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October 2nd, 2010, 02:32 AM  #1 
Newbie Joined: Jun 2009 Posts: 14 Thanks: 0  2 elementary number theory problems.... problem 1: suppose is a prime number such that (P1)/4 and (P+1)/2 are also primes. Show that p = 13. problem 2: let be n numbers such that each is either or . If , then prove that divides . HELP! 
October 2nd, 2010, 09:36 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: 2 elementary number theory problems....
For #1, work mod 4.

October 4th, 2010, 06:31 PM  #3 
Newbie Joined: Sep 2010 Posts: 17 Thanks: 0  Re: 2 elementary number theory problems....
If P, (P+1)/2 and (P1)/4 are prime numbers then P=4x+1, so our numbers are 4x+1, 2x+1 and x. Now, if x=3K then K must be 1, x=3. if x=3K+1 then 2x+1=2(3K+1)+1=3(2K+1). Then 2K+1 must be 1, so x=1 which is not prime. if x=3K+2 then 4x+1=4(3K+2)+1=3(4K+3). Then 4K+3 must be 1, and x isn't even an integer. So the only solution is x=3, P=13. ;] 
October 8th, 2010, 07:42 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,640 Thanks: 959 Math Focus: Elementary mathematics and beyond  Re: 2 elementary number theory problems....
Clearly, p = 2, 3 are not solutions. Assume p is of the form 6n  1. p = 6n  1, (p  1)/4 = (3n  1)/2, (p + 1)/2 = 3n; 3n can never be prime so there are no primes of the form 6n  1 that are solutions. Assume p is of the form 6n + 1. p = 6n + 1, (p  1)/4 = 3n/2, (p + 1)/2 = 3n + 1; 3n/2 can be prime only when n = 2, therefore p = 13. 
October 9th, 2010, 02:19 AM  #5  
Global Moderator Joined: Dec 2006 Posts: 18,142 Thanks: 1417  Quote:
 
October 9th, 2010, 06:54 AM  #6 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361  Re: 2 elementary number theory problems....
Thanks, CRGreathouse, I'll use . I unfortunately can't edit anymore, so i'll replace a part: So we have: And than factor out mod 2 is in the exponent. Would that help? Or is there a "better" alternative? Hoempa 

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