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  1. #1
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    legendre symbol

    show that the legendre symbol of y^3 congruent to 2 mod 7 is -1. I know how to work quadratic problems, but cubic is something i have never come across. Any ideas?? I assume you can use the quadratic law of reciprocity, but i have no idea how to go about doing it.
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    Quote Originally Posted by bigb View Post
    show that the legendre symbol of y^3 congruent to 2 mod 7 is -1. I know how to work quadratic problems, but cubic is something i have never come across. Any ideas?? I assume you can use the quadratic law of reciprocity, but i have no idea how to go about doing it.
    It is +1 not -1.

    If y^3 \equiv 2 (\bmod 7) then \left[ (y/7) \right]^3 = (y^3/7) = (2/7) = 1 \implies (y/7) = 1.
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    Quote Originally Posted by ThePerfectHacker View Post
    It is +1 not -1.

    If y^3 \equiv 2 (\bmod 7) then \left[ (y/7) \right]^3 = (y^3/7) = (2/7) = 1 \implies (y/7) = 1.
    its says in the book the cubic congruence y^3 congruent to 2 mod 7 is not solvable means that 2 is a cubic nonresidue modulo 7 (please verify this), but ur saying it is solvable. I am not sure now.
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    Quote Originally Posted by bigb View Post
    its says in the book the cubic congruence y^3 congruent to 2 mod 7 is not solvable means that 2 is a cubic nonresidue modulo 7 (please verify this), but ur saying it is solvable. I am not sure now.
    2 is not a cubic residue modulo 7 because if y^3 \equiv 2 \mod 7, then \gcd(y,7)=1, and hence 1 \equiv y^6 \equiv 4 \mod 7, which is nonsense!

    (by the way the theory of cubic residues is a mess compared to nice and clean theory of quadratic residues!)
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    Quote Originally Posted by bigb View Post
    its says in the book the cubic congruence y^3 congruent to 2 mod 7 is not solvable means that 2 is a cubic nonresidue modulo 7 (please verify this), but ur saying it is solvable. I am not sure now.
    Your problem was not clear. What I wrote above was my interpertation of your problem because of your title "Legendre symbol". There is no such thing as a Legendre symbol for cubic resides, while there is an analouge called the cubic residue symbol.

    Quote Originally Posted by NonCommAlg View Post
    (by the way the theory of cubic residues is a mess compared to nice and clean theory of quadratic residues!)
    I happen to think that biquadradic (quartic) is easier.
    Maybe it is just because \mathbb{Z}[i] looks nicer than \mathbb{Z}[\omega].
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