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January 29th, 2015, 05:40 PM  #1 
Member Joined: Jan 2015 From: Orlando, Florida Posts: 92 Thanks: 10  Chinese Remainder!!!!!!!!!!!!
Suppose that m and n are integers with 1 ≤ m ≤ 49 and n ≥ 0 such that m divides n^(n+1) + 1. What is the number of possible values of m?
Last edited by skipjack; January 30th, 2015 at 07:16 PM. 
January 29th, 2015, 08:48 PM  #2 
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 
Well, as a last resort, $n^{n+1}+1\equiv (n+k)^{n+k+1}+1$ when $k$ is a multiple of $n\varphi(n),$ so you could step through each m this way and see which work. I get 29, FWIW. 
January 30th, 2015, 12:57 PM  #3 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 
I'm a Chinese. I don't even know what's Chinese remainder theorem. Can anyone show me where's the resource to learn this theorem? I feel shameful. 
January 30th, 2015, 12:59 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,922 Thanks: 1518  
January 30th, 2015, 01:06 PM  #5 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 
I looked out in Wikipedia. But it looks complicated. I don't even know what's mod 
January 30th, 2015, 02:20 PM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,922 Thanks: 1518  
January 30th, 2015, 02:59 PM  #7 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 
Thanks

January 30th, 2015, 04:30 PM  #8 
Member Joined: Jan 2015 From: Orlando, Florida Posts: 92 Thanks: 10  
January 30th, 2015, 06:26 PM  #9  
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  Quote:
So if you have 1 mod 2 2 mod 3 7 mod 11 then the CRT says there is exactly one residue class mod 66 = 2*3*11 which satisfies all three of these, and tells you how to find it. (If the residues aren't coprime, you take the LCM instead of product, but unless the residues are compatible you won't get any residues that work.) In that case you need more help than I can give.  

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