The link between the Riemann Hypothesis and AKS-Primality Test?

[Unity+]

Thanks to iNow, I was able to make this finding(unless this was already found before I found it). I was meaning to post this sometime or other, but now I have the time to do so.

 

When I saw the equation for the prime test, I decided to mess with it. When I took it's derivative, I found that the equation, when solving for x when y = 0, would come up with complex numbers, where the real part is 1/2. I modified the equation(only the exponents) so it would fit the characteristics of the Riemann Zeta function(though an unorthodox method, I thought it would be important).

 

[math]\zeta (x)=(x-1)^{s}-x^{s}[/math]

 

Where s = p+1 and s must be an even number.

 

Now, in the paper iNow provided there didn't seem to be a reference to the relationship presented here. I think it would be interesting to investigate this because of the similarities.

 

One thing to point out is whenever s is odd, x equals 1/2, not 1/2 +/- ti, where t is a real number. I feel that this is not a coincidence and has relevance to the Riemann Hypothesis.

 

EDIT: Forgot to link the paper http://www.cse.iitk.ac.in/users/manindra/algebra/primality_v6.pdf

 

EDIT2: I just found something new after making this topic. I will post it soon.

 

EDIT3: Other properties I have found are:

 

• The amount of solutions, if s is even, that will exist is equal to p-1.
• The product of all roots will be 1/s.
• The sum of all roots will equal to p/2.

Edited March 28, 2014 by Unity+

[imatfaal]

Can you expand on your route Unity? Which equation - most of the prime checkers I know have modulo(target number); how did you amend the equation; you need to be more specific about connexions - my memory of the reimann zeta is the infinite sum of the p-series (ie only one input - you have two).

 

One thing to point out is whenever s is odd, x equals 1/2, not 1/2 +/- ti, where t is a real number.

 

ok - so let s=3,

f(x)

= (x-1)^3 - x^3

= x^3-3x^2+3x-1 - x^3

=-3x^2+3x-1

 

that has no real roots (two complex ones)

I could quite believe that f(x) will always have a real root at 1/2 when s is even - and if I think hard enough I might see why but I dont yet ;-) - but first two s as odd I tried don't work

[Unity+]

Can you expand on your route Unity? Which equation - most of the prime checkers I know have modulo(target number); how did you amend the equation; you need to be more specific about connexions - my memory of the reimann zeta is the infinite sum of the p-series (ie only one input - you have two).

 

 

ok - so let s=3,

f(x)

= (x-1)^3 - x^3

= x^3-3x^2+3x-1 - x^3

=-3x^2+3x-1

 

that has no real roots (two complex ones)

I could quite believe that f(x) will always have a real root at 1/2 when s is even - and if I think hard enough I might see why but I dont yet ;-) - but first two s as odd I tried don't work

The roots are not supposed to be real. What I am pointing out is how the complex roots have 1/2 as the real part of the complex root.

 

The prime-test equation I am referring to is (x-1)^p - (x^p - 1).

 

I fixed the equation so s must be even and bigger than 1.

 

[math]\zeta (x)=\left ( x-1 \right )^{s}+x^{s}[/math]

Edited March 28, 2014 by Unity+

[imatfaal]

The roots are not supposed to be real. What I am pointing out is how the complex roots have 1/2 as the real part of the complex root.

...

 

OK that makes sense - but note that is the opposite of what you said above "x equals 1/2, not 1/2 +/- ti,"

 

I see you have updated will post after reading yr new post

...

 

The prime-test equation I am referring to is (x-1)^p - (x^p - 1).

...

 

OK - for starters (I am being pedantic) that is an expression not an equation.

 

Fermat's Little is very similar but has an all important equivalence sign and a modulo. I cannot see how it works without a mod. Your paper has a nice polynomial version

....

[math]\zeta (x)=\left ( x-1 \right )^{s}+x^{s}[/math]

 

Zeta(x)=Sum k running from 1 to infinity of 1/(k^x) - still dont see how that is equivalent.

[Unity+]

OK - for starters (I am being pedantic) that is an expression not an equation.

 

Fermat's Little is very similar but has an all important equivalence sign and a modulo. I cannot see how it works without a mod. Your paper has a nice polynomial version

 

 

 

Zeta(x)=Sum k running from 1 to infinity of 1/(k^x) - still dont see how that is equivalent.

The equations are not equal(though I am working to see if they are in a certain way), however they have properties that are very similar. This is what I am getting at.

 

EDIT: I forgot to add the subscript i.

 

[math]\zeta (x)_{i}=\left ( x-1 \right )^{s}+x^{s}[/math]

 

 

 

OK - for starters (I am being pedantic) that is an expression not an equation.

I knew that, but thought it could slide.

Edited March 28, 2014 by Unity+

[imatfaal]

Re - Your numberphile video and the expression you are using. As I have been banging on about you need a modulo - just like Fermat's Little - this comes in the test on the video where they check if all the coefficients are divisible by p. Your expression is only the first stage and provides no sieving on whether p is prime or not - it is only when the coefficients are checked to see if they are divisible by p (ie the equal zero mod p ) that you get any separation of primes and composites.

 

And I am still lost on your zeta function connexion. Why are you now subscripting the x in zeta(x) - there is only one solution to zeta(x) not a series of them; unless you have an implied sum somewhere that I am too dense to notice?

[Unity+]

Re - Your numberphile video and the expression you are using. As I have been banging on about you need a modulo - just like Fermat's Little - this comes in the test on the video where they check if all the coefficients are divisible by p. Your expression is only the first stage and provides no sieving on whether p is prime or not - it is only when the coefficients are checked to see if they are divisible by p (ie the equal zero mod p ) that you get any separation of primes and composites.

 

And I am still lost on your zeta function connexion. Why are you now subscripting the x in zeta(x) - there is only one solution to zeta(x) not a series of them; unless you have an implied sum somewhere that I am too dense to notice?

Sorry for the misinterpretation. I am merely focusing on the equation, though.

 

I am sub scripting the function because I am showing that it is different than the zeta function.

 

I am only focusing on one particular solution out of all of the solutions.