Amazing pattern with -12 (was -3)

[paulunderwood]

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.

[paulunderwood]

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

[paulunderwood]

And the pattern is broken with 28027505969 which has 4 r's, each with gcd(r-1,n-1)!=1

[paulunderwood]

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.

[RMLabrador]

one Computer say to me once - Do not believe Computers, they always [illegibly]!

[paulunderwood]

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:

{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])))))}

[RMLabrador]

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)))

[paulunderwood]

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

[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

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

[paulunderwood]

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.

[paulunderwood]

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!