Solving for Solutions in Z12: X^2=4

mikee
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Homework Statement

Find all solutions in Z12 to the equation X^2=4



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The Attempt at a Solution

This is my attempt, since it is in Z12, i wrote X^2=4(modulo 12) and by this the answers are trivial, X= 2, 4,8,... and also -2,-4...
this just seems to easy so i was wondering if i was doing it wrong
 
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That's correct. It can't be any of the odd numbers because the odds squared will yield an odd number, which will still be an odd number mod 12.

0^2 = 0 mod 12 and 6^2 = 36 = 0 mod 12, so your answer is right.

Although you have to be careful with the ..., that could be interpreted as an exponential series, what you're saying is the numbers that are congruent to 2,4,8 and 10 mod 12.
 
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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