$a+b\sqrt{3}=\sqrt{21-12\sqrt{3}}, a,b \in \mathbb {Z}$ Find a+b So far I've reasoned that $\mathbf{a}$ and $\mathbf{b}$ can't be both negative, because $\sqrt{21-12\sqrt{3}}$ cannot be negative. 
Also $\mathbf{a}$ and $\mathbf{b}$ can't be both positive, because $\sqrt{21-12\sqrt{3}}$  is from 0 to 1, thus there is no positive whole numbers which could satisfy that $\mathbf{a}$ plus $\mathbf{b}*\sqrt{3}$ is close to 0 and 1.
At this stage, I don't know what to do. I appreciate any help.

$$\sqrt{21-12\sqrt3}=\sqrt{12-12\sqrt3+9}=\sqrt{(2\sqrt3-3)^2}=2\sqrt3-3,$$
which gives $a=-3$,$b=2$ and $a+b=-1$.
