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Math Help - prove that a congruence is solvable

  1. #1
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    prove that a congruence is solvable

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

    From a theorem, the question is equivalent to showing that
    \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  (8,p-1) since  \{1,2,4,8\} is the exhaustive list of factors of  8 .

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

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

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

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

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