March 2nd, 2017, 06:51 PM  #1 
Newbie Joined: Mar 2017 From: Chile Posts: 5 Thanks: 0  Not primes formula?
Hello, I'm new in this forum and sorry about my English is not the best, English is not my mother language. Okay I was fascinated with Mersenne primes, so I started playing with them. 2^p – 1 = 2^p – (21) = 2^p – (21)^p1 Then I started changing p with n, where n is a integer: (2)^(n)(21)^(n1) I know that this is only a mersenne prime only if n is a prime number and not every prime. Then I change 2 with x just for fun (x)^n(x1)^n1 and long story short (I can explain more but I only want to know if this is a thing because I don’t know how to search this) if n = 6p+2 where p is an integer there is no x that I found there the equation (x)^n(x1)^n1 is prime and if n takes any other value there is a x where the result of the equation is prime I have more information: Formula = (x)^n – (x1)^n1 Números: 1 4^13^0 = 3 is prime 2 2^21^1 = 3 is prime 3 2^31^2 = 7 is prime 4 3^42^3 = is prime 5 2^51^4 = 31 is prime 6 4^63^5 = 3853 is prime 7 2^71^6 = 127 is prime 8 ¿? I have check pass x = 10.000 9 3^92^8 = 19427 is prime 10 3^102^9 = 58537 is prime 11 3^112^10 = 176123 is prime 12 3^122^11= 529393 is prime 13 2^131^12 = 8191 is prime 14 ¿? I have check pass x = 10.000 I have called this not prime formula because every number when n = 6p+2 is not prime so this could help to find number not primes. Last edited by skipjack; March 2nd, 2017 at 10:33 PM. 
March 2nd, 2017, 07:10 PM  #2 
Senior Member Joined: Aug 2012 Posts: 1,887 Thanks: 525 
> (x)^n – (x1)^n1 I'm assuming you mean (x)^n  (x1)^(n1)? $\displaystyle 4^3  3^2 = 64  9  55 = 5 \times 11$. Am I understanding your idea correctly? Last edited by Maschke; March 2nd, 2017 at 07:13 PM. 
March 2nd, 2017, 07:17 PM  #3 
Newbie Joined: Mar 2017 From: Chile Posts: 5 Thanks: 0 
A little bit but can you find a x where the equation is prime if n = 8

March 2nd, 2017, 07:21 PM  #4  
Senior Member Joined: Aug 2012 Posts: 1,887 Thanks: 525  Quote:
$\displaystyle 8^1  7^0 = 7$, prime, and $\displaystyle 8^2  7^1 = 57 = 3 \times 19$, composite. Last edited by Maschke; March 2nd, 2017 at 07:28 PM.  
March 2nd, 2017, 07:28 PM  #5 
Newbie Joined: Mar 2017 From: Chile Posts: 5 Thanks: 0 
Sorry about not being able to explain. I know that some times the equation is prime and sometimes is composite but to simplify the problem try to solve this y = x^8  (x1)^(7) where y is prime and x is greater than 1

March 2nd, 2017, 07:35 PM  #6  
Senior Member Joined: Aug 2012 Posts: 1,887 Thanks: 525  Quote:
Quote:
ps  $\displaystyle x^8  (x1)^7 =(x^2 x + 1)(x^6  6x^4 15x^3 +14x^2 6x +1 )$ so it's always composite. If the geniuses of the past had Wolfram Alpha, think what they could have done! https://www.wolframalpha.com/input/?i=x%5E8++(x1)%5E7 Last edited by Maschke; March 2nd, 2017 at 07:51 PM.  
March 2nd, 2017, 07:47 PM  #7 
Newbie Joined: Mar 2017 From: Chile Posts: 5 Thanks: 0 
I'm a computer scientis and I made a program to check that equation and I have checked until x = 10.000 and no solution I don't know if I have a bug or something or if there are no solution I have a slow computer and the number to check are a bit large so I haven't check bigger numbers

March 2nd, 2017, 08:16 PM  #8  
Senior Member Joined: Aug 2012 Posts: 1,887 Thanks: 525  Quote:
So what I wonder is, how does Wolfram Alpha do these factorizations? Does it use brute force or are there some shortcuts? There are other computer algebra systems out there. I don't know anything about them. But if you need to factor a lot of big polynomials, there is software out there to do it. One more thing, you know of course that for number theory, programs can give clues but can never prove anything. There are examples of patterns that continue to very large numbers but fail in general.  
March 2nd, 2017, 08:34 PM  #9  
Newbie Joined: Mar 2017 From: Chile Posts: 5 Thanks: 0  Quote:
For example, if you want to check if a number is prime you need to see if all number below n are coprimes but if you think a little bit you only need to check the square root of n. Last edited by skipjack; March 2nd, 2017 at 10:27 PM.  
March 2nd, 2017, 10:45 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 18,956 Thanks: 1602  

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