for $n > 1$ : odd , prove that $\Phi_{2n}(x) = \Phi_{n}(-x)$

Question

for $n > 1$ : odd , prove that $\Phi_{2n}(x) = \Phi_{n}(-x)$

Note: $\Phi_n(x)$ is the $n$th cyclotomic polynomial whose roots are the primitive $n$th roots of unity

if n is odd then $-1$ cannot be an $n$th root of unity, easily checked by

$(-1)^n-1 = -1 - 1 = -2$ so need a primitive $n$th root, say $\xi$ such that $(-\xi)^k = (-1)^k(\xi)^k = 1$ and so must have $(-1)^k=1$

this is where I am stuck

Answer 1

Suppose we start from $$\Phi_n(x) = \prod_{d|n} (x^d-1)^{\mu(n/d)}.$$ With $n$ odd we get $$\begin{array}{lcl}\Phi_{2n}(x) & = & \prod_{d|2n} (x^d-1)^{\mu(2n/d)} \\ & = & \prod_{d|n} (x^d-1)^{\mu(2n/d)} \prod_{d|n} (x^{2d}-1)^{\mu(2n/d/2)} \\ & = & \prod_{d|n} (x^d-1)^{-\mu(n/d)} \prod_{d|n} (x^{2d}-1)^{\mu(n/d)} \\ & = & \prod_{d|n} (x^d+1)^{\mu(n/d)}. \end{array}$$ This is $$(-1)^{\sum_{d|n} \mu(n/d)}\prod_{d|n} (-x^d-1)^{\mu(n/d)}.$$ To conclude recall that $\sum_{d|n} \mu(d) = 0$ when $n\gt 1$ and with $d$ odd we have $-x^d = (-x)^d$ which yields $$\prod_{d|n} ((-x)^d-1)^{\mu(n/d)}$$ which is indeed $\Phi_n(-x).$

Answer 2

Hint: The primitive $n$th roots of unity are $e^{2\pi i k/n}$ for $(k,n) = 1$, while the primitive $2n$th roots of unity are $e^{2\pi i \ell/2n}$ for $(\ell,2n)=1$. If $(k,n) = 1$ then $(2k+n,2n)=1$. Therefore $$ \Phi_{2n}(x) = \prod_{(k,n)=1} (x-e^{2\pi i(2k+n)/2n}) = \prod_{(k,n)=1} (x+e^{2\pi i k/n}) = \prod_{(k,n)=1} (-x-e^{2\pi i k/n}) = \Phi_n(-x). $$