What kind of recursion is this?

Consider the sequence of vectors $\dbinom{a_n}{b_n}.$ If we write our recursions as

$\begin{aligned} a_{n+1} &= (\sqrt{6} - \sqrt{2})a_n + (-\sqrt{6} - \sqrt{2})b_n\\ b_{n+1} &= (\sqrt{6} + \sqrt{2})a_n + (\sqrt{6} - \sqrt{2})b_n, \end{aligned}$

we realize that we can represent $\dbinom{a_{n+1}}{b_{n+1}}$ as a product of a matrix and the vector $\dbinom{a_n}{b_n}.$ We have

$\begin{aligned} \binom{a_{n+1}}{b_{n+1}} &= \binom{(\sqrt{6} - \sqrt{2})a_n + (-\sqrt{6} - \sqrt{2})b_n}{(\sqrt{6} + \sqrt{2})a_n + (\sqrt{6} - \sqrt{2})b_n}\\ &= \begin{pmatrix} \sqrt{6} - \sqrt{2} & -\sqrt{6} - \sqrt{2}\\ \sqrt{6} + \sqrt{2} & \sqrt{6} - \sqrt{2} \end{pmatrix} \binom{a_n}{b_n} \\ &= 4 \begin{pmatrix} (\sqrt{6} - \sqrt{2})/4 & (-\sqrt{6} - \sqrt{2})/4 \\ (\sqrt{6} + \sqrt{2})/4 & (\sqrt{6} - \sqrt{2})/4 \end{pmatrix} \binom{a_n}{b_n}. \end{aligned}$

Since $\cos 75^{\circ} = \dfrac{\sqrt{6} - \sqrt{2}}{4}$ and $\sin 75^{\circ} = \dfrac{\sqrt{6} + \sqrt{2}}{4},$

$\binom{a_{n+1}}{b_{n+1}} = 4 \begin{pmatrix} \cos 75^{\circ} & -\sin 75^{\circ} \\ \sin 75^{\circ} & \cos 75^{\circ} \end{pmatrix} \binom{a_n}{b_n}.$

Let $A = \begin{pmatrix} \cos 75^{\circ} & -\sin 75^{\circ} \\ \sin 75^{\circ} & \cos 75^{\circ} \end{pmatrix} .$ Notice that, when a vector is multiplied by $A$ , the vector is rotated $75^{\circ}$ counterclockwise. We have

$\binom{a_{2016}}{b_{2016}} = 2^2 A \binom{a_{2015}}{b_{2015}} = 2^2 A \left (2^2 A \binom{a_{2014}}{b_{2014}} \right )= \cdots = 2^{4032} A^{2016} \binom{a_0}{b_0}.$

$A^{2016}$ is a counterclockwise rotation of $75^{\circ}$ $2016$ times. Since $2016 = 36 \times 2 \times 28$ and $75 = 5 \times 15,$ $2016 \times 75$ is a multiple of $360,$ so $A^{2016}$ is just the identity matrix, and this term vanishes. We are finally left with

$\binom{a_{2016}}{b_{2016}} = 2^{4032} \binom{a_0}{b_0} = \binom{2^{4032}a_0}{2^{4032}b_0}.$

If two vectors are equal, then their components must be equal, so $a_{2016} = 2^{4032}a_0$ and $b_{2016} = 2^{4032}b_0.$ Finally,

$a_{2016}b_{2016} = 2^{4032}a_0(2^{4032}b_0) = 2^{8064}a_0b_0 = 2^{8064},$

so $x = \boxed{8064}.$