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Thread: prove that a congruence is solvable

  1. #1
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    prove that a congruence is solvable

    For all primes p, show that $\displaystyle x^8 \equiv 16 \mbox{ (mod p)}$ is solvable.

    From a theorem, the question is equivalent to showing that
    $\displaystyle \displaystyle{16^{\frac{p-1}{(8, p-1)}} \equiv 1 \mbox{ (mod p)}}$

    And then I am stuck.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    There are four options for $\displaystyle (8,p-1) $ since $\displaystyle \{1,2,4,8\} $ is the exhaustive list of factors of $\displaystyle 8 $.

    If $\displaystyle (8,p-1) = 1 $, it's pretty easy to see $\displaystyle 16^{p-1} \equiv 1 \mod{p} $.

    If $\displaystyle (8,p-1) = 2, \; 16^{\frac{p-1}{2}} = 4^{p-1} \equiv 1 \mod{p} $.

    If $\displaystyle (8,p-1) = 4, \; 16^{\frac{p-1}{4}} = 2^{p-1} \equiv 1 \mod{p} $.

    If $\displaystyle (8,p-1) = 8, \; 16^{\frac{p-1}{8}} = 2^{\frac{p-1}{2}}$.
    In this case, $\displaystyle p \equiv 1 \mod{8} $ and hence $\displaystyle 2 $ is a square modulo $\displaystyle p $, i.e. $\displaystyle \left(\frac{2}{p}\right) = 1 $. Therefore we can choose $\displaystyle x $ such that $\displaystyle x^2 \equiv 2 \mod{p} $. So $\displaystyle 2^{\frac{p-1}{2}} \equiv (x^2)^{\frac{p-1}{2}} \equiv x^{p-1} \equiv 1 \mod{p} $.

    Therefore $\displaystyle x^8 \equiv 16 \mod{p} $ is always solvable.
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