Commutative Rings and units

**dlbsd** · January 14th 2010, 11:13 AM

Let R = { $m+n*\sqrt{2}$ | $m,n \in Z$ }

Q: Show that $1+2\sqrt{2}$ has infinite order in $R^x$

From previous part of the Problem:
$m+n*\sqrt{2}$ is a unit in R if and only if $m^2-2n^2= \pm{1}$

I'm a little confused by this since according to the above theorem $1+2\sqrt{2}$ is not a unit

What am I missing? Is it actually a unit?

**dlbsd** · January 14th 2010, 11:57 AM

disregard this thread. Its a typo in the book

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