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Let $z = a + b \sqrt{3}\, i$ be in ${\mathbb Z}[ \sqrt{3}\, i]\text{.}$ If $a^2 + 3 b^2 = 1\text{,}$ show that $z$ must be a unit. Show that the only units of ${\mathbb Z}[ \sqrt{3}\, i ]$ are 1 and $-1\text{.}$
Note that $z^{-1} = 1/(a + b\sqrt{3}\, i) = (a -b \sqrt{3}\, i)/(a^2 + 3b^2)$ is in ${\mathbb Z}[\sqrt{3}\, i]$ if and only if $a^2 + 3 b^2 = 1\text{.}$ The only integer solutions to the equation are $a = \pm 1, b = 0\text{.}$