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Math Help - solvable congruent

  1. #1
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    solvable congruent

    How do you proof that -1 congruent to x^4 mod p is solvable iff (-1)^(p-1/d) is congruent to 1 mod p where d= gcd(4,p-1)?

    i was trying to use euler theorem that (-1)^(p-1) is congruent to 1 mod p then how else can i continue?
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  2. #2
    Member Sylvia104's Avatar
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    Re: Solvable congruent

    I take it you want p to be prime. The case p=2 is trivial, so we'll assume p to be odd. Then d=2 if p\equiv3\mod4 and d=4 if p\equiv1\mod 4.


    If p\equiv3\mod4 then by Euler's criterion \left(\frac{-1}p\right)=(-1)^{\frac{p-1}2}=-1. In this case x is not a quadratic residue modulo p (and so it certainly can't be a quartic residue modulo p); we also have (-1)^{\frac{p-1}d}=(-1)^{\frac{p-1}2}=-1\not\equiv1\mod p.


    Consider p\equiv1\mod4.

    (i) Suppose x^4\equiv-1\mod p. As d=4 in this case, -1\equiv x^d\mod p \implies (-1)^{\frac{p-1}d}\equiv x^{p-1}\mod p\equiv1\mod p by Fermat's little theorem.

    (ii) Conversely suppose (-1)^{\frac{p-1}d}=(-1)^{\frac{p-1}4}\equiv1\mod p. Then \frac{p-1}4 is even, i.e. 8\mid p-1. Let k be a primitive root \mod p and set x=k^{\frac{p-1}8}. Then x^8\equiv1\mod p \implies p divides x^8-1=\left(x^4+1\right)\left(x^4-1\right). But p cannot divide x^4-1 otherwise k^{\frac{p-1}2}\equiv1\mod p which would contradict the fact that, as a primitive root, k has order p-1 in the multiplicative group of the integers modulo p. Hence p divides x^4+1 and we are done.
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