<s>[INST] Groups of units of $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ On page 230 of Dummit and Foote's Abstract Algebra, they say: the units of $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ are determined by the integers $a,b$ with $a^2+ab+b^2=\pm1$ i.e. with $(2a+b)^2+3b^2=4$, from which is is easy to see the group of units is a group of order $6$ given by $\{\pm1,\pm\rho,\pm\rho^2\}$ where $\rho=\frac{-1+\sqrt{-3}}{2}$.
First, why change the characterization of unit from integers solutions of $a^2+ab+b^2=\pm1$ to integers solutions of $(2a+b)^2+3b^2=4$? How did they arrive at their answer?
 [/INST] Let $\omega=\frac{1+\sqrt{-3}}{2}$.  Notice that the norm of an element $a+b\omega$ in $\mathbb{Z}[\omega]$ is $$(a+b\omega)(a+b\overline{\omega})=a^2+ab+b^2.$$  Since the norm is multiplicative, all units must have norm $1$, otherwise they would not be invertible.  This means we are looking for solutions to $a^2+ab+b^2=1$.  (It can't be $-1$, I don't know why you put $\pm$ for all these quantities.) 
For the second part, the author likely found it nicer to shift the quadratic form, as we need not consider different cases.  Here is a direct way:  To solve $a^2+ab+b^2=1$, split based on whether $ab$ negative or nonnegative.
Case 1: If $ab$ is nonnegative, then $1=a^2+ab+b^2\geq a^2 +b^2$, so we see that $a=\pm 1$, $b=0$ and $a=0$, $b=\pm 1$ are the only solutions.
Case 2: If $ab$ is negative, then both $a$ and $b$ are not zero, and then $1=a^2+ab+b^2=a^2+2ab+b^2-ab=(a+b)^2 -ab$.  This then gives the only other solutions, $a=1=-b$, and $b=1=-a$.
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