2022 AMC 12A Problem 21

Question

Let

P(x)=x^{2022}+x^{1011}+1.

Which of the following polynomials divides

P(x)?

x^2-x+1

x^2+x+1

x^4+1

x^6-x^3+1

x^6+x^3+1

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.

Problem 21 in other years