Working With Polynomials

From $a^4 + a^3 + a^2 + a + 1 = 0$ we find that $a\neq1$ , i.e. $a-1\neq0$ . From $a(a^4 + a^3 + a^2 + a + 1)-(a^4 + a^3 + a^2 + a + 1)$ = $a^5 + a^4 + a^3 + a^2 + a - a^4 - a^3 - a^2 - a - 1 = a^5-1;$ we find that $a^5 = 1$ . Therefore $a^{2000} + a^{2010} + 1 = (a^5)^{400} + a(a^5)^{402} + 1 = 1 + 1 + 1 = 3$ :

$\begin{aligned} a^4+a^3+a^2+a+1 & = 0 & \small \color{#3D99F6} \text{Multiplying both sides by }x \\ a^5+ a^4+a^3+a^2+a & = 0 & \small \color{#3D99F6} \text{Adding } 1 \text{ both sides} \\ a^5+ \color{#3D99F6} a^4+a^3+a^2+a + 1& = 1 & \small \color{#3D99F6} \text{Note that } a^4+a^3+a^2+a+1 = 0 \\ \implies a^5 & = 1 \end{aligned}$

Therefore, $a^{2000}+a^{2010} + 1 = \left(a^5\right)^{400} +\left(a^5\right)^{402} + 1 = 1+1+1 = \boxed{3}$

Relevant wiki: De Moivre's Theorem - Roots

$\text{The Nth Rule theorem says that if}$

$1 + x + x^2 + x^3 + .... x^{n-1}= 0$

$\text{Then}$ $x^n = 1$

$\text{So Here We get}$ $a^5 = 1$ $\text{from the theorem mentioned}$ . $\text{Or x is the 5th root of unity}$

$\text{Hence we}$ $a^{2000} + a^{2010} + 1 = (a^5)^{400} + a(a^5)^{402} + 1 = 1 + 1 + 1 = \boxed{3}$ $\text{which can be done mentally }$