1. ## 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.

2. Originally Posted by bigb
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 $\left[ (y/7) \right]^3 = (y^3/7) = (2/7) = 1 \implies (y/7) = 1$.

3. Originally Posted by ThePerfectHacker
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.

4. Originally Posted by bigb
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!)

5. Originally Posted by bigb
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.

Originally Posted by NonCommAlg
(by the way the theory of cubic residues is a mess compared to nice and clean theory of quadratic residues!)
Maybe it is just because $\mathbb{Z}[i]$ looks nicer than $\mathbb{Z}[\omega]$.