Let be a root of the equation 1+x^2+x^4=0.

Question

Let $\alpha$ be a root of the equation $1+x^2+x^4=0$. Then the value of $\alpha^{1011}+\alpha^{2022}-\alpha^{3033}$ is equal to:

Accepted Answer

$1+x^2+x^4=0$ means $y=x^2$ satisfies $1+y+y^2=0$, so $\alpha^2=\omega$ or $\omega^2$ (a primitive cube root of unity). Take $\alpha^2=\omega$.

$$\alpha^{1011}+\alpha^{2022}-\alpha^{3033} = \alpha(\alpha^2)^{505}+(\alpha^2)^{1011}-\alpha(\alpha^2)^{1516} = \alpha\omega^{505}+\omega^{1011}-\alpha\omega^{1516}.$$

Since $505\equiv1,\ 1011\equiv0,\ 1516\equiv1\pmod3$:
$$=\alpha\omega+1-\alpha\omega=1.$$

Solutions · 1