Solution
= Solution
For odd $n>1$, the primitive $2n$th roots are exactly the negatives of primitive $n$th roots. Since $\varphi(2n)=\varphi(n)$, their monic root <polynomials> satisfy $\Phi_{2n}(X)=\Phi_n(-X)$.
= Solution
For odd $n>1$, the primitive $2n$th roots are exactly the negatives of primitive $n$th roots. Since $\varphi(2n)=\varphi(n)$, their monic root <polynomials> satisfy $\Phi_{2n}(X)=\Phi_n(-X)$.