# legendre symbol

• Dec 3rd 2008, 03:35 PM
bigb
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.
• Dec 3rd 2008, 05:35 PM
ThePerfectHacker
Quote:

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 $\displaystyle y^3 \equiv 2 (\bmod 7)$ then $\displaystyle \left[ (y/7) \right]^3 = (y^3/7) = (2/7) = 1 \implies (y/7) = 1$.
• Dec 3rd 2008, 09:30 PM
bigb
Quote:

Originally Posted by ThePerfectHacker
It is +1 not -1.

If $\displaystyle y^3 \equiv 2 (\bmod 7)$ then $\displaystyle \left[ (y/7) \right]^3 = (y^3/7) = (2/7) = 1 \implies (y/7) = 1$.

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.
• Dec 4th 2008, 03:10 AM
NonCommAlg
Quote:

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 $\displaystyle y^3 \equiv 2 \mod 7,$ then $\displaystyle \gcd(y,7)=1,$ and hence $\displaystyle 1 \equiv y^6 \equiv 4 \mod 7,$ which is nonsense!

(by the way the theory of cubic residues is a mess compared to nice and clean theory of quadratic residues!)
• Dec 4th 2008, 06:01 AM
ThePerfectHacker
Quote:

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.

Quote:

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 $\displaystyle \mathbb{Z}[i]$ looks nicer than $\displaystyle \mathbb{Z}[\omega]$. (Thinking)