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.
$\endgroup$3 Answers
$\begingroup$$$\sqrt{21-12\sqrt3}=\sqrt{12-12\sqrt3+9}=\sqrt{(2\sqrt3-3)^2}=2\sqrt3-3,$$ which gives $a=-3$,$b=2$ and $a+b=-1$.
$\endgroup$ $\begingroup$Write
$$\left(a+b\sqrt{3}\right)^2=a^2+3b^2+2ab\sqrt{3}=21-12\sqrt{3}$$
This means
$$\begin{cases} a^2+3b^2&=21\\ 2ab&=-12 \end{cases}$$
Now substitute $a=-6/b$. You get the following biquadratic
$$b^4-7b^2+12=0$$
Has two solution for $b^2$, $3$ to be discarded because $b$ is an integer and $4$ which means that
$$b=\pm 2\\a=\pm 3$$
The solution to retain is the non negative one ie $-3+2\sqrt{3}$
$\endgroup$ $\begingroup$$$\implies21-12\sqrt3=a^2+3b^2+2\sqrt3ab$$
$$\implies a^2+3b^2=21, ab=-6$$
$$\implies3b^2+\dfrac{36}{b^2}=21\iff b^4-7b^2+12=0$$
As $b \in \mathbb {Z},b^2\ne3\implies b^2=4$
But $\sqrt{21-12\sqrt3}>0$ and $3+\sqrt3(-2)<0$
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