Community

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
Login Register Memberlist Search AoPS Blogs Contests Galleries Forum Index
The time now is Wed Dec 02, 2009 1:16 am
All times are UTC - 8
 		View posts since last visit
View unanswered posts
Home » Olympiad Section » Number Theory » Number Theory Open Questions
 
p_i divides (all others -1)
Moderators: High School Olympiad Moderators, amfulger, Arne, darij grinberg, freemind, harazi, Megus, N.T.TUAN, orl, pbornsztein, ZetaX
Post new topic   Reply to topic View previous topicView next topic
2 Posts • Page 1 of 1
Author Message
bboypa
Riemann Hypothesis
Riemann Hypothesis


Offline
Joined: 12 Feb 2008
Posts: 315
Location: Milan
ItalyEuropean Union
To rate posts you must be logged in
#1
p_i divides (all others -1)
a cyclic divisibility between primes
Let p_1,p_2,\ldots,p_n be pairwise distinct odd primes such that \displaystyle p_i \mid \left(\frac {\displaystyle \prod_{j = 1}^n{p_j}}{p_i}\right) - 1 for all 1 \le i \le n.

I guess that n = 1. Try to (dis-)prove this conjecture.

PostPosted: Wed Jul 01, 2009 4:06 pm  Back to top 
  ProfilePMWWWMSN
maxal
Riemann Hypothesis
Riemann Hypothesis


Offline
Joined: 09 Apr 2005
Posts: 378
To rate posts you must be logged in
#2
Re: p_i divides (all others -1)
a cyclic divisibility between primes
bboypa wrote:
Let p_1,p_2,\ldots,p_n be pairwise distinct odd primes such that \displaystyle p_i \mid \left(\frac {\displaystyle \prod_{j = 1}^n{p_j}}{p_i}\right) - 1 for all 1 \le i \le n.

This condition can be restated as
p_1\cdot p_2\cdots p_n \mid - 1 + p_1\cdot p_2\cdots p_n\cdot \sum_{i = 1}^n \frac {1}{p_i}.
Therefore, for n>1, there exists integer k\geq 1 such that
k\cdot p_1\cdot p_2\cdots p_n = - 1 + p_1\cdot p_2\cdots p_n\cdot \sum_{i = 1}^n \frac {1}{p_i}
or
\sum_{i = 1}^n \frac {1}{p_i} = \frac {k\cdot p_1\cdot p_2\cdots p_n + 1}{p_1\cdot p_2\cdots p_n}.
and, in particular,
\sum_{i = 1}^n \frac {1}{p_i} > k \geq 1.
Minimal set of odd primes satisfying this inequality has n = 9 (for primes 3,5,\dots,29). Therefore, if solution for n > 1 exists then n\geq 9.

PostPosted: Tue Oct 27, 2009 6:22 am  Back to top 
  ProfilePM
Display posts from previous:   Sort by:   
2 Posts • Page 1 of 1
Post new topic   Reply to topic View previous topicView next topic
Home » Olympiad Section » Number Theory » Number Theory Open Questions
 
Jump to:  


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
You cannot attach files in this forum
You can download files in this forum
You cannot post calendar events in this forum


© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us