20221109, 14:58  #1 
Sep 2002
Database er0rr
4669_{10} Posts 
Amazing pattern with 12 (was 3)
Testing Mod(3,n)^((n1)/2) == kronecker(3,n) and Mod(Mod(x,n),x^23^r*x3)^(n+1)==3 for all r reveals some interesting patterns in the pseudoprimes thereof:
To me as a failing mathematician this is a great pattern to observe being churned out by a verification script. I really think I am close to something astounding in the way of a zoomy fast proofs of general primes. I am almost sure that by studying the properties of this system that something very valuable can be gained. The testing will continue overnight to see if the above patterns are broken. Last fiddled with by paulunderwood on 20221109 at 17:41 
20221110, 02:08  #2 
Sep 2002
Database er0rr
7·23·29 Posts 
The results are consistent so far up to 21451730441.
The two pseudoprime indices r1 and r2 are given by: r1=z/4+1 r2=3*z/4+1 resulting in r1+r2=2 Rearranging: 2*r11 = z/2+1 2*r21 = 3*z/2+1 Working over the equation x^23^r*x3 can be transformed to working over y^2(3^(2*r1)2)*y+1. Rearraging terms gives 3^(2*r1) = (y^2+1+2y)/y = (y+1)^2/y. Substituting one of our r's  it works for both  3^(z/2+1) = (y+1)^2/y and squaring both sides gives: 3^(z+2) = 3^2 = (y+1)^4/y^2. Multiplying by y^2 and rearranging terms gives (y+1)^4  (3*y)^2 = (y^2+2*y+1  3*y )*(y^2+2*y+1 + 3*y ) = (y^2y+1)*(y^2+5*y+1) = 0 The left hand of the product of polynomials is cyclotomic and we would be done with it. Working with the right hand side we have y^2+5*y+1 = y^2(3^(z/2+1)2)*y+1. That is 5 = 3^(z/2+1) + 2. I.e. 3^(z/2) == 1 which is a contradiction of z being the multiplicative order, since z is divisible by 4. But we are still along way from the main proof Last fiddled with by paulunderwood on 20221110 at 02:16 
20221110, 04:27  #3 
Sep 2002
Database er0rr
7·23·29 Posts 
And the pattern is broken with 28027505969 which has 4 r's, each with gcd(r1,n1)!=1

20221110, 21:09  #4 
Sep 2002
Database er0rr
11075_{8} Posts 
12
I tried other bases 2,(3),4,6 and 8. but boy oh boy 12 is absolutely great. The test over x^212^r*x12 has no pseudoprimes less than 300,000,000 and consequently no need to take a GCD. Again I let it run overnight.

20221110, 21:26  #5 
"Chereztynnoguzakidai"
Oct 2022
Ukraine, near Kyiv.
3·19 Posts 
one Computer say to me once  Do not believe Computers, they always [illegibly]!

20221110, 21:47  #6  
Sep 2002
Database er0rr
7·23·29 Posts 
Quote:
Here is the code I am running: Code:
{b=12;forstep(n=3,1000000000001,2, if(n%1000000000==1,print(n)); if(gcd(b,n)==1&&!ispseudoprime(n)&&Mod(b,n)^((n1)/2)==kronecker(b,n), z=znorder(Mod(b,n));for(r=1,z, if(Mod(Mod(x,n),x^2Mod(b,n)^r*xb)^(n+1)==b, print([b,n,n%4,kronecker(3,n),gcd(2*r1,n1)==z,z,gcd(2*r1,n1),r])))))} 

20221110, 21:58  #7 
"Chereztynnoguzakidai"
Oct 2022
Ukraine, near Kyiv.
39_{16} Posts 
You know, there is thing that we call a 'forest' or 'woods' there is many trees here, and if we going from one tree to another  there is so many patterns!!! we can think that is some rules in distance or some cycles (as Collatz for example) but is is just tree in the woods)))

20221110, 23:41  #8 
Sep 2002
Database er0rr
7·23·29 Posts 
In the "wood" it looks like I found a clearing where there is no tree!
I could speed up testing by letting a=lift(Mod(12,n)^r) and pretesting with kronecker(a^2+48,n)==1 
20221111, 02:12  #9 
Sep 2002
Database er0rr
7·23·29 Posts 
As well as "12" I am running "+12" overnight. The latter has the following pattern so far:
I am running these tests on a Celeron  if someone wants to put it on a 1024 core system, maybe not in Pari/GP, that would be most appreciated Last fiddled with by paulunderwood on 20221111 at 02:15 
20221111, 09:37  #10  
Sep 2002
Database er0rr
4669_{10} Posts 
Quote:
Testing of 12 is going great guns. Now at n>32,000,000,000, with no output except way points. Last fiddled with by paulunderwood on 20221111 at 09:38 

20221114, 11:47  #11 
Sep 2002
Database er0rr
7×23×29 Posts 
A proof?
We are busy verifying to 10^12, the test for the test over x^212^r*x12 where x^(n+1)==12.
I notice that the discriminant is 1+48 == 49 for r=0, i.e a square. Anyway back to the "proof" and it is for x^23*x3... The expression in x can be transformed into a test over z^2(3^(2*r1)2)*z+1 of (3*z)^((n+1)/2)==3. There are two cases to avoid the cyclotomic z^2(12)*z+1: First: gcd(2*r1,n1)!=1 which we can check for. We choose another r. Secondly, If gcd(2*r1,n1)==1 then we can use Euclid's algorithm to show that 3^1==1, a contradiction for n being odd. For example suppose the order is 47 and r=3 so that 3^47==1 and 3^5==1. the by Euc.Alg. 3^1==1. Done. Well this is halfbaked since there is the other cyclotomic z^2z+1 to be dealt with. Then 3^(2*r1)==3 because 32==1. That is (3)^(2*r2)==1 but we know (3)^((n1)/2)==+1. Hence we avoid gcd(r1,n1). Now we have avoided both cyclotomic equations by choosing r such that gcd((r1)*(2*r1),n1)==1. The above is not quite a proof yet and verification to infinity is impossible! Last fiddled with by paulunderwood on 20221114 at 11:56 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Amazing 6  paulunderwood  Miscellaneous Math  118  20210318 21:53 
Amazing result in P1  Miszka  Information & Answers  2  20140704 17:11 
Amazing academic resource  Xyzzy  Lounge  6  20120325 22:57 
Amazing!!!  R.D. Silverman  Factoring  5  20060126 09:14 
Two Amazing Things  clowns789  Hardware  1  20031227 16:57 