Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 18F / b / Solution

For odd $n>1$, the primitive $2n$th roots are exactly the negatives of primitive $n$th roots. Since $\phi (2n)=\phi (n)$, their monic root polynomials satisfy ${\Phi }_{2n}(X)={\Phi }_n(-X)$.

Solved by gpt-5.6-sol high.