5=p That true ONLY if p is prime. 
[QUOTE=RMLabrador;556735]Theorem 1 also works for composites.
Yes.There is a lot of exeption. For some values u. Carmichael numbers do it even for many of u, but there is not a problem. Not exist such exeption that have BOTH type of resudials for some different u, if so, this numbers is prime. [/quote] [CODE][n,u,w]=[1247, 601, 638];Mod([1,1;1,u],n)^(n)==[u,1;1,1]&&Mod([1,1;1,w],n)^(n)==[1,1;1,w] 1 [/CODE] [quote][quote] Theorem 2 is illdefined. What are a(u) and c(u). Can you give a numerical example how this works?[/QUOTE] a(u), c(u), b(u)  polynomial of u, result of analytical powering of matrix. p=5 u^4+u^3+4u^2+5u+5 (1) 5u+2=7u, substitute u^429u^3+319u^21580u+2980 (2) (1)(2)=5(2u7)(3u^221u+85) (2u7)(3u^221u+85) = g(u) 5=p That true ONLY if p is prime.[/QUOTE] I will have to think more about what you are writing. 
Thank You for advice! You are right! And I'm right too when talk about symmetry of resudials, ok 1247  601 and 638? try 1247601+2 and 1247638+2. For prime numbers, such symmetry exist; Sorry for take Your time, I'm just not search for ordinary exeptions, I'm find out such of them that break the symmetry.

The characteristic equations of the matrices are:
x^2  (1+u)*x + u1 == 0 and x^2  (1+w)*x + w1 == 0 The "residuals" depend on the jacobi symbol of the discriminants of these equations: Jacobi((1+u)^24(u1), n) jacobi((1+w)^24(w1), n) If the Jacobi symbol is 1 then the matrical test is a glorified Fermat PRP test. If the symbol is 1 then it is a lucasbased test. Recieved wisdom says that any Fermat+Lucas test has counterexamples for freely varying parameters (unless maybe the choice of u and w are more restricted  which is what I am currently studying). 
I'm refined condition of my Theorem 2)) If we write this in modulo form, we got Carmichael numbers as exception for some u, in strict form there in no exeption. Modulo form have a flaw...
I guess, You see my point? First, about symmetry and modulo flaw? And please check this too [url]http://romanvmprime.com/?page_id=145[/url] 
See the real test on the page, Theorem 4 below on page. United Magic of Symmetry and Complex Numbers)
romanvmprime 
[QUOTE=RMLabrador;556799]See the real test on the page, Theorem 4 below on page. United Magic of Symmetry and Complex Numbers)
[URL="http://romanvmprime.com"]romanvmprime.com[/URL][/QUOTE] [CODE]n=3225601;A=Mod(Mod([1+x,1;1,0],n),x^2+1);B=Mod(Mod([1,1;1,x],n),x^2+1);R=lift(lift(A^nB^n));print(R) [x, 0; 0, 3225600*x] [/CODE] Your "theorem" is wrong. 
I did notice one unusual situation:
The characteristic polynomial is x^2  (u + 1)*x + (u  1); the discriminant is u^2  2*u + 5. As long as this is nonzero (mod p), there are two eigenvalues, and it is at least not insane to think about diagonalizing the matrix, which might be probative. If p == 1 (mod 4) however, i[sup]2[/sup] == 1 (mod p) has two solutions. Taking u = 1 + 2*i makes the discriminant of the characteristic polynomial 0, so it has a repeated factor: x^2  (2 + 2*i)*x + 2*i = (x  (1 + i))^2. In this case, there is only one eigenvalue, 1 + i, of multiplicity 2. The matrix A  (1+i)I (I = 2x2 identity matrix) is (using PariGP syntax) N = [i,1;1,i] which is nilpotent of index 2. Either column [i;1] or [1;i] of N serves as an eigenvector of A. Thus A = (1 + i)*I + N. Also, I and N commute, so A^p == (1+i)^p*I + N^p (mod p), and N^p is the 2x2 zero matrix. Thus, A^p == (1+i)^p*I == (1 + i)*I (mod p) in this case. Examples: p = 5, i = 2, u = 1 + 2*2 = 0, 1 + i = 3; or i = 3, u = 1 + 2*3 = 2, 1 + i = 4 p = 13, i = 5, u = 1 = 2*5 = 11, 1 + i = 6; or i = 8, u = 1 + 2*8 = 4, 1 + i = 9 
Very nice! I love this place.
I talk that using modulo computation lead to false "exception" and once again do it by themselves)) We must do something with to avoid error like this P*B == 0 mod P situation where part P must be an outcome of the test and give 0. Most of false "exception" arise from this, when B is some polynomial, and B give zero to us for some test condition for nonprime P instead of expected P Can not emphasize this more clearly in my poor English Do You understand me? We can build such test. Symmetry is one of the game changer. The [B]second[/B] way exist too. We can not deal with modulo and we can not (at lest now) do the test without modulo)) Once again, go to the bottom of my page 
All times are UTC. The time now is 08:24. 
Powered by vBulletin® Version 3.8.11
Copyright ©2000  2020, Jelsoft Enterprises Ltd.