## Congruence Class Question

 Quote by Brimley So I should have the following then (for a truly complete explanation): A. $x \equiv 1 (mod \lambda)$, then $x^3 \equiv 1(mod \lambda)^3$. If $x = 1 (mod \lambda)$, then $x = 1 + a*(1 - w) + b*(1 - w)^2.$ $x - 1 = a*(1 - w) + b*(1 - w)^2$ $= (1 - w)*(a + b*(1 - w))$. Clearly, this is divisible by $1 - w = \lambda$. From here, we can see that: $x^3 - 1 = (x - 1)(x - w)(x - w^2)$ is also divisible by $1-w = \lambda$, and hence also divisible by $\lambda$. Since this is divisible by $\lambda$, this means it is also divisble by $\lambda^3$ (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..
 Alright I'm wasting way too much time on this triviality. Here's the complete ****ing proof, leave me alone now: 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 ?

 Quote by hochs 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..
I'm sorry about your frustration hochs! I'm really trying here! (Btw, it takes a ton of time to write it in LaTeX, but I think its easier on the eyes of those reading it herein). I'm putting in the time and effort to understand this, so I do appreciate the patience!

If you remember, I told you in several posts that I had trouble understanding your translation from using $a,b,$ and $w$ to just $x$ and $w$. I'm strill trying to make that connection, then I could better attempt explaining why $\lambda$ divides $x-1,x-w,$ and $x-w^2$.

 Quote by hochs 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 do see it now. I really appreciate it, but please don't get mad at me. I'm just trying to understand what you were doing, and I had most of it but was confused in the translation. Seeing that performed makes sense of it. I do appreciate it and I understand your frustration, but please don't get mad at someone who is just trying to learn mathematics.

I take it to alter this for part B, all I would need to do is flip the + and - signs for part A's?
Is there a quick explanation on why part C's is trivial?

 Quote by Brimley 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.

 Quote by stoolie77 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.
Can anyone help me finish this? Stoolie I appreciate your input, but yes, I'd really appreciate if someone could confirm this or give me a hand finishing this.