Number theory. modulu question.

[cap.r]

Homework Statement

x cong 1(mod m^k) implies x^m cong 1(mod m^(k+1))

Homework Equations

x cong 1(mod m^k) <=> m^k|x-1 <=> ym^k=x-1

The Attempt at a Solution

starting with ym^k=x-1 add one to both sides
ym^k+1=x now rise to the power m.

(ym^k+1)^m=x^m <=> subtract the 1^m from the end of the expansion to get
x^m-1^m-(...)=(y^m)(m^(km)) where (...) is the rest of the binomial expansion.

I am stuck here. somehow I need to get a m^(k+1) on the LHS and say it divides everything on the RHS. I don't think my approach is the best.

 

[Hurkyl]

somehow I need to get a m^(k+1) on the LHS and say it divides everything on the RHS. I don't think my approach is the best.

Wouldn't it be much easier to just reduce everything modulo m^k+1?

 

[cap.r]

you mean reduce everything in the binomial expansion mod m^k+1?

 

[Hurkyl]

Yep.

 

[cap.r]

so i couldn't figure out what you meant or maybe i did... but instead of doing the binomial expansion i factored the x^m-1.

so I have m^(k+1)|x^m-1 and I expanded x^m-1 to give me m^(k+1)|(x-1)(x^m+x^(m-1)+...+x+1).

now (x-1) is m^k so I have m^(k+1)|m^k(x^m+...+x+1) I need to somehow pull an m out of (x^m+...+x+1) to make this work but I don't know how.