2022 AMC 12A 第 21 题

A divisor $D(x)$ works iff every root $\zeta$ of $D$ satisfies $\zeta^{2022}+\zeta^{1011}+1=0$

For $x^6+x^3+1$ the roots are primitive $9$th roots of unity, so $\zeta^3$ is a primitive cube root of unity

If $\zeta$ is a root of a divisor, then $P(\zeta)=\zeta^{2022}+\zeta^{1011}+1=0$ is required.

The roots of $x^6+x^3+1$ are the primitive $9$th roots of unity, so $\zeta^9=1.$ Reducing exponents modulo $9,$ $2022\equiv6$ and $1011\equiv3,$ giving $\zeta^6+\zeta^3+1.$ Here $\omega=\zeta^3$ is a primitive cube root of unity, so this equals $\omega^2+\omega+1=0.$

The other four options fail: substituting their roots yields nonzero values (for instance, the primitive cube roots of unity give $P=3$).

Thus, the correct answer is E.