Integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt3)$

Question

I know that $x = a+b\sqrt{3}$ is a root of $X^2-2aX+a^2-3b^2$.

I found the trace of $a+b\sqrt{3}$ it's $2a = tr \begin{pmatrix} a & b \\ 3b & a \end{pmatrix}$.

Now I see that $a$ should be half-integer for $x$ to be in the closure. The closure contains $a+b\sqrt{3}$ for integer $a,b$.

That is all I can say for now.

What can I do next?