Modulo Reduction Using Fermat's Little Theorem

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Homework Statement



Reduce 34567 modulo 19.

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The Attempt at a Solution


I approached by first reducing 4567 modulo 18.

I got the following: 34567= 318*253+13=(318)253*313 congruent to 12531594323 congruent to 14 mod 19

Is this the correct approach? I am not sure where 14 came from in the end I just guess and checked until I found a value that worked which was 14. Any explanations on how to correctly find 14?
 
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You were exactly right to use Fermat's Little Theorem to show that 34567 = 313 mod 19. From there I'd use binary exponentiation: 313 = 38+4+1 = 3834*3.

So mod 19, 34 = 5, so then 38 = 3434 = 25 mod 19 = 6 mod 19.

Put it together: 3834*3 = 6*5*3 mod 19 = 90 mod 19 = 14.
 
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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