Solving Number Theory: Showing Congruence Has Exactly k Distinct Solutions

teddyayalew
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Homework Statement



http://i43.tinypic.com/fymy3l.jpg

question 22.4 (a)

Homework Equations





The Attempt at a Solution



xk=(xp-1)m = (xp-1-1)(1 + xp-1 +x2(p-1) +...+x(m-1)(p-1))

I know that xp-1-1 = 0 mod p has p-1 solutions but I can't make anything from the geometric sum. Can someone push me in the right direction.

 
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teddyayalew said:

Homework Statement



http://i43.tinypic.com/fymy3l.jpg

question 22.4 (a)

Homework Equations



The Attempt at a Solution



xk=(xp-1)m = (xp-1-1)(1 + xp-1 +x2(p-1) +...+x(m-1)(p-1))

I know that xp-1-1 = 0 mod p has p-1 solutions but I can't make anything from the geometric sum. Can someone push me in the right direction.
(xp-1-1)(1 + xp-1 +x2(p-1) +...+x(m-1)(p-1)) ≠ (xp-1)m

(xp-1-1)(1 + xp-1 +x2(p-1) +...+x(m-1)(p-1)) = (xp-1)m - 1
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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