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Math Help - Odd prime divisors

  1. #1
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    Odd prime divisors

    Prove that the odd prime divisors of the integer n^2 + n + 1 that are different from 3 are of the form 6k+1.

    Proof.

    Let p be an odd prime of n^2 + n + 1 that is not 3, then n^2 + n \equiv -1 \ (mod \ p)

    But I'm stuck in here
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  2. #2
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    If p|n^2+n+1 then x^2+x+1\equiv 0 (\bmod p) is solvable. This means (2x+1)^2 \equiv -3(\bmod p) is solvable. Thus, (-3/p) = 1 this means (3/p)=(-1/p) = 1 or (3/p)=(-1/p) = -1. Can you find the the primes which make it true?
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