20200917, 13:24  #34 
"Roman V. Makarchuk"
Aug 2020
Ukraine
2·17 Posts 
AKS, i'm aware about) Symmetry is much better. Currently, I'm going on to the proof that not exist even false (from modulo flaw) symmetry break in the form of [u,p1;p1,u]

20200917, 15:42  #35 
Aug 2006
2×2,969 Posts 
Has anyone coded this in PARI/GP so we can test various ranges?

20200917, 15:57  #36  
"Sam"
Nov 2016
100111110_{2} Posts 
Quote:


20200919, 08:04  #37 
"Roman V. Makarchuk"
Aug 2020
Ukraine
2·17 Posts 
(10:14) gp > { n=5; A=[1,1;1,u];P=[1,1;1,nu+2];A=P^nA^n; P=[0,1;1,0];A*=P; print((trace(A)))}
60*u^3 + 630*u^2  3170*u + 5950 (10:15) gp > { n=5; A=[1,1;1,u];P=[1,1;1,nu+2];A=P^nA^n; P=[0,1;1,0];A*=P; b=factor(trace(A));print(b)} [2*u  7, 1; 3*u^2  21*u + 85, 1] (10:15) gp > { n=5; A=[1,1;1,u];P=[1,1;1,nu+2];A=P^nA^n; P=[0,1;1,0];A*=P; b=factor(trace(A)/2);print(b)} [2*u  7, 1; 3*u^2  21*u + 85, 1] (10:16) gp > Greting! This Pari/GP bug, or i miss something in factor() parameters? If not, PG eat a factor, clearly from first row to last. Test with almost done, for some numbers symmetry is "broken" for some p, and some u, nu+2, in form [u,p1;p1,1] opposite Carmichael numbers [1,1;1;u] I show you later, with proof of course, that happens only for those that have this at u=0, u=2 values {forstep(p=1,100000000,2,if(isprime(p),, A=Mod([1,1;1,0],p)^p;B=[0,p1;p1,1]; T=lift(lift(A))%p; if(T==B,A=Mod([1,1;1,2],p)^p;B=[2,p1;p1,1];T=lift(lift(A))%p; if(T==B,print([p,lift(lift(A))];)))))} [5777, [2, 5776; 5776, 1]] [10877, [2, 10876; 10876, 1]] [75077, [2, 75076; 75076, 1]] [80189, [2, 80188; 80188, 1]] [100127, [2, 100126; 100126, 1]] [113573, [2, 113572; 113572, 1]] So for proof this (i post full PG code later) stupid test IS correst for all numbers, we must only proof that only Carmichael and this numbers is broke the symmetry in modulo computation This is stupid, as far as i have the proof. that only prime numbers posses the symmetry u <> nu+2. first thing first  what with Pari/GP factoring? Other math system work well, Last fiddled with by RMLabrador on 20200919 at 08:49 
20200919, 12:19  #38 
Feb 2017
Nowhere
3·7·181 Posts 
Here is one proof that b(u) == b(pu+2) (mod p) if p is prime. [I replace 2+pu by 2u since the two are identical (mod p).]
The characteristic polynomial x^2  (u+1)*x + u1 of the matrix A has discriminant Δ = (u1)^2 + 4, which is invariant under the substitution u < 2  u. Therefore, the quadratic character the (Δ/p) remains the same under the substitution, so for p prime the offdiagnal entries of A^{p} (mod p), which by formula are 1,1, or 0 (mod p) for u in Z/pZ^{x}, remain invariant also. Therefore, if b(u) is the offdiagonal polynomial in A^{p}, b(u) = b(2u) has at least p1 roots (mod p). Since b(u) has degree p1, the two polynomials are identical (mod p), so their difference is identically 0 (mod p). I haven't looked at the claim that b(u) is an irreducible polynomial when p is prime. I do know that if m divides n then b(m) divides b(n), so b(n) is not irreducible if n is composite. I don't have a convenient formula for the coefficients of b(u). I therefore also don't know whether (a) computing them can be done as conveniently as the binomial coefficients, or (b) whether anything like the method used in the proof of AKS can be applied to the polynomials b(u). EDIT: My proof as stated isn't quite right. Two polynomials which evaluate the same for as many argument values as their degree need not be identical. This is easily seen for degree 1: two lines can have 1 point in common without being the same line. The simplest correction is to add the hypothesis (satisfied by b(u) and b(2u) (mod p) if p is prime) that the two polynomials have the same leading coefficient. Then their difference is a polynomial in F_{p}[x] with more zeroes than its degree, which is, ipso facto, identically zero. Another correction would be to show that b(u) and b(2  u) are equal for all p values of u (mod p). Note that, if n is composite, the integers mod n do not form a field, so the above argument does not apply. Last fiddled with by Dr Sardonicus on 20200923 at 11:10 
20200922, 09:29  #39 
"Roman V. Makarchuk"
Aug 2020
Ukraine
34_{10} Posts 
Very well! we can easy compute the b(u) value, numerical value
and it seems to be ok. 1) lets try to proof, that only 2 types of pseoudoprimes break a symmetry in the modulo case and a) Carmichael numbers. they have [1,1;1,u]  [1,1;1,nu+2] "break" for some u and can not have a [u,p1;p1,1] case at all. b) pseudoprimes in form n^2+1, that have more than one form a^2+b^2, they have very interesting symmetry when factored as Gaussian prime they always have [u,p1;p1,1] at u=0,2 and do not have [1,1;1,u] form at all I.e. no one of "exception" have no posses the both types of symmetry, piece of the PG code {forstep(p=23,100000000,2, if(Mod(p,1000003)==0, "";print(p)); if(ispseudoprime(p), f1=0;f2=0;f3=0; A=Mod([1,1;1,0],p)^p;B=[0,p1;p1,1]; C=[1,1;1,0];D=[A[1,1],0;0,A[1,1]]; if(A==B,f2++);if(A==C,f1++);if(A==D,f3++); if(f1+f2+f3==0,next); A=Mod([1,1;1,p+2],p)^p;B=[2,p1;p1,1]; C=[1,1;1,2];D=[A[1,1],0;0,A[1,1]]; if(A==B,f2++);if(A==C,f1++);if(A==D,f3++); if(f1+f2+f3<2,next); A=Mod([1,1;1,2],p)^p;B=[2,p1;p1,1]; C=[1,1;1,2];D=[A[1,1],0;0,A[1,1]]; if(A==B,f2++);if(A==C,f1++);if(A==D,f3++); if(f1+f2+f3<3,next); A=Mod([1,1;1,p],p)^p;B=[0,p1;p1,1]; C=[1,1;1,0];D=[A[1,1],0;0,A[1,1]]; if(A==B,f2++);if(A==C,f1++);if(A==D,f3++); if(f1+f2+f3<4,next); k1=f1/4;k2=f2/4; if(f3>1,next); f1=0;f2=0;f3=0;a=1; while(a, u=2+random((p1)/2); A=Mod([1,1;1,u],p)^p;B=[u,p1;p1,1]; C=[1,1;1,u]; if(A==B,f2++);if(A==C,f1++); if(f1+f2+f3==0,break(1)); A=Mod([1,1;1,pu+2],p)^p;B=[pu+2,p1;p1,1]; C=[1,1;1,pu+2]; if(A==B,f2++);if(A==C,f1++); if(f1+f2+f3==0,break(1)); if((k1==0&&f1==2)(k2==0&&f2==2),break); \\a=a+1; \\for some prime numbers, a can grow to high values. \\if(a>10,print(p)); f1=0;f2=0; ); if(isprime(p),,print(p)); \\the p prime is here ) )} 
20200922, 16:26  #40  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
21706_{8} Posts 
not a "Universal primality test" so far, by a mile
You asked if there was a proof for "if p prime, <bla>". That proof exists.
Quote:
Quote:
Where is the proof, "if <bla>, then p is prime"? What others helped you check was that the opposite is not easily shown by contradiction/"exception". But that only goes to show that the set of words at least starts to shape into a conjecture. Where is the proof (or at least a path to it)? 

20200922, 18:09  #41  
Aug 2006
1732_{16} Posts 
Quote:
Code:
th2(p)=my(u='u,b=(Mod([1,1;1,u],p)^p)[1,2]); b==subst(b,u,pu+2) 

20200923, 08:34  #42 
"Roman V. Makarchuk"
Aug 2020
Ukraine
2×17 Posts 
Someone even quietly, without notice, changed the name of the topic)))))
He's an asshole, proven by themselves. 
20200923, 08:59  #43 
Romulan Interpreter
Jun 2011
Thailand
2^{2}×7×11×29 Posts 
Prove him an asshole, by coming with a proof of the fact that your test is more than a PRP test. Otherwise you are just a crank or (worse) a troll.
As Serge asked, prove us the part "if <bla> then p is prime". Last fiddled with by LaurV on 20200923 at 09:00 
20200923, 12:04  #44  
Feb 2017
Nowhere
3801_{10} Posts 
Quote:
FWIW the polynomials trace(A^{n}) are the analogs of the Lucas polynomials, which are defined for the polynomial x^2  u*x  1. Here we have L_{0} = 2, L_{1} = u+1, L_{k+2} = (u+1)*L_{k+1}  (u1)*L_{k}, k = 0, 1, 2, etc and the offdiagonal polynomials b(u) are the analogs of the Fibonacci polynomials: F_{0} = 0, F_{1} = 1, F_{k+2} = (u+1)*F_{k+1}  (u1)*F_{k}, k = 0, 1, 2, etc These polynomials satisfy the identity L_{n}^{2}  Δ*F_{n}^{2} = 4*(u1)^{n} where Δ = (u1)^2 + 4 is the discriminant of the characteristic polynomial of A. As to changing thread titles, it's a common occurrence on this forum. It's childish, but acting childishly does not make someone an a. 

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