For odd $p$, doubling the index alternates the signs: $\Phi_{2p}(x)=\Phi_p(-x)$.Suggest a slogan
For an odd prime \(p\), the cyclotomic polynomial \(\Phi_{2p}\) is \[ \Phi_{2p}(x)=\Phi_p(-x)=1-x+x^2-\cdots+x^{p-1}. \]
Since \(p\) is odd, \(-1\) has order \(2\) and \(\zeta\mapsto-\zeta\) is a bijection from the primitive \(p\)th roots of unity to the primitive \(2p\)th roots of unity. Hence \[ \Phi_{2p}(x)=\prod_\zeta(x+\zeta)=(-1)^{p-1}\prod_\zeta(-x-\zeta)=\Phi_p(-x), \] the products running over the primitive \(p\)th roots of unity \(\zeta\), because \(p-1\) is even. Substituting \(-x\) in \(\Phi_p(x)=1+x+\cdots+x^{p-1}\) (FF-MUJDE) gives the sum.
For \(p=2\) the formula fails: \(\Phi_4(x)=x^2+1\), whereas \(\Phi_2(-x)=1-x\).
What is the cyclotomic polynomial $\Phi_{2p}(x)$?