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Since $x$ is the Golden Ratio ,

$a_1=F_{6006}$

$b_2=F_{12354}$

Then,

$\gcd(a_1,b_2)$

$=\gcd(F_{6006},F_{12354})$

$=F_{gcd(6006,12354)}$

$=F_6$

$=8$

$\begin{aligned} x&=\frac{1+\sqrt5}{2}\\ 2x&=1+\sqrt5\\ 2x-1&=\sqrt5\\ 4x^2-4x+1&=5\\ 4x^2-4x&=4\\ x^2&=x+1 \end{aligned}$

$\begin{aligned} x^3&=x^2x\\ &=(x+1)x\\ &=x^2+x\\ &=(x+1)+x\\ &=2x+1 \end{aligned}$

$\begin{aligned} x^1&=1x+0\\ x^2&=1x+1\\ x^3&=2x+1\\ &\,\,\,\vdots\\ x^n&=F_nx+F_{n-1} \end{aligned}$

$x^{6006}=F_{6006}x+F_{6005}=a_1x+b_1$

$\Rightarrow\left(a_1,b_1\right)=(F_{6006},F_{6005})$

$x^{12355}=F_{12355}x+F_{12354}=a_2x+b_2$

$\Rightarrow\left(a_2,b_2\right)=(F_{12355},F_{12354})$

$\begin{cases} 6006&=2\times3\times7\times11\times13\\ 12354&=2\times3\times29\times71 \end{cases}$

$\begin{aligned} \gcd(a_1,b_2)&=\gcd(F_{6006},F_{12354})\\ &=F_{\gcd(6006,12354)}\\ &=F_{2\times3}\\ &=F_6\\ &=\boxed{8} \end{aligned}$

for the highest common factor for x is the golden mean = 8

for the highest common factor for x is the golden mean = 8