Abstract Algebra: Question About the Elements in U(n)

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Homework Statement


For any integer n>2, show that there are at least two elements in U(n) that satisfy x^2 = 1.


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The Attempt at a Solution


If the definition of the group U(n) is "the set of all positive integers less than n and relatively prime to n" then the group U(5) has elements {1,2,3,4}. Clearly then, 1^2 = 1. But I can't drum up any way to get any other elements in the group to equal 1 when squared.

The set U(n) is a group under modulo n, but I am pretty weak on modular arithmetic, and am trying to spend the day boning up on it. So for now, I don't understand how to find another element such that x^2 = 1.
 
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I'm starting to figure this one out. I chose the case where n = 5. So the set of elements for U(n) = {1,2,3,4}. And under mod 5, 1^2 - 1 and 4^2 = 1.

I'm trying other mod n's, and it is looking like the integers which will equal 1 when squared are 1, and the largest integer in the set.

I don't know how to generalize this for the arbitrary case though. But I'm trying to screw around with a quadratic equation to see where I can go. Since the largest element in U(n) = n-1, I am looking for an mod n such that (n-1)^2 = 1. But I'm not getting very far with this line of thought.
 
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Here's one way to look at it. In the integers, the equation x2 = 1 has two solutions, namely 1 and -1. Well, this is actually true in U(n) as well, since U(n) is really the multiplicative group (\mathbb{Z}/n\mathbb{Z})^{\times}. So how do you write -1 in \mathbb{Z}/n\mathbb{Z}?
 
(n-1)^2=n^2-2n+1. Reduce those three terms mod n.
 
Dick said:
(n-1)^2=n^2-2n+1. Reduce those three terms mod n.

I've been searching through my book and the internet, and I can't figure out how to do this. I never learned modular arithmetic in my proofs class, so I am having to learn it on the fly in my modern algebra class. Is there a step in the right direction you could give me on this one?
 
Two numbers are equal mod n if they differ by an even multiple of n. Can you tell if 0 and n^2 are equal mod n?
 
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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