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 nth cyclotomic polynomial.


We begin with a lemma.

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

Let \zeta be a primitive nth 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 nth root of unity, with n odd, the powers of \zeta are nth 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 2nth 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).

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