Find an equation knowing its solutions

$x=\sqrt{6-3\sqrt{2+\sqrt{3}}} - \sqrt{2+\sqrt{2+\sqrt{3}}} = \sqrt{3}\sqrt{2-\sqrt{2+\sqrt{3}}} - \sqrt{2+\sqrt{2+\sqrt{3}}}$

$\Rightarrow x^2= 3(2-\sqrt{2+\sqrt{3}}) + 2 + \sqrt{2+\sqrt{3}} - 2\sqrt{3}\sqrt{(2-\sqrt{2+\sqrt{3}})(2+\sqrt{2+\sqrt{3}})}$

$\Leftrightarrow x^2= 8-2\sqrt{2+\sqrt{3}} - 2\sqrt{3}\sqrt{2-\sqrt{3}}$

$\Leftrightarrow x^2- 8 = -2(\sqrt{2+\sqrt{3}} +\sqrt{3}\sqrt{2-\sqrt{3}})$

$\Rightarrow (x^2-8)^2 = 4(\sqrt{2+\sqrt{3}} +\sqrt{3}\sqrt{2-\sqrt{3}})^2$

$\Leftrightarrow x^4-16x^2+64 = 4[2+\sqrt{3} + 3 (2-\sqrt{3}) + 2\sqrt{3}\sqrt{(2+\sqrt{3})(2-\sqrt{3})}]$

$\Leftrightarrow x^4 - 16x^2 + 64 = 4(8-2\sqrt{3} + 2\sqrt{3}) = 4\times8 = 32$

$\Leftrightarrow x^4 -16x^2 + 32 = 0$

$\Rightarrow (a,b) = (-16,32)$

Therefore, the answer is $(-16) + 32 = \boxed{16}$

Let $x_1 = \sqrt{6-3\sqrt{2+\sqrt 3}} - \sqrt{2+\sqrt{2+\sqrt 3}}$ . Then

$\begin{aligned} x_1^2 & = 6 - 3\sqrt{2+\sqrt 3} - 2 \sqrt{3\left(2-\sqrt{2+\sqrt 3}\right)\left(2+\sqrt{2+\sqrt 3}\right)}+ 2+\sqrt{2+\sqrt 3} \\ & = 8 - 2\sqrt{2+\sqrt 3} - 2 \sqrt{6-3\sqrt 3} \\ & = 8 - 2\sqrt{\left(\sqrt{2+\sqrt 3} + \sqrt{6-3\sqrt 3}\right)^2} \\ & = 8 -4\sqrt 2 \end{aligned}$

We note that $x_1^2$ is a root of $(x^2)^2 + a x^2 + b = 0$ , a quadratic equation of $x^2$ . Since coefficients $a$ and $b$ are rational, we can expect the other root of the quadratic equation be $x_2^2 = 8 + 4\sqrt 3$ and the quadratic equation is:

$\begin{aligned} \left(x^2 - 8 + 4\sqrt 2\right)\left(x^2 - 8 - 4\sqrt 2\right) & = 0 \\ x^4 - 16x^2 + 32 & = 0 \end{aligned}$

Therefore $a+b = -16 + 32 = \boxed{16}$ .