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April 13th, 2017, 10:13 AM   #1
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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 $p\equiv 1\pmod{3}$
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April 13th, 2017, 07:18 PM   #2
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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.
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April 14th, 2017, 12:44 AM   #3
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Where did this problem come from?
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April 14th, 2017, 06:50 PM   #4
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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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