Proving Equivalence Classes in Modular Arithmetic

rallycar18
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Homework Statement



Suppose [d], \in Z sub n.
 
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The good news here is, that you can perform all operations in Zm by picking any representative and working in the integers.

So if c sits in the conjugacy class of a, you can write
c = a + i m
where 0 <= a < m and i is some integer, similarly
d = b + j m
 
If cd is congruent to ab mod m, then cd lies in [ab] by definition, doesn't it?
If you want a less trivial proof, show that
cd = ab + km
for some integer k.
 
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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