TITLE: Cyclotomic polynomial
QUESTION [5 upvotes]: I want to prove basic results of cyclotomic polynomials over $\mathbb{Q}$.

$(1)\ \Phi_n(0)=1$ for all $n>1$.
$(2)\ \Phi_{2n}(x)=\Phi_n(-x)$ for all odd numbers $n>1$.

I want use this result as definition: For every positive integer $n$,
$$ x^{n}-1=\prod_{d|n} \Phi_{d}(x)$$
where $d$ runs over all positive divisors of $n$.

REPLY [10 votes]: 1. Remark that $n=1$ is an exception, since $\Phi_1(0) = -1$. Since $x^n-1 = \prod_{d \mid n} \Phi_d(x)$, putting $x=0$ on both sides yields $$-1 = \prod_{d\mid n} \Phi_d(0).$$ Now use induction on $n$, and the fact that $\Phi_1(0)$ accounts for a factor of $-1$ on the right-hand side.

2. Remark that $n=1$ is again an exception, because $\Phi_2(x) = x+1 = -\Phi_1(-x)$. Now, for a general odd $n$,
$$\prod_{d \mid 2n} \Phi_d(x) =x^{2n}-1 = (x^n-1)(x^n+1) = -(x^n-1)((-x)^n-1) = -\prod_{d \mid n} \Phi_d(x)\Phi_d(-x).$$
Now we use induction on $n$. The divisors of $2n$ come in pairs $(d, 2d)$, where $d$ is a divisor of $n$. By the induction hypothesis, $\Phi_{2d}(x) = \Phi_d(-x)$ for each proper divisor of $n$, except when $n=1$ when there is an extra minus sign. By comparing both sides it follows that $\Phi_{2n}(x) = \Phi_n(-x)$.