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Cubic reciprocity?

  1. Jan 17, 2008 #1
    [SOLVED] cubic reciprocity?

    I would like to prove the following conjecture:

    If [tex]p \equiv 2\ (mod\ 3)[/tex] is a prime, then the cubing function [tex]x \mapsto x^3 [/tex] is a permutation of [tex]\mathbb{Z}_p [/tex].

    I've tried to find a contradiction to the negation by assuming that if [tex]n \neq m\ (mod\ 3)[/tex], but [tex]n^3 \equiv m^3\ (mod\ 3)[/tex], then since [tex]m^3 - n^3 = (m-n)(m^2-mn+n^2)[/tex], we must have [tex](m^2-mn+n^2) \equiv 0\ (mod\ 3)[/tex] to avoid a contradiction concerning zero-divisors. Now I'm stuck.
  2. jcsd
  3. Jan 17, 2008 #2
    what exactly you mean saying "the cubing function [tex]x \mapsto x^3 [/tex] is a permutation of [tex]\mathbb{Z}_p [/tex]"?

    We know that there are [tex]2^p[/tex] permutations (sub-groups) in [tex]\mathbb{Z}_p [/tex], right?

    So you mean that one of these subgroups represent a cube? If yes, is the sub-group sum or product?
  4. Jan 17, 2008 #3


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    I would have interpreted "the cubing function [tex]x \mapsto x^3 [/tex]is a permutation of [tex]\mathbb{Z}_p [/tex]" to mean that it permutes the members of [tex]\mathbb{Z}_p [/tex]- but that's so trivial, that's probably not what's meant!

    03= 0, 13= 1, 23= 2 mod 3. In fact, the cubing function is just the identity on [tex]\mathbb{Z}_p [/tex]!
  5. Jan 17, 2008 #4


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    3 isn't the only prime. :tongue:

    As for the original poster... what do you know about the multiplicative structure of Z/pZ?
  6. Jan 17, 2008 #5
    I thought "the cubing function [tex]x \mapsto x^3 [/tex] is a permutation of [tex]\mathbb{Z}_p [/tex]" was a pretty unambiguous statement. Maybe it would be better to say it's a "bijection [tex]\mathbb{Z}_p \rightarrow \mathbb{Z}_p[/tex]".

    HallsofIvy: Besides the fact that 3 is not congruent to 2 mod 3, in [tex]\mathbb{Z}_5 [/tex], [tex]2^3 \equiv 3\ (mod\ 5)[/tex] so it is not the identity.

    Hurkyl: It is a group of order p-1, which is of the form 3k+1. I guess the obvious fact that springs to mind is that no element can have order 3 then by Lagrange. This means no cube can be congruent to 1 mod p (except, of course, 1). So I guess this means the kernel of the cubing map is {1} ***alarm bells ringing*** actually this map is a group homomorphism so Ker={1} implies it is injective. Of course this implies it is a bijective too.
  7. Jan 17, 2008 #6
    It can be looked at a little more algebratically: If X^3 == Y^3 Mod P, then there exists an element x/y such that (x/y)^3 == 1 Mod P, which implies 3 divides the order of the group, p-1. Thus, unless x=y, P would be of the form 3k+1. (Which is about what Hello Kitty is saying)
    Last edited: Jan 17, 2008
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