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 \(-1\text{.}\)
Hint.
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 \(a = \pm 1, b = 0\text{.}\)

