$a+b\sqrt{3}=\sqrt{21-12\sqrt{3}}, a,b \in \mathbb {Z}$ Find a+b

Question

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.

Answer 1

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

Answer 2

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 $-3+2\sqrt{3}$

Answer 3

$$\implies21-12\sqrt3=a^2+3b^2+2\sqrt3ab$$

$$\implies a^2+3b^2=21, ab=-6$$

$$\implies3b^2+\dfrac{36}{b^2}=21\iff b^4-7b^2+12=0$$

As $b \in \mathbb {Z},b^2\ne3\implies b^2=4$

But $\sqrt{21-12\sqrt3}>0$ and $3+\sqrt3(-2)<0$