Proving Prime p≥3 Satisfies pr ≡ 1, 5, 7 or 11 (mod 12)

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Homework Statement



If p is a prime and p>3, show that pr\equiv1,5,7 or 11 (mod12)

Homework Equations


The Attempt at a Solution


Do I go about this by knowing that any prime p greater than 3 is of the form 6n+1 or 6n+5? Any direction on how to go about this will be helpful. Thanks.
 
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Amannequin said:

Homework Statement



If p is a prime and p>3, show that pr\equiv1,5,7 or 11 (mod12)

Homework Equations


The Attempt at a Solution


Do I go about this by knowing that any prime p greater than 3 is of the form 6n+1 or 6n+5? Any direction on how to go about this will be helpful. Thanks.

Show any other possibility doesn't work. For example, suppose ##p^r\equiv 2 (mod 12)##. What's wrong with that?
 
If k is an odd number, list the possibilities for k mod 12.
Next if p is a prime > 3, list the possibilities for p mod 12.
Then what conclusion can you draw about ##p^r##?
 
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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