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Old 2022-11-09, 14:58   #1
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Default 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
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Old 2022-11-10, 02:08   #2
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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
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Old 2022-11-10, 04:27   #3
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And the pattern is broken with 28027505969 which has 4 r's, each with gcd(r-1,n-1)!=1
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Old 2022-11-10, 21:09   #4
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Default -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.
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Old 2022-11-10, 21:26   #5
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one Computer say to me once - Do not believe Computers, they always [illegibly]!
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Old 2022-11-10, 21:47   #6
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Quote:
Originally Posted by RMLabrador View Post
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])))))}
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Old 2022-11-10, 21:58   #7
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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)))
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Old 2022-11-10, 23:41   #8
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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

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Old 2022-11-11, 02:12   #9
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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
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Old 2022-11-11, 09:37   #10
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Quote:
Originally Posted by paulunderwood View Post
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
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Old 2022-11-14, 11:47   #11
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Default 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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