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 2022-11-09, 14:58 #1 paulunderwood     Sep 2002 Database er0rr 2×33×83 Posts Amazing pattern with -12 (was -3) Testing Mod(-3,n)^((n-1)/2) == kronecker(-3,n) and Mod(Mod(x,n),x^2-3^r*x-3)^(n+1)==-3 for all r reveals some interesting patterns in the pseudoprimes thereof: g := g(r-1,n-1) != 1. z := multiplicative order of -3 mod n is even. two solutions for r up to z g*4 = z n%4 = 1 n%6 = 5 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 2022-11-09 at 17:41
 2022-11-10, 02:08 #2 paulunderwood     Sep 2002 Database er0rr 2·33·83 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*r1-1 = z/2+1 2*r2-1 = 3*z/2+1 Working over the equation x^2-3^r*x-3 can be transformed to working over y^2-(-3^(2*r-1)-2)*y+1. Rearraging terms gives 3^(2*r-1) = (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^2-y+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 2022-11-10 at 02:16
 2022-11-10, 04:27 #3 paulunderwood     Sep 2002 Database er0rr 2·33·83 Posts And the pattern is broken with 28027505969 which has 4 r's, each with gcd(r-1,n-1)!=1
 2022-11-10, 21:09 #4 paulunderwood     Sep 2002 Database er0rr 10001100000102 Posts -12 I tried other bases -2,(-3),-4,-6 and -8. but boy oh boy -12 is absolutely great. The test over x^2-12^r*x-12 has no pseudoprimes less than 300,000,000 and consequently no need to take a GCD. Again I let it run overnight.
 2022-11-10, 21:26 #5 RMLabrador   "Chereztynnoguzakidai" Oct 2022 Ukraine, near Kyiv. 2·23 Posts one Computer say to me once - Do not believe Computers, they always [illegibly]!
2022-11-10, 21:47   #6
paulunderwood

Sep 2002
Database er0rr

2×33×83 Posts

Quote:
 Originally Posted by RMLabrador one Computer say to me once - Do not believe Computers, they always [illegibly]!
Indeed!

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)^((n-1)/2)==kronecker(-b,n),
z=znorder(Mod(-b,n));for(r=1,z,
if(Mod(Mod(x,n),x^2-Mod(-b,n)^r*x-b)^(n+1)==-b,
print([b,n,n%4,kronecker(-3,n),gcd(2*r-1,n-1)==z,z,gcd(2*r-1,n-1),r])))))}

 2022-11-10, 21:58 #7 RMLabrador   "Chereztynnoguzakidai" Oct 2022 Ukraine, near Kyiv. 2E16 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)))
 2022-11-10, 23:41 #8 paulunderwood     Sep 2002 Database er0rr 106028 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
 2022-11-11, 02:12 #9 paulunderwood     Sep 2002 Database er0rr 2×33×83 Posts As well as "-12" I am running "+12" overnight. The latter has the following pattern so far: g := g(2*r-1,n-1). one solution for r up to z the multiplicative order of 12 mod n g = z n%4 = 3 n%6 = 5 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 2022-11-11 at 02:15
2022-11-11, 09:37   #10
paulunderwood

Sep 2002
Database er0rr

2×33×83 Posts

Quote:
 Originally Posted by paulunderwood As well as "-12" I am running "+12" overnight. The latter has the following pattern so far: g := g(2*r-1,n-1). one solution for r up to z the multiplicative order of 12 mod n g = z n%4 = 3 n%6 = 5
The pattern broke for +12, with n=20935371731 and r=6163, yet gcd(2*r-1,n-1)==145.

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 2022-11-11 at 09:38

 2022-11-14, 11:47 #11 paulunderwood     Sep 2002 Database er0rr 448210 Posts A proof? We are busy verifying to 10^12, the test for the test over x^2-12^r*x-12 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^2-3*x-3... The expression in x can be transformed into a test over z^2-(-3^(2*r-1)-2)*z+1 of (-3*z)^((n+1)/2)==-3. There are two cases to avoid the cyclotomic z^2-(1-2)*z+1: First: gcd(2*r-1,n-1)!=1 which we can check for. We choose another r. Secondly, If gcd(2*r-1,n-1)==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 half-baked since there is the other cyclotomic z^2-z+1 to be dealt with. Then -3^(2*r-1)==3 because 3-2==1. That is (-3)^(2*r-2)==-1 but we know (-3)^((n-1)/2)==+-1. Hence we avoid gcd(r-1,n-1). Now we have avoided both cyclotomic equations by choosing r such that gcd((r-1)*(2*r-1),n-1)==1. The above is not quite a proof yet and verification to infinity is impossible! Last fiddled with by paulunderwood on 2022-11-14 at 11:56

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