My Math Forum For odd n and prime p, (p | (3^n+1)) ⇒ 3|(p-1)

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 April 13th, 2017, 10:13 AM #1 Senior Member   Joined: Dec 2006 Posts: 167 Thanks: 3 For odd n and prime p, (p | (3^n+1)) ⇒ 3|(p-1) 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: 2,043 Thanks: 584 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: 19,712 Thanks: 1805 Where did this problem come from?
 April 14th, 2017, 06:50 PM #4 Member   Joined: Jan 2016 From: Athens, OH Posts: 89 Thanks: 47 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. Thanks from Maschke

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