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July 18th, 2014, 08:38 AM   #1
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Another prime theory

First the facts:

a is element of IN

a^3+(a-1)^3=(2a-1)*(a^2-a+1)

x=a^3+(a-1)^3
y=(2a-1)
z=(a^2-a+1)

x=y*z

Ok, now my new theory:

If (x-1)/3 or (x+1)3 is element of IN,
then x have one primefactor in the form of y or z!

It is similar to:

If the cross sum of x is completely divisible by 9,
then it is the only case were x have none primefactor in the form of y or z!

To see what i mean you can watch the table..

Have fun with prime research^^
Attached Images
 xyz-prime-method.PNG (3.1 KB, 1 views) xyz-prime-table.jpg (21.6 KB, 4 views)

 July 18th, 2014, 09:13 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms What is IN? Is this the integers $\mathbb{Z}$, the positive integers $\mathbb{Z}^+$, the natural numbers $\mathbb{N}=\{0,1,\ldots\}$, or something else? It seems that you are saying: If a^3+(a-1)^3 is not divisible by 3, then either 2a-1 or a^2-a+1 is prime. Is this correct?
 July 18th, 2014, 11:54 AM #3 Member   Joined: Nov 2012 From: Germany Posts: 59 Thanks: 0 IN is the natural numbers. yes or lets say If a or a+1 is divisible by 3, then either 2a-1 or a^2-a+1 is prime. But i found to much composites, often when y is divisible by 5, also when y is p^k Maybe just luck in a small research But this is interesting: When log((a^2-a+1)+1)=log(z+1) is element of IN, then log(z+1) is an Mersenne exponent. There are Mersenne Primes z for a={2;3;6;91;...} Last edited by PerAA; July 18th, 2014 at 12:16 PM.
 July 19th, 2014, 04:45 PM #4 Senior Member   Joined: Aug 2008 From: Blacksburg VA USA Posts: 346 Thanks: 6 Math Focus: primes of course strike the log from both sides, it's irrelevant. Check your claim, I don't think 91 works.
July 19th, 2014, 04:49 PM   #5
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Quote:
 Originally Posted by PerAA IN is the natural numbers. yes or lets say If a or a+1 is divisible by 3, then either 2a-1 or a^2-a+1 is prime. But i found to much composites, often when y is divisible by 5, also when y is p^k Maybe just luck in a small research
Indeed. The conjecture fails infinitely often, indeed on a subset of the naturals with density 1.

Quote:
 Originally Posted by PerAA When log((a^2-a+1)+1)=log(z+1) is element of IN, then log(z+1) is an Mersenne exponent. There are Mersenne Primes z for a={2;3;6;91;...}
Is this for any integers a and z? What is the base of the logarithm?

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