1
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 $a = \\pm 1, b = 0\\text{.}$