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

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 April 13th, 2017, 11:13 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, 08:18 PM #2 Senior Member   Joined: Aug 2012 Posts: 2,426 Thanks: 759 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 08:20 PM. April 14th, 2017, 01:44 AM #3 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 Where did this problem come from? April 14th, 2017, 07:50 PM #4 Member   Joined: Jan 2016 From: Athens, OH Posts: 93 Thanks: 48 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 Tags 3|p1, ⇒, odd, prime Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post momo Number Theory 4 June 3rd, 2017 04:29 PM blind887 Algebra 2 April 12th, 2017 03:35 AM M_B_S Number Theory 59 October 6th, 2014 12:52 AM gaussrelatz Number Theory 4 September 1st, 2012 12:06 PM momo Number Theory 14 September 26th, 2008 08:21 AM

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