## A fact about cyclotomic polynomials

Let $n > 1$ be an odd integer. Prove that $\Phi_{2n}(x) = \Phi_n(-x)$, where $\Phi_n(x)$ denotes the $n$th cyclotomic polynomial.

We begin with a lemma.

Lemma: -1 is not an $n$th root of unity for odd $n$. Proof: $(-1)^n-1=-1-1=-2$. $\square$

Let $\zeta$ be a primitive $n$th root of unity. Now suppose $(-\zeta)^t = 1$, so that $(-1)^t\zeta^t = 1$, so $\zeta^t = (-1)^t$. Since $\zeta$ is a primitive $n$th root of unity, with $n$ odd, the powers of $\zeta$ are $n$th roots of unity. By the lemma, we must have $(-1)^t = 1$, so that $2|t$, and thus $\zeta^t=1$, so that $n|t$. Since $2$ and $n$ are relatively prime, $2n|t$, and so $-\zeta$ is a primitive $2n$th root of unity.

In particular, $\Phi_n(-x)$ divides $\Phi_{2n}(x)$.

Now $\Phi_n(-x)$ has degree $\varphi(n)$, and $\Phi_{2n}(x)$ has degree $\varphi(2n)$, where $\varphi$ denotes the Euler totient. Since $2$ and $n$ are relatively prime, we have $\varphi(2n) = \varphi(2)\varphi(n) = \varphi(n)$. So in fact $\varphi_{2n}(x) = \varphi_n(-x)$.