Ring Theory: Proving $\mathbb{Z} [ \sqrt{2} ]$ has Infinitely Many Units

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Show \mathbb{Z} [ \sqrt{2} ] = \{ a + b \sqrt{2} | a,b \in \mathbb{Z} \} has infinitely many units.

I started by taking an element:

a + b \sqrt{2} \in \mathbb{Z} [ \sqrt{2} ]

and finding an inverse

\left( a + b \sqrt{2} \right) ^{-1}

such that the product gives zero and tried to show any element works. But I'm not sure about doing this.
 
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The product should give 1.
 
And I found the inverse and I didn't see an infinite number of units. Z is the integers, right? What did you get for the inverse?
 
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Dick said:
And I found the inverse and I didn't see an infinite number of units. Z is the integers, right? What did you get for the inverse?

Ooops. My mistake. There are more units than just 1 and -1. Can you find some? Once you've found one that isn't 1 or -1, can you think of a simple way to use it to generate more?
 
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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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