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Math Help - Solving a congruence modulo p

  1. #1
    Ant
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    Solving a congruence modulo p

     p \equiv 2 (mod 3) and  p-1 = 4q with p and q distinct primes.

    Show that:

     3^{2q} \not\equiv 1 (mod p)

    I've tried various thing but I can't seem to find a contradiction (after assuming 3^(2q) == 1 mod p).

    Can anyone point me in the right direction please??
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  2. #2
    Member Sylvia104's Avatar
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    Re: Solving a congruence modulo p

    Since p\equiv1\mod4, quadratic reciprocity gives 1 = (-1)^{\left(\frac{p-1}2\right)\left(\frac{3-1}2\right)} = \left(\frac p3\right)\left(\frac3p\right) = (-1)\left(\frac3p\right) (where \left(\frac**\right) denotes the Legendre symbol). In other words 3 is not a quadratic residue modulo p; by Eulerís criterion, -1 = \left(\frac3p\right) \equiv 3^{\frac{p-1}2} = 3^{2q}\mod p.
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