Congruence Class Question

So I should have the following then (for a truly complete explanation):

A. [itex]x \equiv 1 (mod \lambda)[/itex], then [itex] x^3 \equiv 1(mod \lambda)^3[/itex].

If [itex] x = 1 (mod \lambda)[/itex], then [itex] x = 1 + a*(1 - w) + b*(1 - w)^2. [/itex]
[itex]x - 1 = a*(1 - w) + b*(1 - w)^2[/itex]
[itex] = (1 - w)*(a + b*(1 - w))[/itex]. Clearly, this is divisible by [itex]1 - w = \lambda[/itex]. From here, we can see that:

[itex] x^3 - 1 = (x - 1)(x - w)(x - w^2)[/itex] is also divisible by [itex]1-w = \lambda[/itex], and hence also divisible by [itex]\lambda[/itex]. Since this is divisible by [itex]\lambda[/itex], this means it is also divisble by [itex]\lambda^3[/itex] (This last line seems weak).

Afterwards I'll just explain what I think the mods for B and C should be.

Just wow... are you a native English speaker?

I told you, you need to show separately that x-1, x-w, and x-w^2 are all divisible by lambda. That's 3 statements to show.

my god..

if x = 1 (mod lambda) then x = 1 + a*(1 - w) + b*(1 - w)^2 for some integers a,b.

Then x-1 = a*(1 - w) + b*(1 - w)^2 is divisible by 1-w clearly.

x-w = 1 - w + a*(1 - w) + b*(1 - w)^2 = (1-w)(1 + a + b*(1 - w)) is divisible by 1-w clearly.

Finally,
x-w^2 = 1 - w^2 + a*(1 - w) + b*(1 - w)^2 = (1 - w)(1 + w) + a*(1 - w) + b*(1 - w)^2 = (1 - w)(1 + w + a + b(1 - w)) is divisible by 1-w.

So all of x-1, x-w, and x-w^2 are divisible by lambda. Do you not see that (x - 1)(x - w)(x - w^2) is now divisible by lambda^3 ?

I take it to alter this for part B, all I would need to do is flip the + and - signs for part A's?

interesting thread. ive tried my best to follow along and i believe that this is all you need to do. hochs should confirm it first though as hes beek working with you this whole time.