Монгол Бодлогын Сан

Заавар: $$\alpha^2+a-1=0\Rightarrow \alpha^2+\alpha=1$$

Бодолт: $\alpha^2+\alpha=1$ тул \begin{align*} \text{Илэрх.}&=\alpha^8+\alpha^7+\alpha^5+\alpha^3+\alpha+9\\ &=\alpha^6(\alpha^2+\alpha)+\alpha^5+\alpha^3+\alpha+9\\ &=\alpha^6+\alpha^5+\alpha^3+\alpha+9\\ &=\alpha^4(\alpha^2+\alpha)+\alpha^3+\alpha+9\\ &=\alpha^4+\alpha^3+\alpha+9\\ &=\alpha^2(\alpha^2+\alpha)+\alpha+9\\ &=\alpha^2+\alpha+9=1+9\\ &=10 \end{align*}