20191224, 06:17  #45  
Sep 2002
Database er0rr
3^{2}·7·53 Posts 
Quote:
Solutions: Code:
? [n,a]=[517567487401, 459234922019];print([n,a,ispseudoprime(n),znlog(a,Mod(2,n)),gcd(a^21,n)==1,gcd(a^22,n)==1,kronecker(a^24,n)==1,Mod(Mod(1,n)*2*x,x^2a*x+1)^((n+1)/2)==2*kronecker(2*(a+2),n),Mod(a^24,n)^((n1)/2)==1]) [517567487401, 459234922019, 0, 246436369, 1, 1, 1, 1, 0] ? [n,a]=[517567487401, 470338872881];print([n,a,ispseudoprime(n),znlog(a,Mod(2,n)),gcd(a^21,n)==1,gcd(a^22,n)==1,kronecker(a^24,n)==1,Mod(Mod(1,n)*2*x,x^2a*x+1)^((n+1)/2)==2*kronecker(2*(a+2),n),Mod(a^24,n)^((n1)/2)==1]) [517567487401, 470338872881, 0, 287348929, 1, 1, 1, 1, 0] Last fiddled with by paulunderwood on 20191224 at 06:28 

20191230, 01:00  #46 
Sep 2002
Database er0rr
110100001011_{2} Posts 
I have uploaded what I hope is the very final draft of my paper. It dots some i's and crosses some t's, inserts missing words, clarifies some points and there is a slight change in when "%" is done in the listed Pari/GP script. The code will be superceded by actual implimenters of the code in other languages such as C/C++/GMP/ASSEMBLY etc.who can do shifts and so forth.
What is really hoped by me that it constitutes a proof of primality, or someone can summon up a counterexample. 
20191230, 22:21  #47 
Sep 2002
Database er0rr
110100001011_{2} Posts 
The gift that keep on giving
Since \[2x=a+\sqrt{\Delta}\] and by the Frobenius automorphism \[\frac{2}{x}\equiv a\sqrt{\Delta} \pmod{n,x^2ax+1}\], by subtraction and dividing out 2 I get \[x^2\sqrt{\Delta}x1\equiv 0 \pmod{n,x^2ax+1}\]. If it is checked that \[\Delta^\frac{n1}{2}\equiv 1 \pmod{n}\] then all that remains  apart form the g.c.d.s  is to check that \[(2x)^{n+1}\equiv 4 \pmod{n, x^2+x1}\]. This can only happen for kronecker(5,n)==1. So a test for such n is:
(Note the last mod cannot take advantage of my 2 selfridge method  maybe for these there is another trick for "fibonacci" expressions) If a=3 (or a=7) then \Delta is 5 anyway, The above test works for n<2^64 without the test Mod(a^24,n)^((n1)/2)==1 Last fiddled with by paulunderwood on 20191231 at 11:14 
20191231, 11:10  #48  
Sep 2002
Database er0rr
3^{2}·7·53 Posts 
Quote:
Quote:
Quote:
Last fiddled with by paulunderwood on 20191231 at 11:26 

20191231, 16:45  #49  
Sep 2002
Database er0rr
3339_{10} Posts 
The revised algorithm did not compute for numbers like 17*257. So I had to revise it again. It now says:
Quote:
I hope there no more loopholes. 

20200102, 18:49  #50 
Sep 2002
Database er0rr
3^{2}×7×53 Posts 
Counterexamples galore!
It is trivial to find counterexamples. Here is one:
Code:
{ [n,b,r]=[53317638559, 47290088223,1]; a=lift(Mod(b,n)^r); if(!ispseudoprime(n)&& !issquare(n)&& gcd(a^22,n)==1&& gcd(a^3a,n)==1&& kronecker(a^24,n)==1&& Mod(a^24,n)^((n1)/2)==1&& Mod(a,n)^(n1)==1&& Mod(Mod(1,n)*2*x,x^2*x+1)^((n+1)/2)==2*kronecker(2*(a+2),n), print("counterexample!")) } counterexample! Last fiddled with by paulunderwood on 20200102 at 18:52 
20200102, 21:42  #51 
Sep 2002
Database er0rr
3^{2}×7×53 Posts 
I am trying to patch things up. With r = 1 there are plenty of counterexamples. So we want to avoid any a=b^r=b i.e. b^(r1)=1.
So how about gcd(r1,n^21)==1 as a condition? Edit: Au contraire, a=b^(6*r) seems good, but I will have to do some more experimentation. Last fiddled with by paulunderwood on 20200102 at 23:53 
20200104, 11:08  #52  
Sep 2002
Database er0rr
3^{2}·7·53 Posts 
:
Quote:
Let me put it this way: Without all the mathematical arguments and using my intuition and experimental results, primailty can be established in 6 selfridges by running "the algorithm" with a=b^r and with a=b^(3*r). Whether I can prove all this is another matter... Last fiddled with by paulunderwood on 20200104 at 13:31 

20200105, 00:59  #53  
Sep 2002
Database er0rr
3^{2}·7·53 Posts 
Quote:
Let a=3^r. This has a unique solution for "the algorithm" for composites precisely when r=(o+1)/2 where o is the multiplicative group order of 3 mod n  proof? Now for the clever part: Test suitable r and r+1. They can't both have a solution. i.e. pass the tests, If they did then 2*r1 = o and 2*(r+1)  1= o. Consequently, 2*r1 = 2*r+1. That is o = 2. Then 3^2=1 which is illegal and also n would be even. A contradiction. 

20200105, 03:22  #54 
"Sam"
Nov 2016
454_{8} Posts 
Perhaps upon revising your algorithm, maybe testing it on one of the PRPs I sent you.

20200105, 04:08  #55  
Sep 2002
Database er0rr
110100001011_{2} Posts 
Quote:
Code:
{ b=3;forstep(n=3,100000000000000000,2, if(!ispseudoprime(n)&&Mod(b,n)^(n1)==1, a=b;r=1;while(a!=1, if(kronecker(a^24,n)==1&&gcd(a^21,n)==1&& Mod(Mod(1,n)*b*x,x^2a*x+1)^((n+1)/2)==b*kronecker(b*(a+2),n), z=(znorder(Mod(3,n))+1)/2;print([b,n,a,r,z]); if(z!=r,quit));r++;a=(a*b)%n))) } It just quitted with: Code:
[3, 1479967201, 970736486, 101203, 124201/2] Last fiddled with by paulunderwood on 20200105 at 05:09 

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