July 10th, 2018, 01:22 PM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 232 Thanks: 26  Prime number question
1) Show there are infinitely many odd prime numbers $\displaystyle p=2n1$ 2) Show that number two (2) is the only even prime number $\displaystyle p=2$ Last edited by skipjack; July 17th, 2018 at 10:10 AM. 
July 10th, 2018, 03:55 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,540 Thanks: 591 
There are an infinite number of primes (wellknown proof). All even numbers are divisible by 2, so 2 is the only even prime. Why are you asking? Last edited by skipjack; July 17th, 2018 at 10:11 AM. 
July 17th, 2018, 06:04 AM  #3  
Newbie Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 20 Thanks: 2  Quote:
Every prime number can be expressed as x + (x+1) or y + (y1) (example: 6+7=13 or 8+9=17), and x, (x+1) and y, (y1) can not have common factors because they add up to create prime number which by definition can't have dividers except for itself and 1. Since every prime number can be expressed in this way, we have y + (y1) = 2y 1 and the conclusion is that every odd prime number can be expressed as 2n  1. Similarly, every odd prime number can be expressed as 2n + 1 since x + (x+1) = 2x + 1. 2) If there is an even prime number different from 2, then that number can be expressed as 2*x, which makes it divisible by 2, which implies that the number is composite, and that is contradiction. Last edited by skipjack; July 17th, 2018 at 09:32 PM.  
July 17th, 2018, 08:51 AM  #4  
Senior Member Joined: May 2016 From: USA Posts: 1,052 Thanks: 431  Quote:
$q \text { is an odd number } \iff \exists \ r \in \mathbb Z \text { such that } 2r  1 = q.$ $\therefore p \text { is odd and prime } \implies p \text { is odd } \implies$ $\exists \ n \in \mathbb Z \text { such that } 2n  1 = p.$ In other words, for any odd prime p, 2n  1 = p follows from the definition of odd and is not peculiar to odd primes. To prove that there are an infinite number of odd primes then follows from the proof of the infinitude of primes and the proof that there is only one even prime. Last edited by JeffM1; July 17th, 2018 at 08:56 AM.  
July 17th, 2018, 10:47 AM  #5 
Newbie Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 20 Thanks: 2 
My proof is also true, but yours is simpler. Is there any way to see all the code for these symbols you use? Last edited by skipjack; July 17th, 2018 at 10:53 AM. 
July 17th, 2018, 10:56 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,186 Thanks: 1648 
Either use the "Quote" button for the post or, if you can, right click on the relevant code and explore the menu system generated.

July 17th, 2018, 12:27 PM  #7 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 204 Thanks: 60 Math Focus: Algebraic Number Theory, Arithmetic Geometry  
July 17th, 2018, 12:41 PM  #8 
Newbie Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 20 Thanks: 2 
i just wanted to say that $2n 1$ is the same as $n + n  1 = n + (n  1)$

July 17th, 2018, 09:43 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 19,186 Thanks: 1648 
Question (1) didn't ask for a proof of that. It asked for a proof that there are infinitely many odd primes, and you didn't even attempt to provide that, link to such a proof or indicate how what you posted could contribute to such a proof.

July 18th, 2018, 03:15 AM  #10 
Newbie Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 20 Thanks: 2 
Every odd number can be represented as a sum of n + (n  1), and every prime number is odd number so i don't know what is the issue.


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