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April 13th, 2017, 10:13 AM  #1 
Senior Member Joined: Dec 2006 Posts: 163 Thanks: 3  For odd n and prime p, (p  (3^n+1)) ⇒ 3(p1)
Show that if $p> 2$ is a prime, $n > 1$ is odd and $p\mid (3^n+1)$, then $p\equiv 1\pmod{3}$ 
April 13th, 2017, 07:18 PM  #2 
Senior Member Joined: Aug 2012 Posts: 1,370 Thanks: 321 
Fiendish. It has something to do with the binomial coefficients, that's my current theory. I'm writing $3^n$ as $(2 + 1)^n$ and noting that the sum of the elements of each row of Pascal's triangle are alternately $1$ and $2$ mod $3$. Can't get any farther than that. Anyone else hacking away at this? Last edited by Maschke; April 13th, 2017 at 07:20 PM. 
April 14th, 2017, 12:44 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 17,162 Thanks: 1284 
Where did this problem come from?

April 14th, 2017, 06:50 PM  #4 
Member Joined: Jan 2016 From: Athens, OH Posts: 44 Thanks: 26 
This is almost an immediate consequence of the quadratic reciprocity theorem  link https://en.wikipedia.org/wiki/Quadratic_residue, a lot of information. The following just defines the relevant notation and some preliminary facts before the proof of the statement. 

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3p1, ⇒, odd, prime 
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