
#19
Nov110, 07:26 PM

P: 39

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 x1 = a*(1  w) + b*(1  w)^2 is divisible by 1w clearly. xw = 1  w + a*(1  w) + b*(1  w)^2 = (1w)(1 + a + b*(1  w)) is divisible by 1w clearly. Finally, xw^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 1w. So all of x1, xw, and xw^2 are divisible by lambda. Do you not see that (x  1)(x  w)(x  w^2) is now divisible by lambda^3 ? 



#20
Nov110, 07:26 PM

P: 77

If you remember, I told you in several posts that I had trouble understanding your translation from using [itex]a,b,[/itex] and [itex]w[/itex] to just [itex]x[/itex] and [itex]w[/itex]. I'm strill trying to make that connection, then I could better attempt explaining why [itex]\lambda[/itex] divides [itex]x1,xw,[/itex] and [itex]xw^2[/itex]. 



#21
Nov110, 07:31 PM

P: 77

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? 



#22
Nov110, 11:57 PM

P: 7





#23
Nov510, 11:54 PM

P: 77




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