Prove that x^100 = 1 for all x in U(1000)

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Homework Statement



Prove that x^100 = 1 for all x in U(1000)

Homework Equations





The Attempt at a Solution



U(1000) is all numbers relatively prime to 1000. And that is as far as I have gotten.

Thanks
 
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Are you sure you haven't missed a part of the problem...?

x=1001 is relatively prime to 1000, but 1001^100 is not equal to one.
 
Huh? Well U(1000) is all numbers less than a 1000 and relatively prime to 1000, that changes things, at least with your counter example...forgot to mention that
 
What about 3, 7, etc? These are all relatively prime to 1000 (and also less than 1000), yet neither 3^100 nor 7^100 are equal to 1. You are most likely missing some part of the problem.

Edit: You probably meant this - "Prove that x^100 = 1 mod(1000) for all x in U(1000)." Try using Euler's Theorem.
 
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Well that is the problem how it is stated...but the idiot I am keeps forgetting to mention all the details of U(n)...it is the group of units modulo n (that is the set of integers less than n and relatively prime to n under multiplication modulo n). But what you say is what I want.
I for example take 3^100=1mod(1000) , 1000 should divide (3^100 - 1) but it does not, you end up with a decimal...so I can't even see what the problem looks like when you plug in a number to x
 
nowimpsbball said:
Well that is the problem how it is stated...but the idiot I am keeps forgetting to mention all the details of U(n)...it is the group of units modulo n (that is the set of integers less than n and relatively prime to n under multiplication modulo n). But what you say is what I want.
I for example take 3^100=1mod(1000) , 1000 should divide (3^100 - 1) but it does not, you end up with a decimal...so I can't even see what the problem looks like when you plug in a number to x

It does divide 3^100 - 1. Not sure what you mean by getting a decimal. If you have Windows, try the built-in calculator and calculate 3^100 mod(1000). You'll get 1, which verifies the equation 3^100=1mod(1000). If you calculate 3^100 - 1 mod(1000), you will get 0, which means that 1000 evenly divides 3^100 - 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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