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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 $n$.
The OP does not say what additional information may be used, so here are two answers that use Möbius inversion as in $$\Phi_n(x) = \prod_{d|n} (x^d-1)^{\mu(n/d)}$$ since this is a instant consequence of the formula you were given to work with.
Using this formula we have immediately that $$\Phi_n(0) = \prod_{d|n} (-1)^{\mu(n/d)} = (-1)^{\sum_{d|n} \mu(n/d)}.$$ Now for $n>1$ we have $\sum_{d|n} \mu(n/d) = 0$, and hence $\Phi_n(0) = (-1)^0 = 1.$ For $n=1$ we get $\Phi_1(0) = (-1)^1 = -1.$ For the second part start with $$\Phi_n(-x) = \prod_{d|n} ((-x)^d-1)^{\mu(n/d)}.$$ Now since $d$ is odd we have $(-x)^d = (-1)^d x^d = -x^d$, giving $$ \prod_{d|n} (-x^d-1)^{\mu(n/d)} = \prod_{d|n} \left(\frac{x^{2d}-1}{-x^d+1}\right)^{\mu(n/d)} = \prod_{d|n} (-1)^{\mu(n/d)} \prod_{d|n} \left(\frac{x^{2d}-1}{x^d-1}\right)^{\mu(n/d)}\\ = \prod_{d|n} (-1)^{\mu(n/d)} \prod_{d|n} \left(\frac{x^d-1}{x^{2d}-1}\right)^{\mu(2n/d)} \\= \prod_{d|n} (-1)^{\mu(n/d)} \prod_{d|n} \left(\frac{x^d-1}{x^{2d}-1}\right)^{\mu(2n/d)} \prod_{d|n} \left(\frac{x^{2d}-1}{x^{2d}-1}\right)^{\mu(2n/d/2)}\\ = \Phi_{2n}(x) \prod_{d|n} (-1)^{\mu(n/d)} \prod_{d|n} \frac{1}{(x^{2d}-1)^{\mu(2n/d)}(x^{2d}-1)^{\mu(n/d)}}\\ = \Phi_{2n}(x) \prod_{d|n} (-1)^{\mu(n/d)} \prod_{d|n} \frac{1}{(x^{2d}-1)^{\mu(2n/d)+\mu(n/d)}}.$$ Now since $n>1$ we have $\prod_{d|n} (-1)^{\mu(n/d)}=1$ and $\mu(2n/d)+\mu(n/d)=0$ so that the final answer is $$ \Phi_{2n}(x) \times 1 \times \prod_{d|n} \frac{1}{(x^{2d}-1)^0} = \Phi_{2n}(x).$$
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 $\Phi_{2n}(x) = \Phi_n(-x)$.
Here is another answer that uses the basic definition $$\Phi_n(x) = \prod_{1\le k\le n \atop \gcd(k,n)=1} (x-e^{2\pi i k/n}).$$ This gives $$\Phi_n(0) = (-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1} e^{2\pi i k/n}.$$ Now recall that for $n>1$ we have $$\sum_{1\le k\le n \atop \gcd(k,n)=1} k = \frac{1}{2} \varphi(n) n.$$ This gives $$\Phi_n(0) = (-1)^{\varphi(n)} e^{(2\pi i/n) \times 1/2\times \varphi(n) \times n} = (-1)^{\varphi(n)} e^{\pi i \varphi(n)} = (-1)^{2\varphi(n)} =1.$$
For the second part we get that $$\Phi_n(-x) = \prod_{1\le k\le n \atop \gcd(k,n)=1} (-x-e^{2\pi i k/n}) = (-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1} (x+e^{2\pi i k/n}) \\ = (-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1} (x-e^{\pi i + 2\pi i k/n})= (-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1} (x-e^{2\pi i (k/n+1/2)})\\ = (-1)^{\varphi(n)} \prod_{1\le k\le n \atop \gcd(k,n)=1} (x-e^{2\pi i (2k+n)/(2n)}).$$ This almost concludes the proof. We can ignore the sign because $\varphi(n)$ is even.
We now claim there is a bijection between the primitive roots modulo $2n$ call them of class $\mathcal{P}_1$ given by $e^{2\pi i k/(2n)}$ where $\gcd(k,2n)=1$ and $1\le k\le 2n$ and the values produced by $e^{2\pi i (2k+n)/(2n)}$ call them of class $\mathcal{P}_2$ where $\gcd(k,n)=1$ and $1\le k\le n.$ (These two conditions will be our domain of $k.$)
Let $\mathcal{Q}$ be the class of primitive roots modulo $n$. The $\mathcal{P}_1$ roots can be obtained from these by taking $q$ as is if it is odd and taking $q+n$ if it is even (recall that $n$ is odd). Now to establish a bijection to $\mathcal{P}_2$ we must show that $q\equiv 2k+n\bmod (2n)$ and $q+n\equiv 2k+n\bmod (2n)$ always have a unique solution $k$ in the domain of $k$. In the first case we take $k=(q+n)/2$ which is easily seen to be in this domain. In the second case take $k=q/2,$ also clearly in this domain. Moreover we clearly hit each value of $k$ exactly once. This finishes the proof.
