20200911, 12:35  #1 
"Roman V. Makarchuk"
Aug 2020
Ukraine
2·17 Posts 
A Universally derided "primality test".
Last fiddled with by Uncwilly on 20200922 at 17:50 
20200911, 14:47  #2 
Sep 2002
Database er0rr
2^{4}×3×73 Posts 
Code:
[n,u]=[231, 65];Mod([1,1;1,u],n)^n==[u,1;1,1] 1 Theorem 2 is illdefined. What are a(u) and c(u). Can you give a numerical example how this works? Last fiddled with by paulunderwood on 20200911 at 14:57 
20200911, 15:40  #3 
"Roman V. Makarchuk"
Aug 2020
Ukraine
100010_{2} Posts 
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. Last fiddled with by RMLabrador on 20200911 at 15:44 
20200911, 16:15  #4  
Sep 2002
Database er0rr
6660_{8} Posts 
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 Quote:
Last fiddled with by paulunderwood on 20200911 at 16:15 

20200911, 16:42  #5 
"Roman V. Makarchuk"
Aug 2020
Ukraine
100010_{2} Posts 
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.

20200911, 16:56  #6 
Sep 2002
Database er0rr
2^{4}·3·73 Posts 
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). Last fiddled with by paulunderwood on 20200911 at 16:57 
20200911, 19:51  #7 
"Roman V. Makarchuk"
Aug 2020
Ukraine
100010_{2} Posts 
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 http://romanvmprime.com/?page_id=145 Last fiddled with by RMLabrador on 20200911 at 20:25 
20200912, 12:47  #8 
"Roman V. Makarchuk"
Aug 2020
Ukraine
42_{8} Posts 
See the real test on the page, Theorem 4 below on page. United Magic of Symmetry and Complex Numbers)
romanvmprime Last fiddled with by Uncwilly on 20200912 at 17:02 Reason: You have already posted your URL twice. No need to spam. 
20200912, 15:01  #9  
Sep 2002
Database er0rr
2^{4}×3×73 Posts 
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] 

20200912, 20:05  #10 
Feb 2017
Nowhere
EEB_{16} Posts 
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^{2} == 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 Last fiddled with by Dr Sardonicus on 20200912 at 20:08 Reason: xingif ostpy 
20200913, 15:31  #11 
"Roman V. Makarchuk"
Aug 2020
Ukraine
2×17 Posts 
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 second 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 
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