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July 4th, 2009, 07:46 AM  #1 
Senior Member Joined: Jan 2009 From: Russia Posts: 113 Thanks: 0  Primes inside the arithmetic progression
Prove that among progression 3, 7, 11, 15, 19, ... there is infinitely many primes. 
July 4th, 2009, 09:23 AM  #2 
Senior Member Joined: Jun 2009 Posts: 150 Thanks: 0  Re: Primes inside the arithmetic progression
Call such a number "3 mod 4" ... the rest of the odd numbers are "1 mod 4" ... Note that the product of two "1 mod 4" numbers is again a "1 mod 4" number. If there were no "3 mod 4" primes, then there could be no "3 mod 4" numbers at all. A contradiction. Therefore, there is at least one "3 mod 4" prime. (By itself, this is not very useful, because we can easily write down such a prime.) Now: soup up that argument: assume there are only finitely many "3 mod 4" primes, then get a contradiction from that. 
July 4th, 2009, 09:29 AM  #3 
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  Re: Primes inside the arithmetic progression
It's not hard to modify Euclid's theorem to work in this case. Suppose that the only primes of this form are 3, 7, ..., k. Take their product and add 4. Can this be divisible by any of the numbers on your list? Can it be the product of primes = 1 mod 4?

July 5th, 2009, 05:40 AM  #4  
Senior Member Joined: Jan 2009 From: Russia Posts: 113 Thanks: 0  Re: Primes inside the arithmetic progression Quote:
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July 5th, 2009, 08:59 AM  #5  
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  Re: Primes inside the arithmetic progression Quote:
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July 5th, 2009, 09:52 AM  #6 
Senior Member Joined: Jan 2009 From: Russia Posts: 113 Thanks: 0  Re: Primes inside the arithmetic progression
Can't believe I was so silly. Thanks!


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